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May 24th

May 24th

# Number Theory and Cryptography

Theorem 1.4 If c|a and c|b, then c|(ax + by) for any integers x and y.

Proof: We have a = cq1 and b = cq2, so ax + by = acq1 + bcq2 = c(aq1 + bq2). This gives
the result.

A number of the form ax+by is called a linear combination of a and b . So this result simply
states that if a number divides two numbers a and b, it divides any linear combination of
them. Note that in line with our previous statement , these letters refer to integers only.

For example, if x and y are any integers, 7|(21x − 35y). This is because 7|21 and 7|35.

The Greatest Common Divisor .
When we reduce a fraction to lowest terms , we find a divisor of the numerator and the
denominator (a common divisor) and cancel it. Thus we have the familiar computation

To reduce the fraction a/b to lowest terms, it would be necessary to find the greatest common
divisor of a and b. This is simply called gcd(a, b). This computation is very cumbersome if
the numerator and denominator are large. For example, how would you reduce the fraction
28, 841/33, 043 to lowest terms? The method we now show handles this problem very quickly,
and was first d one by Euclid . The technique is called the Euclidean Algorithm.

Lemma 1.5 Let a = bq + r. Then any divisor of both a and b is also a divisor of b and r.
Conversely, any divisor of both b and r is also a divisor of a and b. Thus gcd(a, b) = gcd(b, r).

Proof: First, suppose d|a and d|b. Then since r = a − bq, we have d|r since r is a linear
combination of a and b. Thus d|b and d|r.

Conversely, if d|b and d|r, then since a = bq + r is a linear combination of b and r, we see
that d|a and so d|a and d|b. This proves the result.

In words, gcd(a, b) can be found by finding the remainder r when a is divided by b and
computing gcd(b, r). We then use the same result to simplify gcd(b, r), and continue until
we find the answer. Let's use this method to find gcd(75, 55). (Of course this can immediately
be done by most students, but let's illustrate the method.) Dividing, we have

These equations show that

gcd(75, 55) = gcd(55, 20) = gcd(20, 15) = gcd(15,5) = gcd(5, 0) = 5

using the above theorem successively, and the fact that gcd(a, 0) = a when a > 0. This is
a general technique. Instead of finding gcd(a, b) we find the remainder r when a is divided
by b and find gcd(b, r). We keep repeating the process until the remainder is 0, and then
the last non- zero remainder is the required gcd. As noted above, this technique is called
the Euclidean Algorithm. For simplicity, we have as sumed that a and b are positive in
this statement
of the Euclidean algorithm, and we will often make this assumption in what
follows.

We can now reduce 28, 841/33, 043 to lowest terms! The following table systematically gives
the numerator, denominator, quotient and remainders in the Euclidean Algorithm.

 Numerator Denominator Quotient Remainder

The gcd is 191. Note that in each line, the numbers move over one to the left, and the
quotients are ignored. Finally, we reduce to lowest terms:

The Euclidean Algorithm computation will be done automatically in the lab, using the Excel

Many facts about number theory were simply told to us at an early age, and so we take
them for granted. For example, suppose you know that 5|7x. Does it follow that 5|x? We
were taught to think somewhat as follows:

7x/5 is a whole number, and so there must be cancelations. There is no cancelation
of 5 with 7, so all the cancelations are with x, and so 5| x.

Here, the result is true, but the reasoning is suspect. Here is a real proof :

We are given that 5|7x. Also 5|5. Therefore by the linear combination theorem,
5|(3 ٠ 7x − 4x ٠ 5), or using algebra, 5|x.

Query: Suppose we are given that 6|15x. Can we say that 6| x? What can we say about
divisors of x?

The correct proof given above is based on the simple observation that 1 = 7 ٠ 3 − 5٠ 4. We
were able to express 1 as a linear combination of 5 and 7. Here gcd(5, 7) = 1. We shall now
show that if d = gcd(a, b), then d is a linear combination of a and b.

Theorem 1.6 Let d = gcd(a, b). Then there are integers x and y such that

d = ax + by

Remark: The proof will show how to compute x and y. In the lab, the spreadsheet GCD
also computes the values of x and y .

Proof: If we divide a by b to get a = bq + r, we know that d = gcd(b, r), and we note that
r = a − qb = a ٠ 1 + b(−q), a linear combination of a and b. If we continue the Euclidean
Algorithm, we divide again we divide b by r to nd b = rq1+r1. Then the second remainder
r1 = b − rq1 is also a linear combination of a and b since b and r are. Continuing in this
way, we can show that every remainder is a linear combination of a and b. In particular d
is, since it is the last non-zero remainder.

We illustrate this technique with a simple numerical example.

Example 1.7 Find d = gcd(92, 17) and express it as a linear combination of 92 and 17.

Method: The Euclidean algorithm gives

92 = 17٠ 5 + 7
17 = 7٠ 2 + 3
7 = 3 ٠ 2 + 1

Starting from the top, we get

We underline the original numbers to keep track of them. Now use the second equation to
get

Finally, using the last equation and the previous expressions for the remainders 7 and 3, we
obtain

Theorem 1.6 has an interesting corollary. We have defined gcd(a, b) as the greatest common
divisor of a and b. Thus, if d|a and d|b, then d ≤ gcd(a, b). We can now state more.

Corollary 1.8 If d|a and d|b, then d|gcd(a, b)

For a proof, we can write gcd(a, b) = xa + yb, so d|gcd(a, b), since gcd(a, b) is a linear
combination of a and b. In words, any common divisor of a and b is also a divisor of their
gcd.

For example, if two numbers have gcd 6, then 4 cannot divide both of these numbers.

Corollary 1.9 gcd(a, b) = 1 if and only if 1 is a linear combination of a and b.

Proof: Theorem 1.6 shows that if gcd(a, b) = 1, then 1 is a linear combination of a and
b. Conversely, if 1 is a linear combination of a and b then any divisor of a and b must also
divide 1, and so must equal 1. So gcd(a, b) = 1.

Theorem 1.10 Let gcd(a, b) = 1 and a|bn. Then a|n.

Proof: The proof parallels the example above, when 5|7x. Since gcd(a, b) = 1, we have
1 = ax + by for some integers x and y. Multiply by n to get n = axn + bny. Now since a|a
and a|bn, we have a|n since n is a linear combination of a and bn.

Definition 1.11 If gcd(a, b) = 1, we say that a and b are relatively prime.
Thus, we can restate the Theorem 1.10. If a and b are relatively prime, and a|bn, then a|n.
A useful alternative is the following result.

Corollary 1.12 If a and b are relatively prime, a|n and b|n, then ab|n.

Proof: Since a|n, we have n = aq for some q. Thus b|aq. Since a and b are relatively prime,
b|q by the above result, and so q = bq1 for some q1. Therefore n = aq = abq1. This shows
that ab|n.

If we \divide out" the gcd of two numbers, the resulting quotients are relatively prime.

Theorem 1.13 Let d = gcd(a, b). Then a/d and b/d are relatively prime.

Proof: Write d = xa + yb. Divide by d to get 1 = x(a/d) + y(a/d). This shows that a/d
and b/d are relatively prime by Corrolary 1.9.

Theorem 1.14 gcd(xa, xb) = x gcd(a, b) for any x > 0.

Proof: Let d = gcd(a, b). Then since d|a and d|b, we have xd|xa and xd|xb, so xd is a
common divisor of xa and xb. Therefore

gcd(xa, xb) ≥ xd

On the other hand, we know that x is a common divisor of xa and xb. Therefore by
Corollary 1.8, gcd(xa, xb) is a multiple of x, say xD. Thus xD|xa and xD|xb, and therefore
D|a and D|b. Since D is a common divisor of a and b, we must have D ≤ d. Thus,

gcd(xa, xb) = xD ≤ xd:

These two inequalities prove the result.

We can give a useful test to decide whether two numbers a and b are relatively prime. Using
Definition 1.11 and Theorem 1.6, we know that if two numbers are relatively prime, then 1
is a linear combination of them. The following theorem is a valid converse.

Theorem 1.15 Let ax + by = 1 for some integers x and y. Then a and b are relatively
prime.

Proof: Let d be a common divisor of a and b. Then, by Theorem 1.4, d divides any linear
combination of a and b. Therefore d|1 and so d = 1 since the only positive divisor of 1 is 1.
Therefore gcd(a, b) = 1 and so a and b are relatively prime.

For example, all that's required to show that 5 and 7 are relatively prime is to observe that
3 ٠ 7 − 4 ٠ 5 = 1.

Here's an example to use some of these results. It will be used in the next section.

Example 1.16 Given: gcd(a, b) = 1. Prove: gcd(a + b, a − b) = 1 or 2.

Method: Let c = a+b and d = a−b. Add to eliminate b . This gives c+d = 2a. Similarly,
by subtracting , we get c−d = 2b. Now let f be any common divisor of c and d. Thus f|c and
f|d. Therefore, by the above two equations, we find f|2a and f|2b. Therefore f|gcd(2a, 2b).
But by Theorem 1.14, gcd(2a, 2b) = 2gcd(a, b) = 2. Therefore, f|2 and so f = 1 or f = 2.

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Basics

The Spark engine uses a DAG (Directed Acyclic Graph) to keep track of transformations and to lazily execute actions. A transformation includes map, ReduceByKey, GroupByKey, JoinByKey, etc. and actions include Count, Take, and ForEach. Transformations just build the DAG and actions lead to the actual execution. If necessary, you can store intermediate transformations in memory using persistance feature of Spark.

Shuffle processes can also occur but can be expensive due to storage costs (it might spill into disk and be slow). Spark can also perform dynamic allocation, allowing resources to be returned to pools.

Spark is designed for parrallelized execution. Issues with parrallelism includes making sure that the code is properly split up and that skew doesn’t occur (operation on a single core). As will be explained, behavior on local host may be different than production.

Useful

Transformations and Actions

1. map(func): Applies func to all elements in the RDD and returns a new RDD. This transforms aa element of length n to another element of length n.
2. flatmap(func): Applies func to all elements in the RDD but can return more than one object. This tranforms an element of length n to one of length 0, 1, or more.
3. filter(func): Keep elements that func returns true for.
4. GroupBy(func): This is one aggregate function that allows you to group rows together based on some column value.
5. take(#): Returns the first # elements in the RDD.

For groupBy() it is common to follow with an agg() function. You can use the PySpark functionals to Manipulations of DataFrames (RDDS)

2. show(): Display all the data
3. withColumnRenamed(curColName,newColName): Renames a column of the data frame
4. select([col1,col2,…]): Returns RDD with the selected columns
5. sort(col): Returns the RDD sorted along the specific (default ascending) to do descending import from pyspark.sql import functions as F and wrap F.desc(colname).
6. withColumn(col,F.col(col).cast(IntegerType())): replace the column with a specifically cast type
7. alias(name): Assign a column produced by the agg() command as a specific alias.

