Solution to Exercise 2.54.

  Problem Statement: In Exercise 13. A_i = $\{$ awarded project i $\}$. for i = 1, 2,3. Use the probabilities given there to compute following probabilities:

(a) P(A₂ | A_l)

Solution:

$$P(A_2 \vert A_l) \; = \; \frac{P( A_2 \cap A_1)}{P(A_1)} \; = \; \frac{.11}{.22} \; = \; .5 .$$

(b) $P(A_2 \cap A_3 \vert A_1 )$

Solution:

$$P(A_2 \cap A_3 \vert A_1 ) \; = \; \frac{P( A_2 \cap A_3 \cap A_1)}{P(A_1)} \; = \; .01/.22 \; = \; .0455 .$$

(c) $P(A_2 \cup A_3 \vert A_1 )$

Solution:

$$P(A_2 \cup A_3 \vert A_1 ) \; = \; \frac{P((A_2 \cup A_3) \cap A_1 )}{P(A_1)} .$$

It seems we didn't compute the probability in the numerator yet, so it's back to the Venn diagrams, shown below.

  

Figure 3: Venn Diagram for part (c). The colored part is $A_2 \cup A_3$. The red part is $(A_2 \cup A_3) \cap A_1$.

Also below is my crude attempt to use Venn diagrams to show that

$$(A_2 \cup A_3) \cap A_1 \; = \; (A_2 \cap A_1) \cup (A_3 \cap A_1) .$$

(There is a distributive law at work here. If you know what it means, you can probably check (or already know) that both union and intersection distribute over each other.)

  

Figure 4: Venn Diagram for part (c) demonstrating that $(A_2 \cup A_3) \cap A_1$ = $(A_2 \cap A_1) \cup (A_3 \cap A_1)$.

Given this identity, we can make the calculation using the information provided in conjunction with the proposition on p. 62:

$$P( (A_2 \cap A_1) \cup (A_3 \cap A_1) ) \; = \; P( A_2 \cap A_1 ) + P(A_3 \cap A_1) - P( (A_2 \cap A_1) \cap (A_3 \cap A_1) )$$

Plugging in the numbers, we have

$$P( (A_2 \cap A_1) \cup (A_3 \cap A_1) ) \; = \; .11 + .05 - .01 \; = \; .15 .$$

Hence,

$$P(A_2 \cup A_3 \vert A_1 ) \; = \; \frac{.15}{.22} \; = \; .6818 .$$

(d) $P(A_1 \cap A_2 \cap A_3 \vert A_1 \cup A_2 \cup A_3)$.Express in words probability you have calculated.

Solution: Well, it's pretty easy to see that

$$(A_1 \cap A_2 \cap A_3) \cap (A_1 \cup A_2 \cup A_3) \; = \; A_1 \cap A_2 \cap A_3 .$$

This follows since

$$A_1 \cap A_2 \cap A_3 \, \subset A_1 \cup A_2 \cup A_3 .$$

Draw Venn diagrams if you have to, but basically the event that all three happen is contained in the event that at least one happens. Hence,

$$P(A_1 \cap A_2 \cap A_3 \vert A_1 \cup A_2 \cup A_3) \; = \; \frac{P(A_1 \cap A_2 \cap A_3 )} {P( A_1 \cup A_2 \cup A_3)} .$$

We already figured out the denominator in part (c) of Problem 13, and the numerator was already given, so

$$P(A_1 \cap A_2 \cap A_3 \vert A_1 \cup A_2 \cup A_3) \; = \; \frac{.01}{.51} \; = \; .0196 .$$

In words, we have calculated the conditional probability that the firm wins all 3 projects given that they win at least one.