Solution to Exercise 2.13.

  Problem Statement: A computer consulting firm presently has bids out on three projects. Let A_i = $\{$ awarded project i $\}$, for i = 1, 2, 3, and suppose that P(A₁) = .22, P(A₂) = .25, P(A₃) = .28, $P(A_l \cap A_2) = .11$,$P(A_l \cap A_3) = .05$, $P(A_2 \cap A_3) = .07$,$P(A_1 \cap A_2 \cap A_3) = .01$. Express in words each of the following events, and compute the probability of each event.

(a) $A_1 \cup A_2$

Solution: In words, $A_1 \cup A_2$ is the event the firm is awarded project 1 or project 2. (Note that a set union is always associated with on ``or''. More explicitly, it is a so called inclusive or, meaning one or the other or both.) Applying the Proposition on p. 62:

$$P( A_1 \cup A_2 ) \; = \; .22 \, + \, .25 \, - \, .11 \; = \; .36 .$$

(b) $A_1^{\prime} \cap A_2^{\prime}$[Hint: $(A_1 \cup A_2)^{\prime} = A_1^{\prime} \cap A_2^{\prime}$ ]

Solution: $A_1^{\prime} \cap A_2^{\prime}$ is the event the firm is not awarded project 1 and not awarded project 2. (A set intersection always goes with ``and.'') Note that the hint is true since the event of not winning at least one of projects 1 or 2 is the same as the event of losing both. Applying the result of part (a) and the first proposition on p. 61,

$$P( A_1^{\prime} \cap A_2^{\prime} ) \; = \; P( (A_1 \cup A_... ...= \; 1 \, - \, P( A_1 \cup A_2 ) \; = \; 1 - .36 \; = \; .64 .$$

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

Solution: $A_1 \cup A_2 \cup A_3$ is the event of wining at least one of the three projects. Applying the formula in the box on p. 63,

$$P( A_1 \cup A_2 \cup A_3 ) \; = \; .22 + .25 + .28 - .11 - .05 - .07 + .01 \; = \; .51 .$$

(d) $A_1^{\prime} \cap A_2^{\prime} \cap A_3^{\prime}$

Solution: This is the event that the firm doesn't win any of the projects. It is the complement of hte event in part (c), so

$$P( A_1^{\prime} \cap A_2^{\prime} \cap A_3^{\prime} ) \; = \; 1 - .51 \; = \; .49 .$$

(e) $A_1^{\prime} \cap A_2^{\prime} \cap A_3$

Solution: This is the even of not winning 1 and not winning 2 and winning 3, i.e. of winning only 3. This one is rather tricky, and there don't seem to be any results in the book one can use to get it directly. A Venn diagram depicting the event as the colored region is shown below.

  

Figure 1: Venn diagram for part (e). The colored part is $A_1^{\prime} \cap A_2^{\prime} \cap A_3$

I claim

$$P( A_1^{\prime} \cap A_2^{\prime} \cap A_3 ) \; = \; P( A_3... ... A_3 ) + P (A_2 \cap A_3) - P( A_1 \cap A_2 \cap A_3 ) \right].$$

Why? Notice that the quantity in square brackets is $P( ( A_1 \cap A_3 ) \cup (A_2 \cap A_3) )$ by the proposition on p. 62. From the Venn diagram, it is clear that this is disjoint from the event we are interested in, and their union is the whole of A₃. So,

$$P( A_1^{\prime} \cap A_2^{\prime} \cap A_3 ) \; = \; .28 - ( .05 + .07 - .01 ) \; = \; .17 .$$

(f) $(A_1^{\prime} \cap A_2^{\prime}) \cup A_3$

Solution: This is the event of losing both projects 1 and 2, or winning project 3. The Venn diagram is shown below.

  

Figure 2: Venn diagram for part (f). The colored part is $(A_1^{\prime} \cap A_2^{\prime}) \cup A_3$The green part is $(A_1 \cup A_2 \cup A_3)^{\prime}$,and the red part is A₃.

From the Venn diagram, it is clear that our event is the union of the disjoint events $(A_1 \cup A_2 \cup A_3)^{\prime}$ and A₃. We have already compupted the pobability of $(A_1 \cup A_2 \cup A_3)^{\prime}$ and A₃ in part (d). Thus, by the third axiom of probability,

$$P( (A_1^{\prime} \cap A_2^{\prime}) \cup A_3 ) \; = \; .49 + .28 \; = \; .77 .$$