Aggregations and Join

1. join()

Using Spark in Python

! pip install pyspark
from pyspark import SparkContext, SparkConf

Initialization

conf = SparkConf().setAppName(appName).setMaster(master)
sc = SparkContext(conf=conf)

The SparkContext tells Spark how to access the cluster. The appName parameter is the name of the application to show on the cluster UI and master is the Spark, Mesos, or YARN cluster URL. You will want to launch the application with spark-submit instead of hard-coding.

Abstractions (Components) in Apache Spark

Spark applications use a driver program that runs the main() which executes various operations in parallel across a cluster.

Resilient Distributed Dataset (RDD)

An RDD is a collection of elements partitioned across nodes of a cluster that can be operated in parallel. They can recover from failure and can be persisted in memory in Spark.

There are two ways to create an RDD:

1. Parrallelize an existing collection in the driver program
2. Referencing a dataset in an external storage system (e.g. HDFS, HBase, or any data source offering a Hadoop InputFormat)
    data = [1, 2, 3, 4, 5]
distData = sc.parallelize(data,10) # 10 is the number of partitions to cut the data set into
distData.reduce(lambda a, b: a + b) # a parallel operation performed on data array

Important: You specify the number of partitions and typically you want 2-4 partitions for each CPU in your cluster.

External Datasets

External Dataset Examples: Amazon S3, HBase, Cassandra, HDFS, etc.

Important Commands

1. textFileCommand(): Read in external file or specific text file.
2. wholeTextFiles(): Read in (filename,content) pairs from a directory of small text files

You can read in a text file using the sc.textFile(filename) command where filename can be replaced by the local path on the machine to the external storage (e.g “hdfs://”). Once created, the dataset can be operated on using dataset operations like map() and reduce(). By default, Spark will allocate 1 partition per 128MB block, but that can be specified by the second parameter.

RDD Operations

There are two types of operations supported by RDDs: transformations (e.g. map) or actions (e.g. reduce). All transformations are lazy in that they do not compute the results right away but the transformations are remembered and computed whenever an action requires a result to be returned to the driver program.

This is useful for efficiency, but each transformed RDD may be computed each time the action is called. To stop, this you can use persist or cache in which Spark will keep the elements around the cluster for a much faster access the next time you query.

Basic example

lines = sc.textFile("data.txt") # load in (but not actually just point)
lineLengths = lines.map(lambda s: len(s)) # add map to computation graph
lineLengths.persist() # designate lineLengths to be saved into memory AFTER the first time that it is used
totalLength = lineLengths.reduce(lambda a, b: a + b) # perform an action, so 1) partition data, 2) load, 3) map and persist, 4) reduce

Functions in Spark

There are 3 recommended ways to pass functions into Spark.

1. Lambda expressions (e.g. lambda s: len(s))
2. Local defs in Spark
3. Top-level functions in a module

#1 and #2 are easy, but when using #3 (like if you have a class whose function you are calling) then you cannot reference the class in the function otherwise the whole object will be sent to the cluster. To avoid this, copy whichever field you need into a local variable instead of accessing it externally and then pass in the function.

Closures in Spark

It is important to understand the lifecycle and scope of variables in Spark.

Local vs Cluster Modes

To execute jobs, Spark firsts breaks up the RDD operations into tasks which are executed by the executor. Prior to execution, Spark computes the task’s closure which are variables and methods that need to be visible for the executor to perform its computations on the RDD. Because the closure is sent to each executor you may encounter issues, such common with mutable variables (i.e. no longer atomic.).

A common example is with a counter that references a global variable. The global variable will be sent to each executor within the closure as a copy, thus, if all executors update the variable, you will see that it actually doesn’t actually count correctly. Instead, an Accumulator should be used.

How to correctly print something

rdd.take(NUMBER_TO_PRINT).foreach(println)

Shared Variables

Shared variables allow the programmer to share information across tasks or between the tasks and the driver programs. There are two types:

1. Broadcast: Cache a value in memory on all nodes
2. Accumulators: Variables that operations apply to (e.g. sum or counters)

Important Snippets

This section covers small, important snippets that could be of use.

Master URLS

Relevant Topics: initialization.

The Master URL can be a Spark, Mesos, or YARN Cluster URL. The following image can be useful as a guide to knowing what is what.

format: plain


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Data handling: Use experiments, simulation and probability distribution to set and explore probability models

# Unit 1: Understand probability and make predictions

Natashia Bearam-Edmunds

### Unit outcomes

By the end of this unit you will be able to:

• Understand the difference between independent and dependent events.
• Understand the difference between mutually inclusive and mutually exclusive events.
• Understand complementary events.
• Identify independent and dependent events using $\scriptsize P~\left( {\text{A }\!\!~\!\!\text{ and }\!\!~\!\!\text{ B}} \right)~=~P~\left( \text{A} \right).P~\left( \text{B} \right)$.
• Use the addition rule for mutually exclusive events $\scriptsize P\left( {\text{A }\!\!~\!\!\text{ or }\!\!~\!\!\text{ B}} \right)=~P\left( \text{A} \right)+~P\left( \text{B} \right)$.
• Use $\scriptsize P(\text{A or B)}=P\text{(A)}+P\text{(B)}-P\text{(A and B)}$ when $\scriptsize P\left( {\text{A}\cap \text{B}} \right)\ne 0$.

## What you should know

There is no prior knowledge required for this unit.

## Introduction

Probability is often referred to as chance. It is the study of how likely it is that some event will happen.

When your favourite sports team plays a game, you don’t know whether they will win or not. When the weather report says there is a $\scriptsize \displaystyle 40\%$ chance of rain tomorrow, you may or may not end up getting wet. Uncertainty presents itself to some degree in every event that occurs around us and in every decision that we make.

Many events cannot be predicted with absolute certainty. The best we can do is to say how likely they are to happen, using the idea of probability.

### Note

You can read about the discovery and history of probability at this link when you have an internet connection.

## Experimental and theoretical probability

If we toss a coin $\scriptsize \displaystyle 100$ times, how many times will ‘heads’ come up? We can calculate this using theoretical probability or we can actually perform the experiment, and toss the coin $\scriptsize \displaystyle 100$ times and record the outcomes.

Theoretical probability uses logic and a formula to calculate the chance of an event. An event is one (or more) outcomes. The total number of possible outcomes is called the sample space and is shown using the symbol $\scriptsize \text{S}$. The number of elements in the sample space is denoted as $\scriptsize n(\text{S})$.

The number of possible outcomes to an event E is denoted $\scriptsize n(\text{E})$.

When we toss a coin there are two possible outcomes, H (heads) or T (tails). Logically we would expect that each outcome has a $\scriptsize \displaystyle 50\%$ chance of happening. We say that there is a theoretical probability of $\scriptsize \displaystyle \frac{1}{2}$ for each outcome. Using theoretical probability, ‘heads’ have half a chance of coming up, so we can expect $\scriptsize \displaystyle 50$ heads in $\scriptsize \displaystyle 100$ coin tosses.

When all possible outcomes of an experiment have an equal chance of occurring, we use the following formula to calculate the theoretical probability:

\scriptsize \displaystyle \begin{align*}\text{P}(\text{E})&=\text{ }\displaystyle \frac{{\text{number of favourable outcomes}}}{{\text{number of possible outcomes}}}\\&=\displaystyle \frac{{n(\text{E})}}{{n(\text{S})}}\end{align*}

Probability values are real numbers between and inclusive of $\scriptsize \displaystyle 0$ and $\scriptsize \displaystyle 1$, measured as a fraction or as a decimal. Probability can also be shown as a percentage between $\scriptsize \displaystyle 0\%$ and $\scriptsize \displaystyle 100\%$. We use words such as impossible, unlikely, possible, chance, likely and certain when we describe probabilities.

Probabilities can range from impossible to certain. We can see the likelihood of an event on a probability scale.

It is impossible, in a test out of $\scriptsize 50$ marks, to get a mark over $\scriptsize 50$.

It is equally likely that a pregnant woman will have a boy or a girl.

It is certain that the sun will rise tomorrow.

We have seen that by using theoretical probability the chance of getting a H when a coin is tossed $\scriptsize \displaystyle 100$ times is $\scriptsize 50$ but when we actually do the experiment we may record $\scriptsize 48$H or $\scriptsize 55$H … or anything really, but in most cases it will be a number close to $\scriptsize 50$. Experimental probability is found from the results of an experiment repeated many times.

The more times the coin is tossed, the more accurate the probability of getting heads becomes and the closer the answer from the experimental probability will be to the theoretical probability.

### Example 1.1

Write down the number of possible outcomes (sample space) $\scriptsize n(S)$ for each event below:

1. Throwing a .
2. Tossing a coin.
3. Choosing a card from a pack of playing cards.

Solutions

1. When throwing a die, there are $\scriptsize 6$ possible outcomes in total $\scriptsize \displaystyle \left( {1;\text{ }2;\text{ }3;\text{ }4;\text{ }5;\text{ }6} \right)$, so $\scriptsize n(\text{S})=6$.
2. A coin has two sides: heads (H) and tails (T), so there are $\scriptsize 2$ possible outcomes $\scriptsize \displaystyle (\text{H; T})$. Therefore, $\scriptsize n(\text{S})=2$.
3. There are $\scriptsize 52$ cards in a pack of cards so there are $\scriptsize 52$ possible outcomes to drawing a card from a pack. A pack of cards has four suits with $\scriptsize 13$ card values in each suit. So there are four cards of each value (i.e. there are $\scriptsize 4$ twos, threes, fours, etc. in a pack). The four suits are diamonds (red), spades (black), hearts (red) and clubs (black). The card values in each suit are $\scriptsize (2;\text{ }3;\text{ }4;\text{ }5;\text{ }6;\text{ }7;\text{ }8;\text{ }9;\text{ }10;\text{ Jack; Queen; King; Ace)}$. Therefore, $\scriptsize n(\text{S})=52$.

### Example 1.2

The letters of the word MATHEMATICS are written on separate cards of the same size. The cards are shuffled and dealt, face down, onto a table. A card is selected at random.

1. How many possible outcomes are there?
2. What is the probability that the card selected is:
1. the letter I?
2. the letter M?
3. the letter O?

