Chapter 5, Problem 75SE

A restaurant serves three fixed-price dinners costing $12, $15, and $20. For a randomly selected couple dining at this restaurant, let X = the cost of the man’s dinner and Y = the cost of the woman’s dinner. The joint pmf of X and Y is given in the following table:

p(x, y)		12	y 15	20
	12	.05	.05	.10
X	15	.05	.10	.35
	20	0	.20	.10

a. Compute the marginal pmf’s of X and Y.

b. What is the probability that the man’s and the woman’s dinner cost at most $15 each?

c. Are X and Y independent? Justify your answer.

d. What is the expected total cost of the dinner for the two people?

e. Suppose that when a couple opens fortune cookies at the conclusion of the meal, they find the message “You will receive as a refund the difference between the cost of the more expensive and the less expensive meal that you have chosen.” How much would the restaurant expect to refund?

To determine

Find the marginal pmf’s of X and Y.

Answer to Problem 75SE

The marginal pmf of X is:

X	12	15	20
$p_X\left(x\right)$	0.20	0.50	0.30

The marginal pmf and Y is:

Y	12	15	20
$p_Y\left(y\right)$	0.10	0.35	0.55

Explanation of Solution

Given info:

In a restaurant serving three fixed-price dinners, X is the cost of a man’s dinner and Y is the cost of a woman’s dinner, for a randomly selected couple. The joint pmf of X and Y are given. The possible values of both X and Y are 12, 15 and 20.

Calculation:

Marginal pmf:

The marginal probability mass function or marginal pmf for each possible value x of a random variable X, when the joint pmf of X and Y is $p\left(x,y\right)$, is given as: $p_X\left(x\right)={\displaystyle \sum _{y:p\left(x,y\right)>0}p\left(x,y\right)}$.

The marginal pmf of X for $x=12$ is:

$\begin{array}{c}{p}_X\left(12\right)={\displaystyle \sum _{y:p\left(x,y\right)>0}p\left(12,y\right)}\\ =p\left(12,12\right)+p\left(12,15\right)+p\left(12,20\right)\\ =0.05+0.05+0.10\\ =\mathrm{0.20.}\end{array}$

This is the same as the row total for the row containing $x=12$, in the given table.

Thus, the marginal probability for each value of X is the row total for the corresponding value of X.

The marginal pmf of Y for $y=12$ is:

$\begin{array}{c}{p}_Y\left(12\right)={\displaystyle \sum _{x:p\left(x,y\right)>0}p\left(x,12\right)}\\ =p\left(12,12\right)+p\left(15,12\right)+p\left(20,12\right)\\ =0.05+0.05+0\\ =\mathrm{0.10.}\end{array}$

This is the same as the column total for the column containing $y=12$, in the given table.

Thus, the marginal probability for each value of Y is the column total for the corresponding value of Y.

Thus, the marginal pmf for X and Y are found similarly as follows:

$p\left(x,y\right)$		y
		12	15	20	Total
x	12	0.05	0.05	0.10	0.20
	15	0.05	0.10	0.35	0.50
	20	0	0.20	0.10	0.30
	Total	0.10	0.35	0.55	1

Thus, the marginal pmf of X and Y are:

X	12	15	20
$p_X\left(x\right)$	0.20	0.50	0.30
Y	12	15	20
$p_Y\left(y\right)$	0.10	0.35	0.55

To determine

Find the probability that each of the man’s and the woman’s dinner cost at most $15.

Answer to Problem 75SE

The probability that each of the man’s and the woman’s dinner cost at most $15 is 0.25.

Explanation of Solution

Calculation:

The probability that each of the man’s and the woman’s dinner cost at most $15 is $P\left(X\le 15,Y\le 15\right)$.

Now, $P\left(X\le 15,Y\le 15\right)=p\left(12,12\right)+p\left(12,15\right)+p\left(15,12\right)+p\left(15,15\right)$.

From the data,

$\begin{array}{l}p\left(12,12\right)=0.05,\\ p\left(12,15\right)=0.05,\\ p\left(15,12\right)=0.05,\\ p\left(15,15\right)=\mathrm{0.10.}\end{array}$

Thus,

$\begin{array}{c}P\left(X\le 15,Y\le 15\right)=0.05+0.05+0.05+0.10\\ =\mathrm{0.25.}\end{array}$

Hence, the value of $P\left(X\le 15,Y\le 15\right)$ is 0.25.

To determine

Decide whether X and Y are independent.

Answer to Problem 75SE

The random variables X and Y are not independent.

Explanation of Solution

Calculation:

Independent random variables:

The pair of random variables, X and Y are independent, if, for every pair of values, x and y, taken by the random variables is such that $p\left(x,y\right)=p_X\left(x\right)\cdot p_Y\left(y\right)$.

