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Chapter 7: Problem 19

A library subscribes to two different weekly news magazines, each of which is supposed to arrive in Wednesday's mail. In actuality, each one could arrive on Wednesday (W), Thursday (T), Friday (F), or Saturday (S). Suppose that the two magazines arrive independently of one another and that for each magazine \(P(\mathrm{~W})=.4, P(\mathrm{~T})=.3\), \(P(\mathrm{~F})=.2\), and \(P(\mathrm{~S})=.1\). Define a random variable \(y\) by \(y=\) the number of days beyond Wednesday that it takes for both magazines to arrive. For example, if the first magazine arrives on Friday and the second magazine arrives on Wednesday, then \(y=2\), whereas \(y=1\) if both magazines arrive on Thursday. Obtain the probability distribution of \(y\). (Hint: Draw a tree diagram with two generations of branches, the first labeled with arrival days for Magazine 1 and the second for Magazine 2.)

Short Answer

Expert verified
The probability distribution of \(y\) is obtained by summing up the joint probabilities for each value of \(y\), with \(y\) as the number of days beyond Wednesday it takes for both magazines to arrive.

Step by step solution

01

Determine all possible joint arrival days

We can start by listing all possible joint arrival days for the two magazines. The possibilities are (Wednesday, Wednesday), (Wednesday, Thursday), (Wednesday, Friday), (Wednesday, Saturday), (Thursday, Wednesday), (Thursday, Thursday) ... up to (Saturday, Saturday). This results in 16 possible outcomes.
02

Determine joint probabilities

Next, calculate the joint probability for each of these outcomes. As the arrivals are independent, the joint probability is the product of the individual probabilities. For example, the joint probability for (Wednesday, Thursday) would be \(P(\mathrm{~W}) \times P(\mathrm{~T}) = 0.4 \times 0.3 = 0.12\). Do this for all 16 outcomes.
03

Assign the number of days for each outcome

We then assign \(y\) as the number of days beyond Wednesday for each outcome. Examples would be: \(y=0\) for (Wednesday, Wednesday), \(y=1\) for (Wednesday, Thursday) or (Thursday, Wednesday), \(y=3\) for (Saturday, Saturday), and so on.
04

Sum up joint probabilities for each \(y\)

The final step is to form the probability distribution of \(y\). We do this by summing up the joint probabilities for each value of \(y\). For example, the probability for \(y=2\) would be the sum of the joint probabilities of (Thursday, Thursday), (Wednesday, Friday), (Friday, Wednesday), (Thursday, Friday) and (Friday, Thursday). Repeat this process for all values of \(y\).

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Most popular questions from this chapter

The time that it takes a randomly selected job applicant to perform a certain task has a distribution that can be approximated by a normal distribution with a mean value of \(120 \mathrm{sec}\) and a standard deviation of \(20 \mathrm{sec}\). The fastest \(10 \%\) are to be given advanced training. What task times qualify individuals for such training?

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