Solutions

1. There are $\scriptsize 11$ possible outcomes. $\scriptsize n(S)=11$.
2. $\scriptsize \displaystyle \text{Probability}=\text{ }\displaystyle \frac{{\text{number of favourable outcomes}}}{{\text{number of possible outcomes}}}$
1. As there is only one letter ‘I’ there is only one favourable outcome.
$\scriptsize \displaystyle \text{P(I)}=\text{ }\displaystyle \frac{1}{{11}}$
2. There are two Ms.
$\scriptsize \displaystyle \text{P(M)}=\text{ }\displaystyle \frac{2}{{11}}$
3. There is no letter ‘O’.
\scriptsize \displaystyle \begin{align*}\text{P(O)}=\text{ }\displaystyle \frac{0}{{11}}\\=0\end{align*}

### Exercise 1.1

1. If you roll a die once:
1. What is the lowest possible score?
2. What is the highest possible score?
3. What do you think is the most likely score?
2. If you roll a die once, what is the probability of throwing:
1. four.
2. a prime number.
3. an odd number.
4. an eight.
5. a factor of six.
6. a number less than eight.
3. A restaurant is having a raffle to raise funds. They sell a total of $\scriptsize \displaystyle 500$ tickets. What is the probability of Andy winning if he bought:
1. five tickets?
2. $\scriptsize 50$ tickets?
3. one ticket?

The full solutions are at the end of the unit.

## Dependent and independent events

Using probability, we can work out the chances of two or more events happening.

Events are dependent if the outcome of one event influences the outcome of the other. For example, if your lunchbox contains two apples and one banana, when you eat one of the fruit, this reduces the number of choices you have when deciding to eat a second fruit.

Independent events are not affected by previous events. For example, if you toss a coin and it comes up tails and you toss it again and it lands on heads, neither outcome influences the other. A coin does not ‘know’ it came up heads before, so each toss of the coin is independent.

You can use the product rule for independent events to test if events are independent and to calculate the probability of independent events.

Two events $\scriptsize \text{A}$ and $\scriptsize \text{B}$ are independent if and only if:
$\scriptsize P(\text{A and B)}=P(\text{A})\cdot P(\text{B})$

### Example 1.3

A bag contains five red balls and five blue balls. We remove a ball from the bag at random, record its colour and put it back into the bag. We then remove another ball from the bag and record its colour.

1. What is the probability that the first ball is red?
2. What is the probability that the second ball is blue?
3. What is the probability that the first ball is red and the second ball is blue?
4. Are the first ball being red and the second ball being blue independent events?

Solutions

1. Since there are a total of $\scriptsize 10$ balls, of which five are red, the probability of getting a red ball is:
\scriptsize \begin{align*}P(\text{R)}&=\displaystyle \frac{5}{{10}}\\&=\displaystyle \frac{1}{2}\end{align*}
Note: always simplify fractions as far as possible when giving the answers to probability questions.
2. Since the first ball is placed back into the bag before we take the second ball this means that when we draw the second ball, there are still a total of $\scriptsize 10$ balls in the bag, of which five are blue. Therefore the probability of drawing a blue ball is:
\scriptsize \begin{align*}P(\text{B)}&=\displaystyle \frac{5}{{10}}\\&=\displaystyle \frac{1}{2}\end{align*}
3. When drawing two balls from the bag, there are four possibilities. We can get:
• a red ball and then another red ball
• a red ball and then a blue ball
• a blue ball and then a red ball
• a blue ball and then another blue ball.
.
We want to know the probability of the second outcome, where we have to get a red ball first. Since there are five red balls and $\scriptsize 10$ balls in total, there are $\scriptsize \displaystyle \frac{5}{{10}}$ ways to get a red ball first. Now we put the first ball back, so there are again five red balls and five blue balls in the bag. Therefore, there are $\scriptsize \displaystyle \frac{5}{{10}}$ ways to get a blue ball second if the first ball was red. This means that there are:
$\scriptsize \displaystyle \frac{5}{{10}}\times \displaystyle \frac{5}{{10}}=\displaystyle \frac{{25}}{{100}}$
ways to get a red ball first and a blue ball second. So, the probability of getting a red ball first and a blue ball second is $\scriptsize \displaystyle \frac{1}{4}$.
4. Events are independent if and only if: $\scriptsize P(\text{A and B)}=P(\text{A})\cdot P(\text{B})$
In this example:

• $\scriptsize P(\text{first ball is red)}=\displaystyle \frac{1}{2}$
• $\scriptsize P(\text{second ball is blue)}=\displaystyle \frac{1}{2}$
• $\scriptsize P(\text{first ball red and second ball is blue)}=\displaystyle \frac{1}{4}$
So we see that:
\scriptsize \begin{align*}P(\text{first ball red and second ball is blue)}&=P(\text{first ball is red)}\cdot P(\text{second ball is blue)}\\&=\displaystyle \frac{1}{2}\cdot \displaystyle \frac{1}{2}\\&=\displaystyle \frac{1}{4}\end{align*} Therefore, the events are independent.

In the example above, we picked a random ball and put it back into the bag before continuing. This is called sampling with replacement. In the next example, we will follow the same process, except that we will not put the first ball back into the bag. This is called sampling without replacement.

### Example 1.4

A bag contains five red balls and five blue balls. We remove a ball from the bag at random, record its colour. Then, without replacing the first ball, we remove another random ball from the bag and record its colour.

1. What is the probability that the first ball is red?
2. What is the probability that the second ball is blue?
3. What is the probability that the first ball is red and the second ball is blue?
4. Are the first ball being red and the second ball being blue independent events?

Solutions

1. Since there are five red balls and $\scriptsize 10$ balls in total, there are:
$\scriptsize \displaystyle \frac{5}{{10}}=\displaystyle \frac{1}{2}$ ways to get a red ball first.
2. There are four possible outcomes when we remove the two balls:
• a red ball and then another red ball
• a red ball and then a blue ball
• a blue ball and then a red ball
• a blue ball and then another blue ball.
.
Since the question asks for the probability that the second ball is blue, we will only consider outcomes two and four.
• a red ball and then a blue ball, AND
• a blue ball and then another blue ball.
.
For the first outcome (a red ball and then a blue ball), we get a red ball first. Since there are five red balls and $\scriptsize 10$ balls in total, there are $\scriptsize \displaystyle \frac{5}{{10}}$ ways to get a red ball first.
.
After we have taken out a red ball, there are now four red balls and five blue balls left so that is nine balls altogether. Therefore there are five ways to get a blue ball second if the first ball was red. Therefore, there are $\scriptsize \displaystyle \frac{5}{9}$ ways to get a blue ball second if a red ball was drawn first.
.
This means that there are $\scriptsize \displaystyle \frac{5}{{10}}\times \displaystyle \frac{5}{9}=\displaystyle \frac{{25}}{{90}}$ ways to get a blue ball second if a red ball was drawn first.
.
For the fourth outcome (a blue ball and then another blue ball), we get a blue ball first. Since there are five blue balls and $\scriptsize 10$ balls in total, there are $\scriptsize \displaystyle \frac{5}{{10}}$ ways to get blue ball first.
.
After we have taken out a blue ball, there are now five red balls and four blue balls left so that is nine balls altogether. Therefore there are four ways to get a blue ball second if the first ball was also blue. Therefore, there are $\scriptsize \displaystyle \frac{4}{9}$ways to get a blue ball second if a blue ball was drawn first.
.
This means that there are $\scriptsize \displaystyle \frac{5}{{10}}\times \displaystyle \frac{4}{9}=\displaystyle \frac{{20}}{{90}}$ ways to get a blue ball second if a blue ball was also drawn first.
.
To determine the probability of getting a blue ball on the second draw, we look at all of the outcomes that contain a blue ball second and add them.
.
\scriptsize \begin{align*}P(\text{R and B)}+\text{P(B and B)}&=\displaystyle \frac{{25}}{{90}}+\displaystyle \frac{{20}}{{90}}\\&=\displaystyle \frac{{45}}{{90}}\\&=\displaystyle \frac{1}{2}\end{align*}
.
This is the same as in the previous example! You might find it surprising that the probability of the second ball being blue is not affected by whether or not we replace the first ball. The reason why this probability is still $\scriptsize \displaystyle \frac{1}{2}$is that we are computing the probability that the second ball is blue without knowing the colour of the first ball. Because there are only two equal possibilities for the second ball (red and blue) and because we don’t know whether the first ball is red or blue, there is an equal chance that the second ball will be one colour or the other.
3. We have already calculated the probability that the first ball is red and the second ball is blue. We saw that there are:
\scriptsize \begin{align*}\displaystyle \frac{5}{{10}}\times \displaystyle \frac{5}{9}&=\displaystyle \frac{{25}}{{90}}\\&=\displaystyle \frac{5}{{18}}\end{align*}
ways to get a blue ball second if a red ball was drawn first.
4. Events are independent if and only if: $\scriptsize P(\text{A and B)}=P(\text{A})\cdot P(\text{B})$
In this example:

• $\scriptsize P(\text{first ball is red)}=\displaystyle \frac{1}{2}$
• $\scriptsize P(\text{second ball is blue)}=\displaystyle \frac{1}{2}$
• $\scriptsize P(\text{first ball red and second ball is blue)}=\displaystyle \frac{5}{{18}}$
.
So we see that:
.
\scriptsize \begin{align*}P(\text{first ball red and second ball is blue)}&\ne P(\text{first ball is red)}\cdot P(\text{second ball is blue)}\\\displaystyle \frac{5}{{18}}&\ne \displaystyle \frac{1}{2}\cdot \displaystyle \frac{1}{2}\end{align*} Therefore the events are not independent, in other words the events are dependent.

### Exercise 1.2

1. $\scriptsize P(\text{M})=0.45;\text{ }P(\text{N})=0.35;\text{ }P(\text{M and N})=0.1575$. Are $\scriptsize \text{M}$ and $\scriptsize \text{N}$ independent events?
2. I toss a coin and roll a die. What is the probability of getting a head and a two?
3. A box contains three black cards and four white cards. Two cards are randomly picked one after the other.
1. Calculate the probability that the first card picked is black.
2. Calculate the probability that the first card picked is white.
3. If a black card is picked first and not put back into the box, what is the probability that a black card will be picked second?
4. If $\scriptsize \text{A}$ and $\scriptsize \text{B}$ are independent events and $\scriptsize P(\text{A})=0.2$ and $\scriptsize P(\text{A and B})=0.06$, find $\scriptsize P(\text{B})$.

The full solutions are at the end of the unit.

## Union and intersection

When more than one event occurs at a time there can be a combination of outcomes, which results in unions and intersections of the events.

