# Thread:

1. ## Mutually exclusive events? need help for sure

P(E_1)=0.25, P(E_2)=0.75, P(FlE_1)=0.05. P(FlE_2)=0.12. P(E_2lF). Ok I'm looking for an area or space between both. As for the text book no good example is close to this one. Ok so 0.25*0.75=.1875, and 0.05*0.12=.006. Now is the hard part what do I do next. I would like to say 0.188 is the answer but I'm not sure. List of possible answers are -

1. 0.188
2. 0.878
3. 0.060
4.0.370

Thanks for any help given.

2. What do you want to find, $P(E_{2}|F)?.$

3. ## opps so sorry about that

it was P(E_2lF)=___?

4. Hello, kwtolley!

I solved it with Bayes' Theorem and a lot of algebra . . .
. . always a dependable method.

$P(E_1) = 0.25,\;\; P(E_2) = 0.75,\;\; P(F|E_1)$ $= 0.05,\;\;P(F|E_2) = 0.12$

Find $P(E_2|F)$

$1)\;0.188\qquad2)\;0.878\qquad3)\; 0.060\qquad4)\;0.370$

Bayes' Theorem: . $P(E_2|F) \;= \;\frac{P(E_2\,\land\,F)}{P(F)}$

We are given: . $P(F|E_1) = 0.05$ . . . This means: . $\frac{P(F\,\land\, E_1)}{P(E_1)} \:=\:0.05$
. . Since $P(E_1) = 0.25:\;\;\frac{P(F\,\land\,E_1)}{0.25} \:=$ $\:0.05\quad\Rightarrow\quad P(F\,\land\,E_1)\:=\:0.0125<br />$

We are given: . $P(F|E_2) = 0.12$ . . . This means: . $\frac{P(F\,\land\,E_2)}{P(E_2)} = 0.12$
. . Since $P(E_2) = 0.75:\;\;\frac{P(F\,\land\,E_2)}{0.75} \:=$ $\:0.12\quad\Rightarrow\quad P(F\,\land\,E_2) = 0.09$

Hence: . $P(F)\:=\:P(F\,\land\,E_1) + P(F\,\land\,E_2) \;= \;0.0125 + 0.09 \;= \;0.1025$

Therefore: . $P(E_2|F) \;=\;\frac{P(E_2\,\land\,F)}{P(F)}\;=\;\frac{0.09} {0.1025} \;=\;0.87804878$ . . . answer (2)

5. ## omg

Man that was a lot to set up for such a small problem if you ask me. I see the steps now. I got it, I will plug this in for the rest of the problems I have to do. thanks again soooo much for showing me the way.
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