A couple goes to a restaurant for dinner. Below is a probability table in which X is the
amount (dollars) on the meal spent by the husband and Y is the amount spent by the
wife.
X = 15
X = 20
X = 25
Y = 15
0.20
0.15
0.05
Y = 20
0.15
0.15
0.10
Y = 25
0.05
0.10
0.05
a. (7 pts) What is the probability that this couple spends 45 dollars or more?
b. (7 pts) Suppose that the restaurant is currently running a promotion with a
10% discount if the total amount spent by a table is 45 dollars or more.
What is expected total amount the couple actually has to pay?

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joaobezerra joaobezerra · Answer 1 · 2021-12-07T10:50:52+01:00

Using the probability table, it is found that:

a) There is a 0.25 = 25% probability that this couple spends 45 dollars or more.

b) The expected amount the couple actually has to pay is $36.85.

From the table, the probabilities of each total cost are:

[tex]P(X = 30) = P(X = 15|Y = 15) = 0.2[/tex]

[tex]P(X = 35) = P(X = 15|Y = 20) + P(X = 20|Y = 15) = 0.15 + 0.15 = 0.3[/tex]

[tex]P(X = 40) = P(X = 15|Y = 25) + P(X = 20|Y = 20) + P(X = 25|Y = 15) = 0.05 + 0.15 + 0.05 = 0.25[/tex]

[tex]P(X = 45) = P(X = 20|Y = 25) + P(X = 25|Y = 20) = 0.1 + 0.1 = 0.2[/tex]

[tex]P(X = 50) = P(X = 25|Y = 25) = 0.05[/tex]

Item a:

This probability is:

[tex]P(X \geq 45) = P(X = 45) + P(X = 50) = 0.2 + 0.05 = 0.25[/tex]

There is a 0.25 = 25% probability that this couple spends 45 dollars or more.

Item b:

The expected value is the sum of each outcome multiplied by it's respective probability.

Hence:

[tex]E(X) = 0.2(30) + 0.3(35) + 0.25(40) + 0.9[0.2(45) + 0.05(50)] = 36.85[/tex]

The expected amount the couple actually has to pay is $36.85.

A similar problem is given at https://brainly.com/question/24855677