Question

A. What is the probability that this couple spends 45 dollars or more?

Answer 1

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.

Item a:

To find the probabilities involving the total cost, we have to add the variables X and Y from the table, then:

$P(X = 30) = P(X = 15|Y = 15) = 0.2$

$P(X = 35) = P(X = 15|Y = 20) + P(X = 20|Y = 15) = 0.15 + 0.15 = 0.3$

$P(X = 40) = P(X = 15|Y = 25) + P(X = 20|Y = 20) + P(X = 25|Y = 15) = 0.05 + 0.15 + 0.05 = 0.25$

$P(X = 45) = P(X = 20|Y = 25) + P(X = 25|Y = 20) = 0.1 + 0.1 = 0.2$

$P(X = 50) = P(X = 25|Y = 25) = 0.05$

The probability involving values of 45 or more is:

$P(X \geq 45) = P(X = 45) + P(X = 50) = 0.2 + 0.05 = 0.25$

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

Item b:

For a discrete distribution, the expected value is the sum of each outcome multiplied by it's respective probability, hence, involving the 10% discount for prices above $45:

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

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

A similar problem is given at brainly.com/question/25782059