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:Step-by-step explanation:Let the events beW = WednesdayT = ThursdayF = FridayS = SaturdayThe corresponding probability areP (W) = 0.3P (T) = 0.4P (F) = 0.2P (S) = 0.1Let Y = number of days beyond wednesday that takes for both magazines to arrive (so possible Y values are 0, 1 , 2 ,3)The possible outcome are 4² = 16(W,W) (W,T) (W,F) (W,S)(T,W) (T,T) (T,F) (T,S)(F,W) (F,T) (F,F) (F, S)(S,W) (S,T) (S,F) (S,S)The values associated for each of the following as follows Y(W,W) =0, Y(W,T) =1, Y(W,F)=2, Y (W,S)=3Y(T,W)=1, Y (T,T)=1, Y (T,F)=2, Y (T,S)=3Y(F,W)=2, Y (F,T)=2, Y (F,F)=2, Y (F, S)=3Y(S,W)=3, Y (S,T)=3, Y (S,F)=3, Y (S,S)=3The probability mass function of Y isP(Y=0)=0.3(0.3)=0.9P(Y=1) = P[(W,T) or (T,W) or (T,T)]= [0.3(0.4) + 0.3(0.4) + 0.4(0.4)]=0.4P(Y = 2) = P[(W,F) or (T,F) or (F,W) or (F,T) or (F,F)]=0.3(0.2) + 0.4(0.2) + 0.2(0.3) + 0.2(0.4) + 0.2(0.2)]=0.32P(Y =3) = P[(W,S) or (T,S) or (F,S) or (S,W) or (S,T) or (S,F) or (S,S)]= [0.3(0.1) + 0.4(0.1) + 0.2(0.1) + 0.1(0.3) + 0.1(0.4) + 0.1(0.2) + 0.1(0.1)]= 0.19