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

b. 6

step-by-step explanation:

we are given the table representing an exponential function.

we can see from the table that, as x is increasing, the value of y is increasing in the power of 6.

i.e. the following relation is obtained,

x = 1, y = 6

x = 2, y = 36 =

x = 3, y = 216 =

x = 4, y = 1296 =

x = 5, y = 7776 =

so, the geometric series obtained is .

thus, the common ratio is given by = = 6.

hence, the common ratio is 6.