A library subscribes to two different weekly news magazines, each of which is su

Question

A library subscribes to two different weekly news magazines, each of which is supposed to arive 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.) = 0.27, P(Thurs) = 0.38, P(Fr.) = 0.18, and P(Sat.) = 0.17. 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.) ?@ 0 3 P(n)

Explanation / Answer

Ans The first question requires to calculate the probability that Y is zero that is P(Y = 0) both the magazines arrive at Wednesday is equal to P(1st magazine arrives on Wednesday and 2nd magazine arrives on Wednesday) = P(Wed) * P(Wed) (since arrival of both magazines are independent events)

                     = 0.0729

Second Part is to calculate the P(Y = 1) is probability that one of the magazine comes on wednesday and the other comes on thursday or both comes on Thursday = P( 1st on Wed and 2nd on Thurs) + P(1st on Thurs and 2nd on Wed) + P(Both comes on Thursday) = P(Wed) * P(Thurs) + P(Thurs) * P(Wed) + P(Thurs) * P(Thurs) = 0.27 * 0.38 + 0.38 * 0.27 + 0.38 * 0.38 = 0.3496

P(Y=1) = 0.3496

Third Part is to calculate P(Y = 2) , now Y = 2 has following options :-

1 . Either Wednesday and Friday or 2 . Friday and Wednesday or 3 . Thursday and Friday or 4. Friday and Thursday 5 . Friday and Friday

P(Wed) * P(Fri) + P(Fri) * P(Wed) + P(Thurs) * P(Fri) + P(Fri) * P(Thurs) + P(Fri) * P(Fri) = 0.27 * 0.18 + 0.18 * 0.27 + 0.38 * 0.18 + 0.18 * 0.38 + 0.18 * 0.18 = 0.2664

Therefore P(Y=2) = 0.2664

Fourth part requires P(Y = 3) and Y = 3 can have the possible options : -

1 . Wednesday and Saturday or 2 . Saturday and Wednesday or 3 . Thursday and Saturday or         4. Saturday and Thursday 5 . Saturday and Friday 6. Friday and Saturday 7. Saturday and Saturday

P(Y = 3) = P(Wed) * P(Sat) + P(Sat) * P(Wed) + P(Thurs) * P(Sat) + P(Sat) * P(Thurs) + P(Fri) * P(Sat) + P(Sat) * P(Fri) + P(Sat) * P(Sat)

               = 0.27 * 0.17 + 0.17 * 0.27 + 0.38 * 0.17 + 0.17 * 0.38 + 0.18 * 0.17 + 0.17 * 0.18 +

                   0.17 * 0.17 = 0.3111

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