A union of events is the set of all outcomes that occur in the events, written as:
‘$\scriptsize \displaystyle \text{A}$ or $\scriptsize \displaystyle \text{B}$’ or $\scriptsize \displaystyle \text{A}\cup \text{B}$ (A union B).

For example, the union of the sets $\scriptsize \text{A}=\{1;\text{ }3;\text{ }5;\text{ }7\}$ and $\scriptsize \text{B}=\{3;\text{ 6};\text{ 9};\text{ 12}\}$ is the new set $\scriptsize \text{A}\cup \text{B}=\{1;\text{ }3;\text{ }5;\text{ }6;\text{ }7;\text{ 9; 12}\}$ which contains all the elements from both sets with no elements repeated. Here $\scriptsize n(A)=4$ and $\scriptsize n(B)=4$, and $\scriptsize n(A\cup B)=7$.

The intersection of events is the set of all outcomes that occur in all of the events. It is written as:

‘$\scriptsize \displaystyle \text{A}$and $\scriptsize \displaystyle \text{B}$’ or $\scriptsize \displaystyle \text{A}\cap \text{B}$ (A intersection B).

For example, the intersection of the sets $\scriptsize \text{A}=\{1;\text{ }3;\text{ }5;\text{ }7\}$ and $\scriptsize \text{B}=\{3;\text{ 6};\text{ 9};\text{ 12}\}$ is the new set $\scriptsize \text{A}\cap \text{B}=\{3\}$, which contains all the elements that are common to both the sets. In other words, the intersection contains only the elements repeated in the sets. Here, $\scriptsize n(A\cap B)=1$, meaning that there is only one element that is in both set $\scriptsize \text{A}$ and set $\scriptsize \text{B}$.

We can show the relationships of unions and intersections of sets by using Venn diagrams. We will cover Venn diagrams in detail in unit 2 of this subject outcome.

### Note

Venn diagram (Duration: 03.31)

We can calculate the probability of the union of two events using the addition rule for any two events:

$\scriptsize P(\text{A}\cup \text{B)}=P(\text{A})+P(\text{B})-P(\text{A}\cap \text{B)}$

## Mutually exclusive events

Two events are called mutually exclusive or disjoint if they cannot occur at the same time. For example, you cannot be $\scriptsize 21$-years-old and $\scriptsize 25$-years-old at the same time.

Another way of saying this is that the two events, $\scriptsize \displaystyle \text{A}$ and $\scriptsize \displaystyle \text{B}$, cannot have any elements in common, so $\scriptsize n(A\cap B)=0$. A set with no elements is called ‘the empty set’, and denoted by $\scriptsize \varnothing$. With no possible events in the sample space, it is clear that $\scriptsize \displaystyle P(\text{A}\cap \text{B)}=0$ .

The addition rule for two mutually exclusive events is $\scriptsize P(\text{A}\cup \text{B)}=P(\text{A})+P(\text{B})$. This rule is a special case of the previous rule because the events are mutually exclusive, so $\scriptsize \displaystyle P(\text{A}\cap \text{B)}=0$.

Mutually inclusive events are the opposite of mutually exclusive events and can occur at the same time. For example, you could choose a number that is both less than five and odd. There is a possibility of multiple outcomes, so mutually inclusive events cannot happen independently. The probability of the intersection of mutually inclusive events is greater than zero.

$\scriptsize \displaystyle P(\text{A}\cap \text{B)} \gt 0$

## Complementary events

The complement of a set of events $\scriptsize \text{A}$ is a separate set that consists of all elements that are not in $\scriptsize \text{A}$. The complement of any event is the event not ($\scriptsize \text{A}$). We write the complement of an event as $\scriptsize \displaystyle A'$. Since every element in the sample space ($\scriptsize \text{S}$) is in either $\scriptsize \text{A}$ or $\scriptsize \displaystyle A'$, the union of complementary events covers the entire sample space.

$\scriptsize A\cup A'=S$

Complementary events occur when there are only two outcomes. For example, passing an exam or not passing an exam.

Since every element in $\scriptsize \text{A}$ is not in $\scriptsize \displaystyle A'$, we know that complementary events are mutually exclusive:

$\scriptsize \displaystyle \text{A}\cap A'=0$. All complementary events are mutually exclusive, but not all mutually exclusive events are complementary.

The probabilities of complementary events sum to $\scriptsize 1$.

$\scriptsize \begin{array}{l}\text{P}(A)+\text{P}({A}')&=\text{P}(A\cup {A}')\\&=1\\\therefore \text{P}(A)&=1-\text{P}({A}')\end{array}$

### Example 1.5

State if the following are mutually exclusive, mutually inclusive or complementary.

1. Day and night.
2. Getting a head and getting a tail on a single coin toss.
3. Getting a queen and getting a heart in a single draw from a pack of cards.
5. Getting a salary increase and not getting a salary increase.
6. Rolling a die and getting a five or six.

Solutions

1. It cannot be day and night at the same time. These are mutually exclusive and complementary events.
2. Getting a head and getting a tail on a single coin toss are mutually exclusive and complementary. You can get either a head or a tail on a single coin toss, but not both at the same time.
3. Getting a queen and getting a heart in a single draw from a pack of cards is mutually inclusive. You can get a queen of hearts.
5. Getting a salary increase and not getting a salary increase are mutually exclusive and complementary events.
6. Rolling a die and getting a five or six are mutually exclusive events but NOT complementary since there are more than two possible outcomes.

### Example 1.6

On a single roll of a die, find the probability of getting:

1. one or two .
2. A number less than three or a number greater than three.
3. An even number.
4. An odd number.
5. What type of events are those in questions 3 and 4 above called?

Solutions

1. The probability of rolling a one or two are mutually exclusive.
\scriptsize \begin{align*}P(1\text{ or }2)&=P(1)+P(2)\\&=\displaystyle \frac{1}{6}+\displaystyle \frac{1}{6}\\&=\displaystyle \frac{2}{6}\\&=\displaystyle \frac{1}{3}\end{align*}
2. A number less than three or a number greater than three are mutually exclusive. There are two numbers less than three and three numbers greater than three.
\scriptsize \begin{align*}P(\text{less than }3\cup \text{greater than }3)&=P(\text{less than }3)+P(\text{greater than }3)\\&=\displaystyle \frac{2}{6}+\displaystyle \frac{3}{6}\\&=\displaystyle \frac{5}{6}\end{align*}
3. There are three even numbers.
\scriptsize \begin{align*}P(\text{even number})&=\displaystyle \frac{3}{6}\\&=\displaystyle \frac{1}{2}\end{align*}
4. There are three odd numbers.
\scriptsize \begin{align*}P(\text{odd number})&=\displaystyle \frac{3}{6}\\&=\displaystyle \frac{1}{2}\end{align*}
5. Even and odd numbers are complementary events.

### Example 1.7

A car dealership has $\scriptsize 200$ cars on its website for sale. If $\scriptsize 50$ of the cars are Toyotas, what is the probability that the next car to be sold is:

1. a Toyota?
2. not a Toyota?

Solutions

1. .
\scriptsize \begin{align*}P(\text{ }\!\!~\!\!\text{ Toyota})&=\displaystyle \frac{{50}}{{200}}\\&=\displaystyle \frac{1}{4}\end{align*}
2. .
\scriptsize \begin{align*}P(\text{ }\!\!~\!\!\text{ Not Toyota})&=1-\displaystyle \frac{1}{4}\\&=\displaystyle \frac{3}{4}\end{align*}

### Example 1.8

A bowl contains three apples, four lemons and three bananas. If a fruit is selected at random, what is the probability that:

1. it is either a lemon or a banana?
2. it is not an apple?
3. it is a yellow fruit?

Solutions

1. These are mutually exclusive events.
\scriptsize \begin{align*}P(\text{lemon or banana})&=\displaystyle \frac{4}{{10}}+\displaystyle \frac{3}{{10}}\\&=\displaystyle \frac{7}{{10}}\end{align*}
2. .
\scriptsize \begin{align*}P(\text{not an apple})&=1-\displaystyle \frac{3}{{10}}\\&=\displaystyle \frac{7}{{10}}\end{align*}
3. Both the lemons and bananas are yellow and these are mutually exclusive.
\scriptsize \begin{align*}P(\text{yellow fruit)}&=P(\text{banana})+P(\text{lemon})\\&=\displaystyle \frac{7}{{10}}\end{align*}

### Example 1.9

A card is drawn at random from an ordinary pack of $\scriptsize \displaystyle 52$ playing cards. Find the probability that the card drawn is:

1. the three of diamonds
2. the three of diamonds or any heart
3. a diamond or a three

Solutions

1. There is only one three of diamonds in a pack of cards.
$\scriptsize \displaystyle P(3\text{ of diamonds})=\displaystyle \frac{1}{{52}}$
2. There are $\scriptsize 13$ hearts in a pack of cards and one three of diamonds. These are mutually exclusive events.
\scriptsize \displaystyle \begin{align*}P(3\text{ of diamonds or any heart})&=P(3\text{ of diamonds})+P(\text{heart)}\\&=\displaystyle \frac{1}{{52}}+\displaystyle \frac{{13}}{{52}}\\&=\displaystyle \frac{{14}}{{52}}\\&=\displaystyle \frac{7}{{26}}\end{align*}
3. There are four cards of each suit with a value of three. We must subtract the probability of getting a three of diamonds to avoid ‘over counting’.
.
\scriptsize \displaystyle \begin{align*}P(\text{diamond or 3})&=P(\text{diamond})+P(3\text{)}-P(\text{diamond and 3)}\\&=\displaystyle \frac{{13}}{{52}}+\displaystyle \frac{4}{{52}}-\displaystyle \frac{1}{{52}}\\&=\displaystyle \frac{{16}}{{52}}\\&=\displaystyle \frac{4}{{13}}\end{align*}

### Exercise 1.3

1. A bowl has pink and white sweets. The probability of taking out a pink sweet is $\scriptsize \displaystyle \frac{4}{9}$ . What is the probability of taking out a white sweet?
2. You flip a coin two times.
1. Write down the sample space.
2. What is the probability of getting two heads?
3. What is the probability of getting heads first and then tails?
3. Donald has $\scriptsize 14$ loose socks in a drawer. Six of these are white and two are red. Calculate the probability that the first sock taken out at random is:
1. white
2. red
3. not red
4. white or red
5. neither white nor red
4. A box contains six orange balls, five green balls and five yellow balls. You randomly pick balls one after the other.
1. Calculate the probability of drawing two green balls if the first ball is put back into the box before drawing the second ball.
2. Calculate the probability of drawing two green balls if the first ball is NOT put back into the box before drawing the second ball.
3. Calculate the probability of drawing a yellow ball and an orange ball if the first ball is put back into the box before drawing the second ball.
5. There are $\scriptsize 15$ male (M) and $\scriptsize 20$ female (F) learners in an online lecture. The lecturer randomly chooses a learner to answer a question.
1. Determine $\scriptsize P(\text{M})$ learner is chosen.
2. Determine $\scriptsize P(\text{F})$.
3. If he chooses two learners determine the probability that one is male and the other is female.

