Ch 1 - 15, 27

15

Question

A consulting firm presently has bids out on three projects. Let $A_{i} = {awarded project i}$ for $i = 1, 2, 3$ and suppose that

$P (A_{1}) = .22, P (A_{2}) = .25, P (A_{3}) = .28$
$P (A_{1} \cap A_{2}) = .11, P (A_{1} \cap A_{3}) = .05, P (A_{2} \cap A_{3}) = .07$
$P (A_{1} \cap A_{2} \cap A_{3}) = .01$
Express in words each of the following events, and compute the probability of each event:
a. A1 ∪ A2
b. A′ 1 ∩ A′ 2 (Hint: use De Morgan's Law)
c. A1 ∪ A2 ∪ A3
d. A′ 1 ∩ A′ 2 ∩ A′ 3
e. A′ 1 ∩ A′ 2 ∩ A 3
f. (A′ 1 ∩ A′ 2) ∪ A3

Proof
a. $P (A_{1} \cup A_{2}) = P (A_{1}) + P (A_{2}) - P (A_{1} \cap A_{2}) = .22 + .25 - .11 = .36$
b. $P (A_{1}^{'} \cap A_{2}^{'}) = 1 - P (A_{1} \cup A_{2}) = 1 - .36 = .64$
c.

\begin{aligned} P (A_{1} \cup A_{2} \cup A_{3}) & = P (A_{1}) + P (A_{2}) + P (A_{3}) \\ - P (A_{1} \cap A_{2}) - P (A_{2} \cap A_{3}) - P (A_{3} \cap A_{1}) \\ + P (A_{1} \cap A_{2} \cap A_{3}) \\ = .22 + .25 + .28 - .11 - .05 - .07 + .01 \\ = .75 - .23 + .01 \\ = .53 \end{aligned}

d. $P (A_{1}^{'} \cap A_{2}^{'} \cap A_{3}^{'}) = 1 - P (A_{1} \cup A_{2} \cup A_{3}) = 1 - .53 = .47$
e.

\begin{aligned} P (A_{1}^{'} \cap A_{2}^{'} \cap A_{3}) & = P (A_{1}^{'} \cap A_{2}^{'}) - P (A_{1}^{'} \cap A_{2}^{'} \cap A_{3}^{'}) \\ = .64 - .47 \\ = .17 \end{aligned}

\begin{aligned} P [(A_{1}^{'} \cap A_{2}^{'}) \cup A_{3}] & = P (A_{1}^{'} \cap A_{2}^{'} \cap A_{3}^{'}) + P (A_{3}) \\ = .47 + .28 \\ = .75 \end{aligned}

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27

Question

The three major options on a certain car model are automatic transmission (A), a sunroof (B), and an upgraded sound system (C). If 84% of all purchasers request A, 78% request B, 68% request C, 97% request A or B, 97% request A or C, 86% request B or C, and 98% request A or B or C, compute the probabilities of the following events.
a. The next purchaser will request at least one of the three options.
b. The next purchaser will select none of the three options.
c. The next purchaser will request only an automatic trans- mission and neither of the other two options.
d. The next purchaser will select exactly one of these three options.

Proof
Givens:
$P (A) = .84$ , $P (B) = .78$ , $P (C) = .68$
$P (A \cup B) = .97$ , $P (A \cup C) = .97$ , $P (B \cup C) = .86$
$P (A \cup B \cup C) = .98$

a. It's given $P (A \cup B \cup C) = .98$ .
b. $P (A^{'} \cap B^{'} \cap C^{'}) = 1 - P (A \cup B \cup C) = 1 - .98 = .02$ .
c.

\begin{aligned} P (A \cap B^{'} \cap C^{'}) & = P (A \cup B \cup C) - P (B \cup C) \\ = .98 - .86 \\ = .12 \end{aligned}

d.
We want to find some unrelated regions first:

\begin{aligned} P (A^{'} \cap B \cap C^{'}) & = P (A \cup B \cup C) - P (A \cup C) \\ = .98 - .97 \\ = .01 \end{aligned}

\begin{aligned} P (A^{'} \cap B^{'} \cap C) & = P (A \cup B \cup C) - P (A \cup B) \\ = .98 - .97 \\ = .01 \end{aligned}

So for the event in question:

\begin{aligned} P (A \cap B^{'} \cap C^{'}) + P (A^{'} \cap B \cap C^{'}) + P (A^{'} \cap B^{'} \cap C) & = .12 + .01 + .01 \\ = .14 \end{aligned}

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