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
