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Chapter 5: Q75SE (page 243)

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

a. Compute the marginal \({\rm{pmf's}}\)of \({\rm{X and Y }}\)

b. What is the probability that the man鈥檚 and the woman鈥檚 dinner cost at most \({\rm{15}}\)each?

c. Are \({\rm{X and Y }}\)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 鈥淵ou will receive as a refund the difference between the cost of the more expensive and the less expensive meal that you have chosen.鈥� How much would the restaurant expect to refund?

Short Answer

Expert verified

a)

b) \({\rm{p(15,15) = 0}}{\rm{.25 = 25\% }}\)

c) Not independent

d) \({\rm{33}}{\rm{.35}}\)

e) \({\rm{3}}{\rm{.85}}\)

Step by step solution

01

Definition of probability

the proportion of the total number of conceivable outcomes to the number of options in an exhaustive collection of equally likely outcomes that cause a given occurrence.

02

Calculating the marginal pmf’s of X and Y

Given:

The row totals of the given table represent the marginal \({\rm{pmf}}\)of \({\rm{X}}\):

\(\begin{array}{*{20}{c}}{}&{{{\rm{p}}_{\rm{X}}}{\rm{(12) = 0}}{\rm{.05 + 0}}{\rm{.05 + 0}}{\rm{.10 = 0}}{\rm{.20}}}\\{}&{{{\rm{p}}_{\rm{X}}}{\rm{(15) = 0}}{\rm{.05 + 0}}{\rm{.10 + 0}}{\rm{.35 = 0}}{\rm{.50}}}\\{}&{{{\rm{p}}_{\rm{X}}}{\rm{(20) = 0}}{\rm{.05 + 0}}{\rm{.20 + 0}}{\rm{.10 = 0}}{\rm{.30}}}\end{array}\)

The column totals of the given table are the marginal \({\rm{pmf}}\) of \({\rm{Y}}\):

\(\begin{array}{*{20}{c}}{{{\rm{p}}_{\rm{Y}}}{\rm{(12) = 0}}{\rm{.05 + 0}}{\rm{.05 + 0 = 0}}{\rm{.10}}}\\{{{\rm{p}}_{\rm{Y}}}{\rm{(15) = 0}}{\rm{.05 + 0}}{\rm{.10 + 0}}{\rm{.20 = 0}}{\rm{.35}}}\\{{{\rm{p}}_{\rm{Y}}}{\rm{(20) = 0}}{\rm{.10 + 0}}{\rm{.35 + 0}}{\rm{.10 = 0}}{\rm{.55}}}\end{array}\)

03

Step 3. Calculating the man’s and the woman’s dinner cost at most $15 each 

The likelihood that both will cost at least \({\rm{15}}\) is

\(\begin{array}{*{20}{c}}{{\rm{P(X拢 15\;and\;Y拢 15) = p(12,12) + p(12,15) + p(15,12) + p(15,15)}}}\\{{\rm{ = 0}}{\rm{.05 + 0}}{\rm{.05 + 0}}{\rm{.05 + 0}}{\rm{.1 = 0}}{\rm{.25}}}\end{array}\)

04

Justifying that X and Y are independent 

Because the rows of the specified table do not have the same values, \({\rm{X and Y}}\) are not independent.

05

Step 5. Calculating the expected total cost of the dinner for the two people

The overall cost of a two-person supper is equal to the sum of the costs of the man's and the woman's dinners, thus \({\rm{X + Y}}\)

The anticipated outcome (or mean) \({\rm{x}}\). The total of each possibility's products is \({\rm{\mu }}\) . Given the probability of \({\rm{P(x)}}\)

\(\begin{array}{*{20}{c}}{{\rm{E(X + Y) = {a\circ}(x + y)p(x,y) = (12 + 12) \times 0}}{\rm{.05 + (12 + 15) \times 0}}{\rm{.05 + (12 + 20) \times 0}}{\rm{.10}}}\\{{\rm{ + (15 + 12) \times 0}}{\rm{.05 + (15 + 15) \times 0}}{\rm{.10 + (15 + 20) \times 0}}{\rm{.35}}}\\{{\rm{ + (20 + 12) \times 0 + (20 + 15) \times 0}}{\rm{.20 + (20 + 20) \times 0}}{\rm{.10 = 33}}{\rm{.35}}}\end{array}\)

As a result, the total anticipated cost is \({\rm{33}}{\rm{.35}}\)

06

Step 6. Calculating How much would the restaurant expect to refund

The reimbursement is equal to the difference between the more costly and less expensive means, or \({\rm{|X - Y|}}\).

The sum of the products of each possibility is the expected value (or mean) \({\rm{\mu }}\). \({\rm{P(x)}}\) is the probability of \({\rm{x}}\):

\(\begin{array}{*{20}{c}}{{\rm{E(X + Y) = {a\circ} (x + y)p(x,y) = |12 - 12| \times 0}}{\rm{.05 + |12 - 15| \times 0}}{\rm{.05 + |12 - 20| \times 0}}{\rm{.10}}}\\{{\rm{ + |15 - 12| \times 0}}{\rm{.05 + |15 - 15| \times 0}}{\rm{.10 + |15 - 20| \times 0}}{\rm{.35}}}\\{{\rm{ + |20 - 12| \times 0 + |20 - 15| \times 0}}{\rm{.20 + |20 - 20| \times 0}}{\rm{.10 = 3}}{\rm{.85}}}\end{array}\)

As a result, the anticipated reimbursement is \({\rm{3}}{\rm{.85}}\)

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