Basic intersection/union probabilities.

caffeine

Probability self-study question (please see attached png for diagram).

In the following diagram, A, C, and F have a 50% chance for success. B, D, and E have a 70% chance for success. What is the overall probability for success?

Here's what I've done:

[tex]
A \cap \left[ C \cup \left( E \cap \left[ B \cup D \right] \right) \right] \cap F
[/tex]

plugging numbers,

[tex]
.5 \times \left[ .5 + \left( .7 \times \left[ .7 + .7 \right] \right) \right] \times .5
[/tex]

My calculator says .37. The book says .20. Where did I go wrong?

Attached Images

diagram.png (2.3 KB, 14 views)

EnumaElish

ACF = 0.5 0.5 0.5 = 0.125
ABEF = 0.5 0.7 0.7 0.5 = 0.25 0.49 = 0.1225
ADEF = ABEF = 0.1225

Sum = 0.37

caffeine

So you're implicitly saying the book's answer is wrong?

D H

Do a quick sanity check on your work. Look at the diagram. The probability of success is P(A)*P(success on path from A to F)*P(F). Since P(success on path from A to F) <= 1, P(success) <= P(A)*P(F) = 0.25. Your answer (0.37) cannot be correct.

What you have done wrong is to not take into account (for example) B and D both succeeding.

EnumaElish

For the book to be correct you need P(success between A and F) = 0.8.

D H

It is, more-or-less. The exact value is 0.8185, making the end-to-end probability of success 0.204625. The book or the original poster must rounded that to two significant digits.
I gave a hint on how to get to the correct probability: make sure not to exaggerate success on parallel paths. To see why this must be the case, consider the first half of the upper path between A and F: the parallel branch B and/or D. It is incorrect to compute the probability of B and/or D being successful just by adding the probabilities. (Sanity check again: these sum to 1.4, which is not a valid probability). In set theoretic terms, the correct calculation is
[tex]
\begin{align*}
P(B \cup D) &= P(B) + P(D) - P(B \cap D) \\
&= P(B) + P(D) - P(B)*P(D) \\
&= 0.91
\end{align*}
[/tex]