1. ## Evaluating conditional proabability

I think I figured this out but I was hoping someone could check my answers. Thanks in advance

P(A fails)= .20 ; P(B fails alone)= .15 ; P(A and B both fail)= .15

Evaluate P(A fails | B fails) and evaluate P(A fails alone)

P(A fails | B fails) = [P(A fails and B fails)]/[P(B fails] = 0.15/P(B fails)
P(B fails) = P(B fails alone) + P(B fails and A fails) = 0.15 + 0.15 = 0.30
This would mean P(A fails|B fails) = 0.15/0.30 = 1/2

P(A fails alone) = P(A fails & B does not)
=P(B does not fail|A fails)*P(A fails)
=(1-P(A and B fails))*0.2
=(1-0.15)*0.2
=(0.85)*0.2
= 0.17

2. ## Re: Evaluating conditional proabability

You notice, I presume, that saying "A fails" and "B fails is really no different from saying "A succeeds" and "B succeeds"?

In any case imagine 100 incidents. In 20 of those A fails, and in 15 of those B also fails. In 15 B fails and A does not. So there are 15+ 15= 30 incidents which B fails and in 15 of those A fails. The probability that A will fail given that B fails, is 15/30= 1/2= 0.50.

There are 20 incidents in which A fails and in 15 of those B also fails so there are 20- 15= 5 in which A fails but B does not. The probability that A fails and B does not is 5/20= 1/4= 0.25.

(P(B does not fail given that A fails) is NOT equal to 1-P(A and B both fail). 1- P(A and B) is the probability that at least one of A and B fail and that includes "Both A and B fail" as well as "B fails and A does not".

3. ## Re: Evaluating conditional proabability

Hello, crownvicman!

P(A fails) = 0.20 ; .P(B fails alone) = 0.15 ; .P(A and B both fail) = 0.15

Evaluate: P(A fails | B fails)
Evaluate: P(A fails alone)

The given data can be placed in a chart.

$\begin{array}{c||c|c||c|} & B & \;B'\; & \text{Total} \\ \hline \hline A & 0.15 && 0.20 \\ \hline A' & 0.15 && \\ \hline \hline \text{Total} &&& 1.00 \\ \hline \end{array}$

The remaining cells can be filled.

$\begin{array}{c||c|c||c|} & B & \;B'\; & \text{Total} \\ \hline \hline A & 0.15 & 0.05 & 0.20 \\ \hline A' & 0.15 & 0.65 & 0.80 \\ \hline \hline \text{Total} & 0.30 & 0.70 & 1.00 \\ \hline \end{array}$

Now you should be able to answer the questions.

4. ## Re: Evaluating conditional proabability

Thank you both!

5. ## Re: Evaluating conditional proabability

I believe Halls actually calculated P[A fails alone | A fails]

P[A fails alone] = P[A fails alone | A fails] P[A fails] = 0.25 * 0.2 = 0.05 as in Soroban's table.