# 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:

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

plugging numbers,

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

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

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EnumaElish
Homework Helper
ACF = 0.5 0.5 0.5 = 0.125
ABEF = 0.5 0.7 0.7 0.5 = 0.25 0.49 = 0.1225

Sum = 0.37

caffeine
EnumaElish said:
ACF = 0.5 0.5 0.5 = 0.125
ABEF = 0.5 0.7 0.7 0.5 = 0.25 0.49 = 0.1225

Sum = 0.37
So you're implicitly saying the book's answer is wrong?

D H
Staff Emeritus
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
Homework Helper
For the book to be correct you need P(success between A and F) = 0.8.

D H
Staff Emeritus
\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*}