Question

Bob and Anna are planning to meet for lunch at Sally's Restaurant, but they forgot to schedule a time. Bob and Anna are each going to randomly choose from either 1\text{ p.m.}1 p.m.1, start text, space, p, point, m, point, end text, 2\text{ p.m.}2 p.m.2, start text, space, p, point, m, point, end text, 3\text{ p.m.}3 p.m.3, start text, space, p, point, m, point, end text, or 4\text{ p.m.}4 p.m.4, start text, space, p, point, m, point, end text to show up at Sally's Restaurant. They must both choose exactly the same time in order to meet. Bob has a "buy one entree, get one entree free" coupon that he can only use if he meets up with Anna. If he successfully meets with Anna, Bob's lunch will cost him \$5$5dollar sign, 5. If they do not meet, Bob's lunch will cost him \$10$10dollar sign, 10. What is the expected cost of Bob's lunch?

Answer 1

Answer:

The expected cost is $8.75

Step-by-step explanation:

Given

$Time = \{1pm, 2pm, 3pm, 4pm\}$

$C_1 = \ 5$ --- If Bob and Anna meet

$C_2 = \ 10$ --- If Bob and Anna do not meet

Required

The expected cost of  Bob's meal

First, we list out all possible time both Bob and Anna can select

We have:

$(Bob,Anna) = \{(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(2,4),(3,1),(3,2),(3,3)$$,(3,4),(4,1),(4,2),(4,3),(4,4)\}$

$n(Bob, Anna) = 16$

The outcome of them meeting at the same time is:

$Same\ Time = \{(1,1),(2,2),(3,3),(4,4)\}$

$n(Same\ Time) = 4$

The probability of them meeting at the same time is:

$Pr(Same\ Time) = \frac{n(Same\ Time)}{n(Bob,Anna)}$

$Pr(Same\ Time) = \frac{4}{16}$

$Pr(Same\ Time) = \frac{1}{4}$

The outcome of them not meeting:

$Different = \(Bob,Anna) = \{(1,2),(1,3),(1,4),(2,1),(2,3),(2,4),(3,1),(3,2)$

$,(3,4),(4,1),(4,2),(4,3)\}$

$n(Different) = 12$

The probability of them meeting at the same time is:

$Pr(Different) = \frac{n(Different)}{n(Bob,Anna)}$

$Pr(Different) = \frac{12}{16}$

$Pr(Different) = \frac{3}{4}$

The expected cost is then calculated as:

$Expected = C_1 * P(Same) + C_2 * P(Different)$

$Expected = \ 5 * \frac{1}{4} + \ 10 * \frac{3}{4}$

$Expected = \frac{\ 5}{4} + \frac{\ 30}{4}$

Take LCM

$Expected = \frac{\ 5+\ 30}{4}$

$Expected = \frac{\ 35}{4}$

$Expected = \ 8.75$

The expected cost is $8.75