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.) = 0.3, P(Thurs.) = 0.4, P(Fri.) = 0.2, and P(Sat.) = 0.1. 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). Required:
Compute the pmf of Y.

Step-by-step explanation:

Let the events be

W = Wednesday

T = Thursday

F = Friday

S = Saturday

The corresponding probability are

P (W) = 0.3

P (T) = 0.4

P (F) = 0.2

P (S) = 0.1

Let 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)=3

Y(T,W)=1, Y (T,T)=1, Y (T,F)=2, Y (T,S)=3

Y(F,W)=2, Y (F,T)=2, Y (F,F)=2, Y (F, S)=3

Y(S,W)=3, Y (S,T)=3, Y (S,F)=3, Y (S,S)=3

The probability mass function of Y is

P(Y=0)=0.3(0.3)=0.9

P(Y=1) = P[(W,T) or (T,W) or (T,T)]

= [0.3(0.4) + 0.3(0.4) + 0.4(0.4)]

=0.4

P(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)]

=0.32

P(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

£56

step-by-step explanation:

u divide 24 by 3 and u get 8 so if u do 8 time 7 or 8+8+8+8+8+8+8 u get 56 and 56 divided by 7 is 8.