The full solutions are at the end of the unit.

## Summary

In this unit you have learnt the following:

• How to define probability.
• How to calculate theoretical probability.
• How to identify dependent events.
• How to identify independent events.
• How to use the product rule for independent events.
• How to differentiate between mutually exclusive and mutually inclusive events.
• How to use the probability identity $\scriptsize P(\text{A or B)}=P\text{(A)}+P\text{(B)}-P\text{(A and B)}$.
• How to calculate the probability of mutually exclusive events.
• How to calculate the probability of complementary events.

# Unit 1: Assessment

#### Suggested time to complete: 35 minutes

1. You take all the hearts from a deck of cards. You then select a random card from the set of hearts.
1. What is the sample space?
2. Calculate the probability that the card is the ace of hearts.
3. Calculate the probability that the card is a prime number.
4. Calculate the probability that the card has a letter on it.
2. A and B are two events in a sample space where $\scriptsize \displaystyle P(\text{A)}=0.3;\text{ }P(\text{A or B)}=0.8$ and $\scriptsize P(\text{B})=k$. Determine the value of $\scriptsize k$ if:
1. A and B are mutually exclusive.
2. A and B are independent.
3. You roll two six-sided dice and are interested in the following two events:
A: the sum of the numbers on the dice equals eight
B: at least one of the dice shows a one
Show that these events are mutually exclusive.
4. The surface of a soccer ball is made up of $\scriptsize \displaystyle 32$ faces. $\scriptsize \displaystyle 12$ faces are regular pentagons, each with a surface area of about $\scriptsize \displaystyle 37\text{ c}{{\text{m}}^{2}}$. The other $\scriptsize \displaystyle 20$ faces are regular hexagons, each with a surface area of about $\scriptsize \displaystyle \text{56 c}{{\text{m}}^{2}}$. You roll the soccer ball. What is the probability that it stops with a pentagon touching the ground?
5. A pack of $\scriptsize 20$ batteries contains two defective batteries. If two batteries are randomly chosen one after the other, what is the probability that:
1. only one of the batteries will be defective?
2. both batteries will be defective?
3. neither battery will be defective?
1. Calculate the probability of choosing a red bead and then a black bead if the first bead is put back into the bag.
2. Calculate the probability of choosing two black beads if the first bead is NOT replaced into the bag.

The full solutions are at the end of the unit.

# Unit 1: Solutions

### Exercise 1.1

1. If you roll a die, the total possible outcomes are $\scriptsize \displaystyle \left( {1;\text{ }2;\text{ }3;\text{ }4;\text{ }5;\text{ }6} \right)$.
1. The lowest possible score is one
2. The highest possible score is six.
3. Are all outcomes just as likely or will some happen more often? All outcomes are equally likely so all have a probability of $\scriptsize \displaystyle \frac{1}{6}$ and no outcome is more likely than another to occur.
2. If you roll a die, what is the probability of throwing:
1. $\scriptsize \displaystyle P(4)=\displaystyle \frac{1}{6}$
2. There are three prime numbers $\scriptsize (2;\text{ }3;\text{ }5)$.
\scriptsize \displaystyle \begin{align*}P(\text{prime number})&=\displaystyle \frac{3}{6}\\&=\displaystyle \frac{1}{2}\end{align*}
3. There are three odd numbers $\scriptsize (1;\text{ }3;\text{ }5)$.
\scriptsize \displaystyle \begin{align*}P(\text{odd number})&=\displaystyle \frac{3}{6}\\&=\displaystyle \frac{1}{2}\end{align*}
4. .
\scriptsize \displaystyle \begin{align*}P(8)&=\displaystyle \frac{0}{6}\\&=0\end{align*}
5. There are four factors of $\scriptsize \displaystyle 6$, $\scriptsize (1;\text{ }2;\text{ }3;\text{ }6)$ .
\scriptsize \displaystyle \begin{align*}P(\text{factor of }6)&=\displaystyle \frac{4}{6}\\&=\displaystyle \frac{2}{3}\end{align*}
6. All the numbers are less than $\scriptsize \displaystyle 8$ so this is a certainty.
\scriptsize \displaystyle \begin{align*}P(\text{number less than 8})&=\displaystyle \frac{6}{6}\\&=1\end{align*}
3. .
1. .
\scriptsize \begin{align*}P(\text{winning)}&=\displaystyle \frac{5}{{500}}\\&=1\%\end{align*}
2. .
\scriptsize \begin{align*}P(\text{winning)}&=\displaystyle \frac{{50}}{{500}}\\&=10\%\end{align*}
3. .
\scriptsize \begin{align*}P(\text{winning)}&=\displaystyle \frac{1}{{500}}\\&=0.2\%\end{align*}

Back to Exercise 1.1

### Exercise 1.2

1. $\scriptsize P(\text{M})=0.45;\text{ }P(\text{N})=0.35;\text{ }P(\text{M and N})=0.1575$
\scriptsize \begin{align*}P(\text{M})\times P(\text{N})&=0.45\times 0.35\\&=0.1575\end{align*}
Since $\scriptsize P(\text{M})\times P(\text{N})=P(\text{M and N})$, $\scriptsize \text{M}$ and $\scriptsize \text{N}$ are independent events.
2. .
Tossing a coin and rolling a die are independent events.
$\scriptsize P(\text{H)}=\displaystyle \frac{1}{2}$
$\scriptsize P(\text{2)}=\displaystyle \frac{1}{6}$
\scriptsize \begin{align*}P(\text{H and }2)&=\displaystyle \frac{1}{2}\times \displaystyle \frac{1}{6}\\&=\displaystyle \frac{1}{{12}}\end{align*}
3. .
1. .
\scriptsize \begin{align*}P(\text{B)}&=\displaystyle \frac{{\text{number of black cards}}}{{\text{total possible cards}}}\\&=\displaystyle \frac{3}{7}\end{align*}
2. .
\scriptsize \begin{align*}P(W\text{)}&=\displaystyle \frac{{\text{number of white cards}}}{{\text{total possible cards}}}\\&=\displaystyle \frac{4}{7}\end{align*}
3. If a black card is picked first and not put back into the box, then there will only be two black cards left in the box and six cards in total.
\scriptsize \begin{align*}P(\text{B}\text{ first and B second)}&=\displaystyle \frac{3}{7}\times \displaystyle \frac{2}{6}\\&=\displaystyle \frac{6}{{42}}\\&=\displaystyle \frac{1}{7}\end{align*}
4. Since $\scriptsize \text{A}$ and $\scriptsize \text{B}$ are independent $\scriptsize P(\text{A and B})=P(\text{A})\times P(\text{B})$.
$\scriptsize P(\text{A})=0.2;\text{ }P(\text{A and B})=0.06$
\scriptsize \begin{align*}P(\text{A and B})&=P(\text{A})\times P(\text{B})\\0.2\times P(\text{B})&=0.06\\\therefore P(\text{B})&=0.3\end{align*}

Back to Exercise 1.2

### Exercise 1.3

1. .
\scriptsize \begin{align*}P(\text{W})&=1-P(\text{P})\\&=1-\displaystyle \frac{4}{9}\\&=\displaystyle \frac{5}{9}\end{align*}
2. If you flip a coin two times there are four possible outcomes.
1. $\scriptsize \text{S}=\{(\text{H;H), }(\text{H;T), }(\text{T;T), (T;H) }\!\!\}\!\!\text{ }$
2. The probability of getting two heads:
$\scriptsize P(\text{H;H)}=\displaystyle \frac{1}{4}$
3. The probability of getting heads first and then tails:
$\scriptsize P(\text{H;T)}=\displaystyle \frac{1}{4}$
3. .
1. .
\scriptsize \begin{align*}P(\text{W)}&=\displaystyle \frac{6}{{14}}\\&=\displaystyle \frac{3}{7}\end{align*}
2. .
\scriptsize \begin{align*}P(\text{R)}&=\displaystyle \frac{2}{{14}}\\&=\displaystyle \frac{1}{7}\end{align*}
3. .
$\scriptsize \begin{array}{l}P({R}')&=1-P(R)\\&=1-\displaystyle \frac{1}{7}\\&=\displaystyle \frac{6}{7}\end{array}$
4. .
\scriptsize \begin{align*}P(\text{W}\cup \text{R)}&=P(\text{W)}+P(\text{R)}\\&=\displaystyle \frac{3}{7}+\displaystyle \frac{1}{7}\\&=\displaystyle \frac{4}{7}\end{align*}
5. .
\scriptsize \begin{align*}P(\text{neither W nor R})&=1-P(\text{W or R)}\\&=1-\displaystyle \frac{4}{7}\\&=\displaystyle \frac{3}{7}\end{align*}
4. .
1. .
\scriptsize \begin{align*}P(\text{G and G)}&=\displaystyle \frac{5}{{16}}\cdot \displaystyle \frac{5}{{16}}\\&=\displaystyle \frac{{25}}{{256}}\end{align*}
2. The probability of two green balls if the first ball is not replaced:
\scriptsize \begin{align*}P(\text{G and G)}&=\displaystyle \frac{5}{{16}}\cdot \displaystyle \frac{4}{{15}}\\&=\displaystyle \frac{1}{{12}}\end{align*}
3. .
\scriptsize \displaystyle \begin{align*}P(\text{Y or O)}&=P(\text{Y;O)}+P(\text{O;Y)}\\&=\left( {\displaystyle \frac{5}{{16}}\cdot \displaystyle \frac{6}{{16}}} \right)+\left( {\displaystyle \frac{6}{{16}}\cdot \displaystyle \frac{5}{{16}}} \right)\\&=\displaystyle \frac{{15}}{{64}}\end{align*}
5. .
1. .
\scriptsize \begin{align*}P(\text{M})&=\displaystyle \frac{{15}}{{35}}\\&=\displaystyle \frac{3}{7}\end{align*}
2. .
\scriptsize \begin{align*}P(\text{F})&=\displaystyle \frac{{20}}{{35}}\\&=\displaystyle \frac{4}{7}\end{align*}
3. .
\scriptsize \begin{align*}P(\text{M}\cap \text{F})&=P(\text{M})\cdot P(\text{F})\\&=\displaystyle \frac{3}{7}\cdot \displaystyle \frac{4}{7}\\&=\displaystyle \frac{{12}}{{49}}\end{align*}

