Question

Problem 4. Two four-sided dice are rolled simultaneously.

Answer 1

(a)

Following is the sample space when 4 sided dice twice rolled and also shows the sum of outcome:

$S=\begin{Bmatrix} (1,1) =2&(1,2) =3&(1,3)=4 & (1,4)=5 \\ (2,1)=3 &(2,2)=4 &(2,3)=5 & (2,4)=6 \\ (3,1)=4 &(3,2)=5 &(3,3)=6 & (3,4)=7 \\ (4,1)=5 &(4,2)=6 &(4,3)=7 & (4,4)=8 \end{Bmatrix}$

Let X shows the sum of two fair die. So X can take values 2, 3, 4,...8. Following table shows the probability distribution of X:

X	P(X=x)
2	1/16
3	2/16
4	3/16
5	4/16
6	3/16
7	2/16
8	1/16

Following table shows the calculations for mean and variance of X:

Pgex) XP(X=x) 2116 6/16 12/16 20/16 18/16 14/16 2116 3/16 3/16 2/16 1/16 5 Total

The expected value of X is

(b)

Let Y shows the amount you win. So you have

Y = 5X - A

The expected value of Y is

E(Y) = E(5X-A) = 5E(X) - A =5*5 - A = 25 - A

For break even expected value of Y must be equal to 0 so

E(Y) = 0

25 -A =0

A = 25

Answer: A = $25

(C)

Following table shows the calculations:

X	P(X=x)	P(X=x)	X^2	P(X^2=x^2)	X^2*P(X^2=x^2)
2	1/16	0.0625	4	0.0625	0.25
3	2/16	0.125	9	0.125	1.125
4	3/16	0.1875	16	0.1875	3
5	4/16	0.25	25	0.25	6.25
6	3/16	0.1875	36	0.1875	6.75
7	2/16	0.125	49	0.125	6.125
8	1/16	0.0625	64	0.0625	4
Total		1			27.5

The expected value of X^2 is

Let Y shows the amount you win. So you have

The expected value of Y is

For break even expected value of Y must be equal to 0 so

E(Y) = 0

137.5-A =0

A = 137.5

Answer: A = $137.5