Consider the pair of values $\left\{X=12\text{ and }Y=12\right\}$.

From the marginal pmf table, $p_X\left(12\right)=0.20$ and $p_Y\left(12\right)=0.10$.

Thus,

$\begin{array}{c}{p}_X\left(x\right)\cdot p_Y\left(y\right)=0.20\times 0.10\\ =\mathrm{0.02.}\end{array}$

But from the data, $p\left(12,12\right)=0.05$.

Thus, $p\left(12,12\right)\ne p_X\left(12\right)\cdot p_Y\left(12\right)$.

Hence, the random variables X and Y are not independent.

To determine

Find the expected total cost of the dinner for the two people.

Answer to Problem 75SE

The expected total cost of the dinner for the two people is $33.35.

Explanation of Solution

Calculation:

The total cost of the dinner for the two people is $X+Y$. Thus, the relevant expected total cost will be $E\left(X+Y\right)$.

Expectation of sum of random variables:

The expectation of sum of 2 random variables X and Y, having joint pmf $p\left(x,y\right)$, is given as:

$E\left(X+Y\right)={\displaystyle \sum {\displaystyle \sum \left(x+y\right)p\left(x,y\right)}}$

The calculation for $E\left(X+Y\right)$ involves the sum $\left(x+y\right)$ of pairs of values $\left(x,y\right)$, their corresponding, probabilities $p\left(x,y\right)$ and the product $\left(x+y\right)\cdot p\left(x,y\right)$.

The following table shows the necessary calculations:

$\left(x,y\right)$	$\left(x+y\right)$	$p\left(x,y\right)$	$\left(x+y\right)\cdot p\left(x,y\right)$
$\left(12,12\right)$	24	0.05	1.2
$\left(12,15\right)$	27	0.05	1.35
$\left(12,20\right)$	32	0.1	3.2
$\left(15,12\right)$	27	0.05	1.35
$\left(15,15\right)$	30	0.1	3
$\left(15,20\right)$	35	0.35	12.25
$\left(20,12\right)$	32	0	0
$\left(20,15\right)$	35	0.2	7
$\left(20,20\right)$	40	0.1	4
			${\displaystyle \sum _x{\displaystyle \sum _y\left(x+y\right)p\left(x,y\right)}}=33.35$

Hence, the expected total cost of the dinner for the two people is $33.35.

To determine

Find the amount of money the restaurant would expect to refund, if in the fortune cookies at the end of the meal, a couple finds the message “You will receive as a refund the difference between the cost of the more expensive and the less expensive meal that you have chosen.”

Answer to Problem 75SE

The amount of money the restaurant would expect to refund under the given scenario is $3.85.

Explanation of Solution

Justification:

The difference between the cost of the more expensive and the less expensive meal is $\left|X-Y\right|$. Thus, the expected amount the restaurant has to refund is $E\left(\left|X-Y\right|\right)$.

Expectation of a function of random variables:

The expectation of a function, $h\left(X,Y\right)$ of 2 random variables X and Y, having joint pmf $p\left(x,y\right)$, is given as:

$E\left(h\left(X,Y\right)\right)={\displaystyle \sum {\displaystyle \sum h\left(x,y\right)p\left(x,y\right)}}$

Thus, the expectation of $\left|X+Y\right|$ will be $E\left(\left|X-Y\right|\right)={\displaystyle \sum {\displaystyle \sum \left|x-y\right|p\left(x,y\right)}}$.

The calculation for $E\left(\left|X-Y\right|\right)$ involves the quantity $\left|x-y\right|$ of pairs of values $\left(x,y\right)$, their corresponding, probabilities $p\left(x,y\right)$ and the product $\left|x-y\right|\cdot p\left(x,y\right)$.

The following table shows the necessary calculations:

$\left(x,y\right)$	$\left|x-y\right|$	$p\left(x,y\right)$	$\left|x-y\right|\cdot p\left(x,y\right)$
$\left(12,12\right)$	0	0.05	0
$\left(12,15\right)$	3	0.05	0.15
$\left(12,20\right)$	8	0.1	0.8
$\left(15,12\right)$	3	0.05	0.15
$\left(15,15\right)$	0	0.1	0
$\left(15,20\right)$	5	0.35	1.75
$\left(20,12\right)$	8	0	0
$\left(20,15\right)$	5	0.2	1
$\left(20,20\right)$	0	0.1	0
			${\displaystyle \sum _x{\displaystyle \sum _y\left|x+y\right|p\left(x,y\right)}}=3.85$

Hence, the expected total cost of the dinner for the two people is $3.85.