Back to Exercise 1.3

### Unit 1: Assessment

1. .
1. $\scriptsize \text{S}=\{\text{A; 2; 3; 4; 5; 6; 7; 8; 9; 10; J; K; Q}\}$
2. $\scriptsize P(\text{A})=\displaystyle \frac{1}{{13}}$
3. $\scriptsize P(2)+P(3)+P(5)+P(7)=\displaystyle \frac{4}{{13}}$
4. $\scriptsize P(\text{A})+P(\text{J})+P(\text{K})+P(\text{Q})=\displaystyle \frac{4}{{13}}$
2. .
1. .
\scriptsize \displaystyle \begin{align*}P(\text{A or B)}&=0.8\\P(\text{A}\cap \text{B})&=0\text{ since A and B are mutually exclusive}\\P(\text{A or B)}&=P(\text{A})+\text{P(B)}\\0.3+k&=0.8\\\therefore k&=0.5\end{align*}
2. .
\scriptsize \displaystyle \begin{align*}P(\text{A and B)}&=P(\text{A)}\cdot P(\text{B) since A and B are independent}\\&=0.3\times k\\P(\text{A or B)}&=0.8\\P(\text{A)}+P(\text{B)}-P(\text{A and B})&=0.8\\0.3+k-0.3k&=0.8\\0.7k&=0.5\\\therefore k&=\displaystyle \frac{5}{7}\end{align*}
3. The sum of the dice must equal eight, so it is not possible for one die to show a one, since the other die will have to show a seven, which is not possible.
$\scriptsize n(1\cap 8)=0$ so these events are mutually exclusive.
4. The total surface area of $\scriptsize \displaystyle 12$ faces of pentagons is $\scriptsize 444\text{ c}{{\text{m}}^{2}}$. The total surface area of $\scriptsize \displaystyle 20$ hexagons is $\scriptsize 1\text{ }120\text{ c}{{\text{m}}^{2}}$. The total surface area of the soccer ball is $\scriptsize 1\text{ 564 c}{{\text{m}}^{2}}$.
\scriptsize \begin{align*}P(\text{pentagon surface})&=\displaystyle \frac{{444}}{{1\text{ }564}}\\&=\displaystyle \frac{{111}}{{391}}\\&=0.28\end{align*}
5. $\scriptsize \text{S}=\{(D;D),\text{ }(D;{D}'),\text{ }({D}';D),\text{ }({D}';{D}')\text{ }\!\!\}\!\!\text{ }$  where $\scriptsize \text{D}$ is defective and $\scriptsize D'$ is not defective.
1. .
$\scriptsize \begin{array}{l}P(D\text{ }and\text{ }{D}')=P({{1}^{{st}}}D)\cdot P({{2}^{{nd}}}{D}')+P({{1}^{{st}}}{D}')\cdot P({{2}^{{nd}}}D)\\=\displaystyle \frac{2}{{20}}\cdot \displaystyle \frac{{18}}{{19}}+\displaystyle \frac{{18}}{{20}}\cdot \displaystyle \frac{2}{{19}}\\=\displaystyle \frac{{18}}{{95}}\end{array}$
2. .
\scriptsize \displaystyle \begin{align*}P(\text{defective and defective)}&=P(\text{defective})\cdot P(\text{defective})\\&=\displaystyle \frac{2}{{20}}\cdot \displaystyle \frac{1}{{19}}\\&=\displaystyle \frac{1}{{190}}\end{align*}
3. .
$\scriptsize \begin{array}{l}P({D}')=1-P(D\text{ and }D)\\=1-\displaystyle \frac{1}{{190}}\\=\displaystyle \frac{{189}}{{190}}\end{array}$
6. .
1. .
\scriptsize \begin{align*}P(\text{R}\text{ and B)}&=\displaystyle \frac{4}{{20}}\cdot \displaystyle \frac{9}{{20}}\\&=\displaystyle \frac{9}{{100}}\end{align*}
2. .
\scriptsize \begin{align*}P(\text{B}\text{ and B)}&=\displaystyle \frac{9}{{20}}\cdot \displaystyle \frac{8}{{19}}\\&=\displaystyle \frac{{18}}{{95}}\end{align*}

Back to Unit 1: Assessment


text:
2019
Том 71
№ 1

# Volume 51, № 7, 1999

Article (Russian)

### On the almost-everywhere convergence of the Riesz means of double orthogonal series

Ukr. Mat. Zh. - 1999. - 51, № 7. - pp. 867–880

We establish coefficient conditions of the classical type that guarantee the almost-everywhere summability of double orthogonal series by the Riesz methods of nonnegative order. We also prove certain equiconvergence theorems.

Article (Russian)

### Optimization of approximate integration of monotone functions of two variables

Ukr. Mat. Zh. - 1999. - 51, № 7. - pp. 881–889

We solve the problem of optimization of approximate integration of functions of two variables defined on a rectangle and monotonic in each variable by using the quadrature formulas with nodes at points of a rectangle net.

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### On Hölder continuity of solutions of doubly nonlinear parabolic equations with weight

Ukr. Mat. Zh. - 1999. - 51, № 7. - pp. 890–903

We prove the Hölder regularity of bounded weak solutions of doubly nonlinear degenerate parabolic equations with measurable coefficients.

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### On zeros of functions of given proximate formal order analytic in a half-plane

Ukr. Mat. Zh. - 1999. - 51, № 7. - pp. 904–909

We describe sequences of zeros of functionsf≢0 that are analytic in the half-plane ℂ+={z:Rez> and satisfy the condition $$\forall \varepsilon > 0\exists c_1 \in (0; + \infty )\forall z \in \mathbb{C}_{\text{ + }} :\left| {f(z)} \right| \leqslant c_1 \exp \left( {(\sigma + \varepsilon )\left| z \right|\eta (\left| z \right|)} \right)$$ where 0≤σ<+∞ and η is a positive function continuously differentiable on [0; +∞) and such thatxη′(x)/η(x)→0 asx→+∞.

Article (Russian)

### Impulsive boundary-value problems for weakly nonlinear systems with control

Ukr. Mat. Zh. - 1999. - 51, № 7. - pp. 910–917

For weakly nonlinear impulsive differential systems with control, we obtain necessary and sufficient conditions for the existence of control and the corresponding solutions of differential systems with general boundary conditions.

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### Distribution of the supremum of random processes from quasi-Banach $K_{σ}$-spaces

Ukr. Mat. Zh. - 1999. - 51, № 7. - pp. 918-930

We study random processes from quasi-Banach $K_{σ}$-spaces of random variables whose domain of definition is not necessarily a compact set. We establish conditions under which the supremum of a properly normalized process belongs to the same space as the process itself. We also obtain estimates for the norm of this supremum.

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### Gaussian and non-Gaussian limit distributions of estimates of the regression coefficients of a long-memory time series

Ukr. Mat. Zh. - 1999. - 51, № 7. - pp. 931–939

We obtain Gaussian and non-Gaussian distributions of estimates of regression coefficients of a long-memory time series.

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### Approximation of fractional-order integrals by algebraic polynomials. II

Ukr. Mat. Zh. - 1999. - 51, № 7. - pp. 940–951

We investigate the approximation of functions that are fractional-order integrals of bounded functions by algebraic polynomials.

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### Asymmetric approximations in the space $L_{p(t)}$

Ukr. Mat. Zh. - 1999. - 51, № 7. - pp. 952-959

We introduce the notion of $(α,β)$-norm in the space $L_{p(t)}$ of functions $x(t)$ for which $$\int\limits_0^1 {\left| {x(t)} \right|^{p(t)}< \infty }$$ where $p(t)$ is a positive measurable function. We establish a criterion for the element of the best $(α,β)$-approximation in the space $L_{p(t)}$. We obtain inequalities of the type of duality relations.

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### The theory of the numerical-analytic method: Achievements and new trends of development. VI

Ukr. Mat. Zh. - 1999. - 51, № 7. - pp. 960–971

We analyze the application of the numerical-analytic method proposed by A.M. Samoilenko in 1965 to multipoint boundary-value problems.

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### Stability of semi-Markov risk processes in schemes of averaging and diffusion approximation

Ukr. Mat. Zh. - 1999. - 51, № 7. - pp. 972–979

We investigate the asymptotic stability of semi-Markov risk processes with probability one in schemes of averaging and diffusion approximation.

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### Generalization of the Lidskii theorem on the localization of the spectrum of a product of Hermitian operators

Ukr. Mat. Zh. - 1999. - 51, № 7. - pp. 980–988

We consider productsC=AB of Hermitian operators in ann-dimensional unitary space. Two equivalent localization theorems are proved in the case where one of the factorsA andB is positive definite.

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### Functional polystability of some nonautonomous quasilinear differential systems

Ukr. Mat. Zh. - 1999. - 51, № 7. - pp. 989–995

For quasilinear differential systems with a boundary matrix of coefficients of the system of the first approximation, we obtain sufficient conditions of functional polystability, which generalizes the notion of exponential polystability.

Brief Communications (Ukrainian)

### One version of the projection-iterative method based on the method of chords

Ukr. Mat. Zh. - 1999. - 51, № 7. - pp. 996–1000

We consider the problem of application of one version of the projection-iterative method to nonlinear integral equations. Sufficient conditions for the convergence of this method are established.

Brief Communications (Ukrainian)

### On minimal prime ideals of commutative Bezout rings

Ukr. Mat. Zh. - 1999. - 51, № 7. - pp. 1001–1005

We study the spectrum of minimal prime ideals of commutative Bezout rings. We apply the results obtained to the problem of diagonal reduction of matrices over rings of this sort.

Brief Communications (Ukrainian)

### An attractor of a semiflow generated by a system of phase-field equations without the uniqueness of a solution

Ukr. Mat. Zh. - 1999. - 51, № 7. - pp. 1006–1009

We prove the existence of a global compact attractor for a multivalued semiflow generated by a system of phase-field equations with conditions on nonlinearity that do not guarantee the uniqueness of a solution.


text:
Asian Journal of Probability and Statistics https://www.journalajpas.com/index.php/AJPAS
Asian Journal of Probability and Statistics (ISSN: 2582-0230) aims to publish high-quality papers (Click here for Types of paper) in all areas of ‘Probability and Statistics’. By not excluding papers on the basis of novelty, this journal facilitates the research and wishes to publish papers as long as they are technically correct and scientifically motivated. The journal also encourages the submission of useful reports of negative results. This is a quality controlled, OPEN peer-reviewed, open access INTERNATIONAL journal.
 Asian Journal of Probability and Statistics en-US Asian Journal of Probability and Statistics 2582-0230 Examining the Factors Affecting the Success of the Establishment of Village-Owned Enterprises https://www.journalajpas.com/index.php/AJPAS/article/view/30358
Aims: This study aims to analyze the influence of Village Government Policies, Village Financial Institutions, Resources, and Community Factors on the Success of the Establishment of Village-Owned Enterprises (VOE) with Village Government Support as moderating variables.

