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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?

Sagot :

MrRoyal

Answer:

The expected cost is $8.75

Step-by-step explanation:

Given

[tex]Time = \{1pm, 2pm, 3pm, 4pm\}[/tex]

[tex]C_1 = \$5[/tex] --- If Bob and Anna meet

[tex]C_2 = \$10[/tex] --- 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:

[tex](Bob,Anna) = \{(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(2,4),(3,1),(3,2),(3,3)[/tex][tex],(3,4),(4,1),(4,2),(4,3),(4,4)\}[/tex]

[tex]n(Bob, Anna) = 16[/tex]

The outcome of them meeting at the same time is:

[tex]Same\ Time = \{(1,1),(2,2),(3,3),(4,4)\}[/tex]

[tex]n(Same\ Time) = 4[/tex]

The probability of them meeting at the same time is:

[tex]Pr(Same\ Time) = \frac{n(Same\ Time)}{n(Bob,Anna)}[/tex]

[tex]Pr(Same\ Time) = \frac{4}{16}[/tex]

[tex]Pr(Same\ Time) = \frac{1}{4}[/tex]

The outcome of them not meeting:

[tex]Different = \(Bob,Anna) = \{(1,2),(1,3),(1,4),(2,1),(2,3),(2,4),(3,1),(3,2)[/tex]

[tex],(3,4),(4,1),(4,2),(4,3)\}[/tex]

[tex]n(Different) = 12[/tex]

The probability of them meeting at the same time is:

[tex]Pr(Different) = \frac{n(Different)}{n(Bob,Anna)}[/tex]

[tex]Pr(Different) = \frac{12}{16}[/tex]

[tex]Pr(Different) = \frac{3}{4}[/tex]

The expected cost is then calculated as:

[tex]Expected = C_1 * P(Same) + C_2 * P(Different)[/tex]

[tex]Expected = \$5 * \frac{1}{4} + \$10 * \frac{3}{4}[/tex]

[tex]Expected = \frac{\$5}{4} + \frac{\$30}{4}[/tex]

Take LCM

[tex]Expected = \frac{\$5+\$30}{4}[/tex]

[tex]Expected = \frac{\$35}{4}[/tex]

[tex]Expected = \$8.75[/tex]

The expected cost is $8.75

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