Question

A library subscribes to two different weekly news magazines, each of which is supposed to arrive in Wednesday's mail. In actuality, each one may arrive on Wednesday, Thursday, Friday, or Saturday. Suppose the two arrive independently of one another, and for each one P(Wed.) = 0.28, P(Thurs.) = 0.38, P(Fri.) = 0.21, and P(Sat.) = 0.13. Let Y = the number of days beyond Wednesday that it takes for both magazines to arrive (so possible Y values are 0, 1, 2, or 3). Compute the pmf of Y. [Hint: There are 16 possible outcomes; Y(W,W) = 0, Y(F,Th) = 2, and so on.] (Enter your answers to four decimal places.)

Answer 1

See pmf of Y below

Explanation:

Let's abbreviate the days by using the first letter, W, T, F and S.

The following array shows the possible probabilities for Y.

Since the probabilities are independent Y(a,b) = P(a)*P(b) for any day a,b.

The array shows the pmf of Y, that is, the probability that the number of days beyond TUESDAY (as we are considering W) that it takes for both magazines to arrive after Wednesday equals to or is greater than 0.

`\bf \left[\begin{array}{ccccc}Day&W&T&F&S\\W&0.28^2&0.28*0.38&0.28*0.21&0.28*0.13\\T&0.38*0.28&0.38^2&0.38*0.21&0.38*0.13\\F&0.21*0.28&0.21*0.38&0.21^2&0.21*0.13\\S&0.13*0.28&0.13*0.38&0.13*0.21&0.13^2\end{array}\right]`

computing the probabilities, we get the array

`\bf \left[\begin{array}{ccccc}Day&W&T&F&S\\W&0.0784&0.1064&0.0588&0.0364\\T&0.1064&0.1444&0.0798&0.0494\\F&0.0588&0.0798&0.0441&0.0237\\S&0.0364&0.0494&0.0273&0.0169\end{array}\right]`

The right way to read Y is Y(row,column). For example, the probability that magazine 1 arrives on Thursday and magazine 2 arrives on Friday is Y(T,F) = 0.0798.