Study Design: SEM WarpPLS.

Place: Sumberputih Village, East Java, Indonesia.

Methodology: This research is quantitative research. The research instrument used a questionnaire. The research was conducted in Sumberputih Village, East Java, Indonesia. The sampling process used a simple random sampling technique and obtained 100 respondents. Data analysis using SEM WarpPLS.

Results: The results showed that the Village Government Policy, Village Financial Institutions, Resources, and Community Factors had a significant effect on the success of the establishment of VOE (Y). Meanwhile, Village Government Support cannot moderate the influence of the four variables on the success of VOE establishment.
 . Solimun Indah Yanti Adi Kusumaningrumi Agus Wahyu Widodo ##submission.copyrightStatement## 2021-11-16 2021-11-16 1 9 10.9734/ajpas/2021/v15i430358 Split Domination In Interval-valued Fuzzy Graphs https://www.journalajpas.com/index.php/AJPAS/article/view/30360
Aims / Objectives: In this paper, we introduced and investigated the concept of split domination in interval-valued fuzzy graph and denoted by γs. We obtained many results related to γs. We investigated and study the relationship of γs with other known parameters in interval-valued fuzzy graph. Finally we calculated γs(G) for some standard interval valued fuzzy graphs.
 Nojood A. AL-Khadari Mahiuob M. Q. Shubatah ##submission.copyrightStatement## 2021-11-18 2021-11-18 10 20 10.9734/ajpas/2021/v15i430360 Bayesian Regression of Government Expenditure on Revenue in Nigeria https://www.journalajpas.com/index.php/AJPAS/article/view/30361
The relationship between government expenditure and its revenue is generating serious debate among researchers. Similarly, their has been a controversy between the classical and the bayesian modelling. Therfore, this study examined the relationship between the government expenditure and its revenue in Nigeria using the bayesian approach. The finance data extracted from the Central Bank of Nigeria statistical bulletin from 1989 to 2018 were considered for the study. Bayesian linear regression was used to fit the model. Normal distribution was fit for the likelihood. Thus, normal-gamma prior was elicited for the bayesian regression parameters. The result showed that the Bayesian estimates with elicited normal-gamma prior produced a better posterior mean of 0.536 for the Total Revenue with a smaller posterior standard deviation of 0.00001 when compared with the OLS standard deviation of 0.05256. Similarly, the total revenue explained 78% variations in the Total expenditure. The constructed model fit was: Total Expenditure = 98.57128 + 0.53630* Total Revenue. This showed that a naira unit of the total expenditure will always be increased by 0.54 of the total revenue. Forecast of 30 years for the total expenditure using both OLS and Bayesian (normal gamma prior) were increasing as the years were progressing. Government should look for a way to increase its revenue in order to sustain the future expenses of the government since expenditure increases yearly.
 Olawale Basheer Akanbi ##submission.copyrightStatement## 2021-11-20 2021-11-20 21 37 10.9734/ajpas/2021/v15i430361 Probabilistic Method for Estimating the Level of Reliability of Solar Photovoltaic Systems for Households in Ghana https://www.journalajpas.com/index.php/AJPAS/article/view/30362
Renewable Energy Resources have been identified among the most promising sources of harnessing power for industrial and household consumption but their power generations highly uctuate so building renewable power systems without critical reliability analysis might result in frequent blackouts in the power system. Therefore, in this paper, a robust, effective and ecient design approach is proposed to handle the reliability issues. The study involves a Mathematical modelling strategy of the PV system to estimate the total PV power produced and the Bottom-Up approach for predicting the household load demand. The reliability is defined in terms of Loss of Load Probability. The design methodology was validated with a University Household. The data used for the analysis consists of daily average global solar irradiance and load profiles. The results revealed that throughout the year, November-February is where the system seems to be more reliable. Also, the results indicated that without buck-up systems, the system would experience an average annual power loss of 17.8753% and thus, it is recommended that either solar batteries or the grid are used as backup system to achieve a complete level of reliability.
 Ali Abubakar Anas Musah Frank Kofi Owusu Isaac Afari Addo ##submission.copyrightStatement## 2021-11-22 2021-11-22 38 53 10.9734/ajpas/2021/v15i430362 The Second Hyper-Zagreb Index of Complement Graphs and Its Applications of Some Nano Structures https://www.journalajpas.com/index.php/AJPAS/article/view/30364
In chemical graph theory, a topological descriptor is a numerical quantity that is based on the chemical structure of underlying chemical compound. Topological indices play an important role in chemical graph theory especially in the quantitative structure-property relationship (QSPR) and quantitative structure-activity relationship (QSAR). In this paper, we present explicit formulae for some basic mathematical operations for the second hyper-Zagreb index of complement graph containing the join G1 + G2, tensor product G1 $$\otimes$$ G2, Cartesian product G1 x G2, composition G1 $$\circ$$ G2, strong product G1 * G2, disjunction G1 V G2 and symmetric difference G1 $$\oplus$$ G2. Moreover, we studied the second hyper-Zagreb index for some certain important physicochemical structures such as molecular complement graphs of V-Phenylenic Nanotube V PHX[q, p], V-Phenylenic Nanotorus V PHY [m, n] and Titania Nanotubes TiO2.
 Mohammed Alsharafi Yusuf Zeren Abdu Alameri ##submission.copyrightStatement## 2021-11-30 2021-11-30 54 75 10.9734/ajpas/2021/v15i430364 An Application of the Growth Model to HIV/AIDS Patients on Antiretroviral Therapy with Coinfections in a District Hospital in Ghana https://www.journalajpas.com/index.php/AJPAS/article/view/30365
Lethal opportunistic diseases like Tuberculosis and Hepatitis C are deeply ingrained complications for patients diagnosed with human immunodeficiency virus (HIV).  The effect of Highly Active Antiretroviral Therapy (HAART) on Hepatitis C and Tuberculosis in HIV patients in Ghana continues to be unpredictable, especially in younger patients. This study aimed to describe the patient survival time distribution on antiretroviral treatment using Statistical Growth model. A retrospective cohort of 634 patients aged between 22 to 73 years were selected from the District Health Information Management System 2 (DHIMS 2), a secondary source, using a random sampling approach. These patients were diagnosed with HIV and started antiretroviral therapy between 2000 and 2019 at St. Martins Catholic Hospital in Amansie South District of the Ashanti Region. The probability of survival for almost all of the risk factors decreases gradually at different clinical states, i.e., from state 1 through to state 4. Hepatitis C or Tuberculosis can also be diagnosed chronically in approximately one in ten patients. Age, sex and the CD4 cell count of patients substantially (p- value =0.001 in log-rank tests) contributed to the prevalence of human immunodeficiency virus. Survival of infants, aged <1year, after treatment was of negative effect. The statistical growth analytical approach offers a good estimate of survival rate ( 79.82%) among major risk factors for infants, aged <1yearon ART with proportion of survival growth of 0.95, hence the survival time of infants, aged <1yearon HAART is negatively affected irrespective of the treatment initiation period.
 Michael Fosu Ofori Aliyu Mohammed Kofi Mensah ##submission.copyrightStatement## 2021-12-01 2021-12-01 76 87 10.9734/ajpas/2021/v15i430365


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# BMI Calculator

## What is BMI?

BMI is the abbreviation for the body mass index & this is also known as the Quetelet index. Essentially, this BMI is a value which has been received from the mass or the weight as well as the height of the individuals. In simple terms, one can state that this BMI or the body mass index is the body mass which is usually divided by the square of body height. Moreover, this is usually expressed in terms of kilograms for mass apart from the height in metres. However, one will also be able to find the other units of measurements when BMI is in question.

## What is the use of BMI?

BMI has been brought to life so that an individual will be able to understand as to what will be the tissue mass which is inclusive of the muscle / bone / fat. And this has been accordingly categorized in the following terms which are – (1) underweight (2) normal weight (3) overweight (4) obese.

## Metric BMI Calculation

 Height: meters Weight: kilograms Your BMI: kg/m2

1,017 total views, 7 views today

$$Metric \quad BMI=\frac{weight\quad (kg)}{height^{2}\quad (m^{2})}$$
$$USA\quad BMI=703\times \frac{weight\quad (lb)}{height^{2}\quad (m^{2})}$$


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July 14, 2020

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1. Sep 21, 2004

### Evil Knievel

Hi. I have an arc (Stick) welder with a 1200 Watt output. In South Africa we run 220v single phase electricity. The current should not be above 20 Amps.

In order to weld thinner materials (and maybe even aluminum ) than possible with an AC welder, I am attaching a Wheatstone bridge (Each diode having a 10 Amp/1000 Volt capacity) to the output to obtain DC. How would you suggest that I control the output or is the different settings of the welder OK? I am not able to commercially find a reostat with a capacity of over 300 Watts. Excuse my ignorance, but where can I find (free) info on construction of a suitable reostat?

Knievel

2. Sep 21, 2004

### Cliff_J

A wheatstone bridge is used to find an unknown resistance (or impedance).

You're constructing a half-wave (2 diodes) or full-wave (4 diodes) rectifier.

Welding with DC helps but you have an simple problem with the large electrodes used in stick welding. You might adjust the amperage enough to weld 14 gauge steel but getting thinner than that and sticking the rod from too low a current or burning through from too much current are going to within a very narrow margin.

The small wire in a MIG wire-feed welder helps tremendously to get the heat very localized to the arc, this allows enough heat to get good penetration in the weld area but without so much total heat as to burn up the surrounding metal. Regardless, when you get to really thin sheet metal and butt joints, its tough. I haven't had a chance to use a TIG yet (prices are outrageous as I'm sure you know) but this is suppossed to be where they really shine is on thinner metals.

I could say that you would be better off constructing a resistor bank to drop the power delivered to the electrode, but anything with that kind of capacity would cost more than just purchasing a wire-feed welder. Some of the flux-core wire feeds are quite resonable in price here in the states, might be worth checking out.

Cliff

3. Sep 22, 2004

### Evil Knievel

Yeah, I hear you.

