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.) = 0.26, P(Thurs.) = 0.39, P(Fri.) = 0.25, and P(Sat.) = 0.10. 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.)

y                 0             1              2          3

P(Y=y)   0.0676   0.3549   0.3875   0.19

Step-by-step explanation:

P(Wed) = 0.26

P(Thurs) = 0.39

P(Fri) = 0.25

P(Sat) = 0.10

Y = No. of days beyond Wednesday it takes for both magazines to arrive i.e. 0,1,2,3

Y=0 means the magazines will arrive on Wednesday

Y=1 means the magazines will arrive till Thursday

Y=2 means the magazines will arrive till Friday

Y=3 means the magazines will arrive till Saturday

The possible combinations for Y are

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

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

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

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

So, we can classify these possible outcomes as Y=0,1,2,3.

Y(0) = Y(W,W) (both magazines take 0 days to arrive beyond Wednesday)

Y(1) = Y(W,T), Y(T,T), Y(T,W) (both magazines take 1 day to arrive beyond Wednesday)

Y(2) = Y(W,F), Y(T,F), Y(F,F) Y(F,W) Y(F,T) (both magazines arrive till Friday)

Y(3) = Y(W,S), Y(T,S), Y(F,S), Y(S,W), Y(S,T), Y(S,F), Y(S,S) (both magazines arrive till Saturday)

To calculate the PMF, we need to calculate the probability for each of the points in Y(0,1,2,3).

Y(0) = Y(W,W)

       = 0.26 x 0.26

Y(0) = 0.0676

Y(1) = Y(W,T) + Y(T,T) + Y(T,W)

      = (0.26 x 0.39) + (0.39 x 0.39) + (0.39 x 0.26)

      = 0.1014 + 0.1521 + 0.1014

Y(1) = 0.3549

Y(2) = Y(W,F) + Y(T,F) + Y(F,F) + Y(F,W) + Y(F,T)

  =(0.26 x 0.25) + (0.39 x 0.25) + (0.25 x 0.25) + (0.25 x 0.26) + (0.25 x 0.39)

  = 0.065 + 0.0975 + 0.0625 + 0.065 + 0.0975

Y(2) = 0.3875

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

      = (0.26 x 0.10) + (0.39 x 0.10) + (0.25 x 0.10) + (0.10 x 0.26) + (0.10 x 0.39) + (0.10 x 0.25) + (0.10 x 0.10)

       = 0.026 + 0.039 + 0.025 + 0.026 + 0.039 + 0.025 + 0.010

Y(3) = 0.19

y                 0             1              2          3

P(Y=y)   0.0676   0.3549   0.3875   0.19

The PMF plot is attached as a photo here.

hey there!

if she's saving for six months, let's divide the total number we need for the trip by six, dividing it into the amount that needs to be saved for each month.

500 ÷ 6 =

she needs to save at least this amount each month, so we should use this symbol: ≥

let x be the amount she saves each month.

your final answer is x ≥

hope this !