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

(a) Let X be the sum of the two rolls. Calculate the PMF and the expected value of . (b) Someone proposes to give you in dollars five times the amount of the sum X that you roll, if you pay A dollars in advance. What should be the amount A in order for you to expect to break even? (c) Repeat parts (a) and (b) for the case where X is the square of the sum of the two rolls.

math   Statistics-And-Probability   
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Answer #1

(a)

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

(1,1)=2 (1,2)-3 (1,3)-4 (1,4)一5 (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

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

E(X)=sum xP(x)=5

(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

E(X^{2})=sum x^{2}P(X^{2}=x^{2})=27.5

Let Y shows the amount you win. So you have

Y = 5X^{2} - A

The expected value of Y is

E(y) = E(5X2-A) = 5E(X2) _ A = 5 . 27.5-A = 137.5-A

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

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