Cliff

Hi. Thanx for your reply. I was actually wondering about the rods but the copper wire seems like a good idea. Now for the next problem: Flux/atmosphere control. Building a wire feed with gas supply will most certainly be complicated and expensive. I will have a quick look at obtaining a secondhand unit. The other alternative is to simply build a full wave rectifier and see how far I get with off-the-shelf equipment, welding so as to reduce heat buildup. Hehe... As for the resistor bank, yes, that sounds like a good idea. I will post an article on this from a local magazine when I get permission.

Go well all
Evil

4. Sep 22, 2004

Evil Knievel,
Does your AC 'buzz box' have any adjustment for controlling output amperage?
I’ve worked with budget equipment before but my mind is not quick to recollect something like that.

5. Sep 23, 2004

### Evil Knievel

Boulder

Hi. Yes, it has different electrodes that you can attach the live (stick) cable to. Do you think that will be sufficient output control? Also, what thickness is 14 gauge mild steel? We do not use that measurement here.

Many thanx
Knievel

6. Sep 23, 2004

Hello Knievel,
Yes, I believe that will provide you with sufficient output control for most purposes, although I’m struggling to picture the integration of your diodes into this type of welding rig, but perhaps it will come to my later.

Here is a chart to help make sense of it;
http://www.coasteltools.com/tech_steel_gauge_chart.htm [Broken]

Last edited by a moderator: May 1, 2017
7. Sep 27, 2004

### Evil Knievel

Welder output connection

Boulder

Hi. Well, the way how I am going to do it is to build a separate box for the full wave rectifier with 2 buzz-box like input electrodes on the input and the same for the output of the rectifier. Then I simply make two short welder cables with eyelets on both ends that go from the output of the welder to the input of the rectifier and then my normal welding cables go from the output of the rectifier to th eworkpiece.

8. Sep 27, 2004

Ah, I was thinking along those terms myself because the multi-tapped transformer this model likely uses does not lend itself readily for convenient rectification the way a variable inductor type would. Well, be sure and let us know how it works!

Last edited by a moderator: Sep 27, 2004
9. Sep 27, 2004

### Gokul43201

Staff Emeritus
I didn't know that you could weld aluminum with DC. Would that require much higher currents than AC welding ?

10. Sep 27, 2004

I don't know (never tried it with AC, haha).

11. Sep 27, 2004

### Gokul43201

Staff Emeritus
You're kidding ?? You weld aluminum with DC ? Don't you simply make an oxide layer that just never, ever melts ?

12. Sep 28, 2004

### GENIERE

$105.00 will buy you a 90amp (DC) mig type welder that uses flux core wire from Ebay. The flux core prevents oxidizing. About$15.00 buys 2 lbs. of Aluminum wire with a flux core.

13. Sep 28, 2004

### Evil Knievel

Well, I don't know yet but will be sure to update you. It's been a few years since I've seen a textbook but the theory behind this whole concept is that with DC you can maintain a spark at a lower current because the current is not going through zero on every cycle. Lower current = lower temperature at the same resistance. There are 2 problems with welding Aluminum: 1: It oxidixes very, very quickly. 2: It melts at a much lower temperature than normal arc welding temperatures. I hope that, with the DC current, I am able to maintain the arc at aluminum melting point.

The second problem is the oxide layer. If one can find flux coated aluminum welding electrodes the problem is theoretically solved. Also see Cliff_J's reply to my original question. Basically, the thinner the better. This would probably be the better solution as I believe such electrodes are available.
Another way of solving this problem would be to weld using CO2 wire but because this is not flux coated one would have to augment the setup with a CO2 wire feeder and CO2 gas supply to maintain an inert atmosphere around the weld. I have not yet looked at the cost of such a setup but am sure that a secondhand one can be had.

Aha, here comes Geniere with a solution. I see the US is more friendly to it's consumers. In South Africa, an ARC welder costs around R 1000 whereas a MIG/TIG welder costs around R 20 000. The rectifier can be built for around R20, which is why I started looking at this option. At US prices, I would suggest simply buying a MIG welder. Maybe I should have a look at Ebay..

Last edited: Sep 28, 2004
14. Sep 28, 2004

### GENIERE

Also 2 or 3 automobile 12v batteries can work if you can come up with the flux core wire feed mechanism, maybe as a spare part.

15. Sep 28, 2004

### Cliff_J

If the current is required to flow-stop-jump the gap-flow-stop... like with AC then the problem can come from the jumping the gap part and it seems to dance around more, esp on low current where you're holding the electrode closer. My welds look awesome on DC and just above average on AC although its been years so I'd likely just use MIG to bolster my ego.

The aluminum needs to be super clean (sand it or steel wool surface) or don't bother and just braze the parts. Oh and the toughest part is the puddle is near impossible to see as it doesn't glow red like steel.

If you can weld aluminum with a stick welder, you will have developed some serious welding skills. Can you even get the electrodes for it?

And its pure argon for aluminum welding and no CO2 at all.

Cliff

16. Sep 28, 2004

I'll take that as a compliment !
Yes, you can definitely obtain the sticks. I had purchased a quantity through the [then] local Mennonite farm store but have since used them all up and forgotten the number for them in the process.

17. Sep 28, 2004

### Gokul43201

Staff Emeritus
I'll have to second that compliment. I think DC welding aluminum is crazy and stupid...but if that's your bread 'n' butter...

18. Sep 28, 2004

You do what must be done as dictated by circumstance. Not all welding options are available at any given time or place. I have, for instance, successfully performed more than a single welding operation using two 12V batteries wired in series. From my perspective, to limit oneself only to convention is absurd. This is no doubt because unlike so many, I have spent a great deal of my life in remote areas where you either get the job done yourself or it simply will never get done, often resulting in penalty. As may happen, one discovers new techniques which can be effective and refines them. I don’t really consider ability to stick-weld aluminum using DC much more than a useful [survival] skill. Try testing your hand at maintaining an arc by practicing with a coathanger, now that's an insane survival skill, haha!

19. Sep 29, 2004

### Evil Knievel

OK, I think we have a winner here... Coat hanger?! Hehe..

While it is not ideal, one could even build a generator using a normal DC engine/petrol engine and a car alternator to recharge your car batteries quickly. By the way, is an alternator DC or AC? And a generator?

I will build the rectifier this weekend and will hopefully have some feedback on Monday. Wish me luck.

20. Sep 29, 2004

### Cliff_J

A generator is effectively a permanent magnet DC motor being spun to generate DC using the brushes to keep it DC.

An alternator produces AC in 3 phases and the rectifier inside it converts it back to DC for output.

The generator varies in its output based on the speed. With the alternator the output is regulated by controlling the electromagnetic fields. Regulation is very nice to have to avoid overcharging the batteries and possibly ruining equipment.

Good luck Knievel, may your landing be soft and on the other side of those cars you're jumping.

Cliff


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### Home > CCA2 > Chapter 5 > Lesson 5.1.1 > Problem5-9

5-9.

Antonio’s function machine is shown at right. Homework Help ✎

1. What is A(2)?

2. If 81 came out, what was dropped in?

3. If 8 came out, what was dropped in? Be accurate to two decimal places.

9

$3^x=81$

Since $9·9=81$ and $3·3=9$, how many threes will you multiply to make 81?

See part (b).


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## Entrainment by oscillations

In the book of Goldbeter which I have mentioned in several recent posts a concept which occurs repeatedly is that of entrainment. While looking for some more information about this topic I found a paper of Russo, di Bernardo and Sontag (PloS Computational Biology 4, e1000739) which gives an insightful treatment of the subject. The basic idea is to consider two systems which are coupled in some way and to consider the influence of oscillations in one system on the behaviour of the other. It is easy to see how this might be translated into a problem expressed in terms of dynamical systems. A classical example related to this is contained in a story about Christiaan Huygens who was, among other things, the inventor of the pendulum clock in the mid 17th century. Apparently he did not construct clocks himself but had them made by others according to his plans. The well-known story is that he noticed that when two pendulum clocks were placed next to each other the phase of their oscillations became synchronized with, say, one always at the leftmost point of its swing when the other was at the rightmost. Another example is that of the circadian rhythm. There is a 24 hour rhythm in our body and it is interesting to know whether it comes from an intrinsic oscillator or not. Experiments with subjects isolated from the usual rhythm of day and night show that there is an intrinsic oscillator but that its period is closer to 25 hours. Under normal circumstances its period is brought to 24 hours due to the cycle of day and night by entrainment.

The particular mathematical set-up considered in the paper of Russo et. al. is the following. Consider an autonomous dynamical system containing some parameters. Now replace one or more of those parameters by functions of time with period $T$. If solutions of the original system have a suitable tendency to converge to a stationary solution for a given choice of the parameters then solutions of the resulting non-autonomous system converge to periodic solutions of period $T$. In the papers there are nice plots of numerical simulations which give a striking picture of this behaviour. The central result of the paper is a theorem which guarantees this type of behaviour under certain hypotheses. As pointed out in the paper verifying these hypotheses has some similarity to finding a Lyapunov function for an autonomous system. The positive side is that if it can be done it is possible to get strong conclusions. The negative side is that verifying the hypotheses is generally a matter of trial and error. There is no algorithm available for doing that.

The criterion is dependent on the choice of a matrix norm. This is used to define a quantity called the matrix measure $\mu(A)$ of a matrix $A$. The criterion is that the Jacobian of the function defining the dynamical system should have a matrix measure which is bounded above by a negative constant. In that case the system is said to be infinitesimally contracting. The matrix measure is defined by a limiting procedure, $\mu(A)=\lim_{h\to 0}\frac{1}{h}(\|I+hA\|-1)$, but for particular choices of the matrix norm it is possible to calculate in a purely algebraic way. I have no intuitive feeling for what this definition means.

### 2 Responses to “Entrainment by oscillations”

1. Juliette Hell Says:

Hi!
I did not look up in the paper you quote, but with the few details you wrote here, I have the following intuition…
This matrix measure is the derivative of the function which maps a matrix on its matrix norm, taken at the identity in direction of A. The condition says the slope in this direction is negative, meaning going this way, you’ll find contractions. Furthermore, the flow near an equilibrium for small t has leading order id+tA, i.e. it runs in the direction mentioned above.
My intuition wisely ignored the non-autonomous thing, and why you require the uniform negativity …
Another thing is: if you think of A diagonal, µ(A) negative just means that it has only stable eigenvalues. (at least with the usual matrix norm)

And last but not least, I had to think about the experience Bernold showed in class. The one with the metronomes sitting on a board, the board on two rolling cans, and (anti)synchronization taking place or not, depending on the original tuning of the metronomes. I love it!

2. hydrobates Says:

Hi Juliette,

Thanks for your comment. The class demonstration you mention sounds very interesting. I should talk to Bernold about this subject when I see him again.