Question

Suppose that \(5 \%\) of cereal boxes contain a prize and the other \(95 \%\) contain the message, "Sorry, try again." Consider the random variable \(x\), where \(x=\) number of boxes purchased until a prize is found. a. What is the probability that at most two boxes must be purchased? b. What is the probability that exactly four boxes must be purchased? c. What is the probability that more than four boxes must be purchased?

Short answer

The probability that at most two boxes must be purchased is 0.095. The probability that exactly four boxes must be purchased is approximately 0.043. The probability that more than four boxes must be purchased is approximately 0.814.

Step-by-step solution

Calculate the probability of at most two boxes must be purchased.

The probability that at most two boxes must be purchased is the sum of the probability that exactly one box and exactly two boxes must be purchased. Using the formula of geometric distribution, we have: \(P(x \leq 2) = P(x = 1) + P(x = 2) = q^{1-1}p + q^{2-1}p = 0.05 + 0.95*0.05 = 0.095\).

Calculate the probability that exactly four boxes must be purchased.

The probability that exactly four boxes must be purchased can be calculated as: \(P(x = 4) = q^{4-1}p = 0.95^{3} * 0.05 \approx 0.043\).

Calculate the probability that more than four boxes must be purchased.

The probability that more than four boxes must be purchased is the complement of the probability that at most four boxes must be purchased. We have: \(P(x > 4) = 1 - P(x \leq 4) = 1 - ( P(x = 1) + P(x = 2) + P(x = 3) + P(x = 4)) = 1 - (0.05 + 0.95*0.05 + 0.95^{2}*0.05 + 0.95^{3}*0.05) \approx 0.814\).