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Eddi Din [679]
4 years ago
13

A library subscribes to two different weekly news magazines, each of which is supposed to arrive in Wednesday's mail. In actuali

ty, each one may arrive on Wednesday, Thursday, Friday, or Saturday. Suppose the two arrive independently of one another, and for each one P(Wed.) = 0.26, P(Thurs.) = 0.34, P(Fri.) = 0.17, and P(Sat.) = 0.23. 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.)
Mathematics
1 answer:
____ [38]4 years ago
5 0

Answer:

Step-by-step explanation:

  • Given the following ; P(Wed.) = 0.26, P(Thurs.) = 0.34, P(Fri.) = 0.17, and P(Sat.) = 0.23.
  • Let Y = the number of days beyond Wednesday
  1. The first part asks us to calculate the P(Y = 0)

= P (first magazine arrive on wednesday and 2nd magazine arrives on wednesday) = 0.26 x 0.26 = 0.0676

  • Second part is to calculate P(X = 1) ;

= Probability that one magazine comes on wednesady, the other comes on Thursdays = P( first on WED, 2nd on Thur) or P(2nd on WED, First on THUR) or P(both comes on Thursday)

= 0.26 x 0.34 + 0.34 x 0.26 + 0.34 x 0.34

= 0.2924

  • The next part is to find P( Y = 2) ; There are five possibilities in this case i.e WED and FRI or FR and WED or THUR and FRI or FRI and THUR or Both FRI

= 0.26 x 0.17 + 0.17 x 0.26 + 0.34 x 0.17 + 0.17 x 0.34 + 0.17 x 0.17

= 0.1478

The next part is to find P(Y=3) ; There are seven possibilities here i.e WED and SAT or SAT and WED or THUR and SAT or SAT and THUR or SAT and FRI or FRI and SAT or Both SAT

= 0.26 x 0.23 + 0.23 x 0.26 + 0.34 x 0.23 + 0.23 x 0.34 + 0.17 x 0.23 + 0.23 x 0.17 + 0.23 x 0.23 = 0.3172

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