# Calculating probability of project tasks, and most likely ending date?

Say I have a project (you know the typical one you can represent using a Gantt Chart )

If I where to assign a particula probability to each of the tasks in the Gantt... say for example:

• Task1 (2 days, i am 75% sure of that)
• Task2 (4 days, i am 50% sure of that) for it to start, Task 1 most be finished
• Task3 (8 days, i am 25% sure of that) for it to start, Task 2 most be finished

How can I calculate how likely it is that I will finish this 14 day project?

(In case you are wondering why I am asking this, I am a software developer, and I am constantly creating project plans and I would like to learn more on this subject so that I can project estimation better)

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1
One conventional approach to this problem, which has been around for at least a quarter century (it was in Primavera in the late '80s), is to specify Beta distributions for task durations in the natural but somewhat indirect way of giving an "expected" (modal) duration, a minimum, a maximum, and (perhaps) additional information. (Without additional information, many people assume triangular distributions.) You should be able to read about this in the project management literature, such as routledge.com/books/details/9781420083170. – whuber♦ Jun 18 '12 at 21:15

## 2 Answers

The joint probability of two events is just their product: $P(A \& B) = P(A) \cdot P(B)$, assuming the events are independent. While the probabilities of each of your events aren't independent, the conditional probabilities P(A) and P(B|A) presumably are, so you can multiply them together.

In this case, you need to complete all three tasks on time, so the probability of success is $$P(\textrm{Success}) = P(\textrm{Task1}) \cdot P(\textrm{Task2} | \textrm{Task1}) \cdot P(\textrm{Task3} | \textrm{Task1 & Task2}).$$ You've already provided these numbers, so just plug them in and you get $0.75 \cdot 0.5 \cdot 0.25=0.09375$, or roughly 9 percent.

Alternately, we can instead enumerate the ways in which you might fail.The probability of failing is $1-P(success)$, since you can only either succeed at a task or fail it. For the first task, $$P(\textrm{Task1}=fail)= 1 - P(\textrm{Task1=success})=0.25$$ In order to fail Task2, you first have to succeed at Task1. There's a 75% chance you've succeeded at Task1 and thus get to attempt Task2, coupled with a 50% chance you've failed Task2, which gives you $$P(\textrm{Task2}=fail) = P(\textrm{Task1}=success) \cdot P(\textrm{Task2}=failure) = 0.75 \cdot 0.5=0.375$$ To compute the probability of failing Task3, you must succeed at Task1 and Task2, then fail Task3: the probability of therefore is $0.75 \cdot 0.5 \cdot 0.75 = 0.28125$. Since failing any one task derails the entire project, we add the probability of failures together, and get $P(failure) = 0.25 + 0.375 + 0.28125 = 0.90625$. The probability of success is $1-P(failure)$, which is 0.09375, matching the earlier answer.

I should also note that, in the general case, you haven't quite provided enough information. I've assumed that you can't finish a task early. Furthermore, given your constraints, there's absolutely no "slack" in the schedule: failing one task fails the project. If neither of these are true, then you need more information to solve this problem. Specifically, you need to specify a distribution of times (e.g., 25% chance it takes 1 day, 50% chance it takes 2, 25% it takes 3).

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We both get the same answer but as I said we do need to assume that if p is the probability that task i takes x days then 1-p is the probability that it will take longer than x days. – Michael Chernick Jun 18 '12 at 19:25

You can't compute the probability because you haven't been specific enough. You say task 1 will be finished in 2 days with probability 0.75 but do not specify the distribution of the number of days required with the remaining probaility 0.25 when it can't be done in two days. The same is true for the way you specify how long tasks 2 and 3 will take. Based on your assumptions the project will be done in exactly 14 days if task 1 takes 2 days, task 2 takes 4 days after that and task 3 takes 8 days after that. If the time for each task is independent of the time it takes to do the others then this probability will be the product namely 0.75x0.5x0.25=3/32=0.09375. Now this will be the answer but only if otherwise the tasks will take longer which may be what you intended to say but did not specify.

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