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Chapter 7: Problem 61

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

Expert verified
The probability that at most two boxes must be purchased is 0.0925 or 9.25%. The probability that exactly four boxes must be purchased is 0.043 or 4.3%. The probability that more than four boxes must be purchased is 0.8645 or 86.45%.

Step by step solution

01

Probability for at most two boxes

For at most two boxes to be opened, either the prize can be found in the first box or in the second box. The probability that prize is found in the first box is simply \(0.05 \) and the probability for the second box is the probability of not getting a prize in the first box and getting it in the second one, which amounts to \( (1-0.05) \cdot 0.05 \). Adding these two gives us: \(0.05 + (0.95 \cdot 0.05) = 0.0925 \) or \(9.25 \% \)
02

Probability for exactly four boxes

Next, we need to calculate the probability that the prize is found exactly in the fourth box. This implies not finding the prize in the first three boxes and finding it in the fourth box. Therefore, the probability is \((1-0.05)^3 \cdot 0.05 = 0.043 \) or \(4.3 \% \)
03

Probability for more than four boxes

To find the probability for more than four boxes to be purchased, we need to subtract the probability for getting the prize in four or less boxes from total probability (which is 1). The probability for getting the prize in four or less boxes is the sum of probabilities calculated in steps 1 and 2. Hence, the required probability is \(1 - (0.0925 + 0.043) = 0.8645\) or \(86.45 \% \)

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Most popular questions from this chapter

Consider the population of all 1 -gallon cans of dusty rose paint manufactured by a particular paint company. Suppose that a normal distribution with mean \(\mu=\) \(5 \mathrm{ml}\) and standard deviation \(\sigma=0.2 \mathrm{ml}\) is a reasonable model for the distribution of the variable \(x=\) amount of red dye in the paint mixture. Use the normal distribution model to calculate the following probabilities: a. \(P(x<5.0)\) b. \(P(x<5.4)\) c. \(\quad P(x \leq 5.4)\) d. \(P(4.64.5)\) f. \(P(x>4.0)\)

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Example \(7.18\) described a study in which a person was asked to determine which of three t-shirts had been worn by her roommate by smelling the shirts ("Sociochemosensory and Emotional Functions," Psychological Science [2009]: \(1118-1123\) ). Suppose that instead of three shirts, each participant was asked to choose among four shirts and that the process was repeated five times. Then, assuming that the participant is choosing at random, \(x=\) number of correct identifications is a binomial random variable with \(n=5\) and \(p=1 / 4\). a. What are the possible values of \(x\) ? b. For each possible value of \(x\), find the associated probability \(p(x)\) and display the possible \(x\) values and \(p(x)\) values in a table. c. Construct a probability histogram for the probability distribution of \(x\).
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