Question

Suppose you roll two 4 sided dice. Let the probabilities of the first die be represented by random variable X and those of the second die be represented by random variable Y. Let random variable Z be (X+Y). The range of Z will be from 2-8. 1.(2.5 points) Find E(Z). 2. (2.5 points) Find P(Z = 71Z > 5). 3.(2.5 points) What is the probability that you have to play the game 4 times before you roll Z=7? 4. (2.5 points) What is the probability that when you play the game 10 times, you get Z=4 5 times out of the 10 rolls?

Answer 1

The value of Z for each X, Y here is obtained as:

X	Y	Z
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

Using this the PDF for Z here is obtained as:

z	P(z)
2	0.0625
3	0.125
4	0.1875
5	0.25
6	0.1875
7	0.125
8	0.0625

a) The expected value of Z here is computed as:

$E(Z) = \sum zP(Z = z) = 2*0.0625 + 3*0.125 + ... + 7*0.125 + 8*0.0625$

Therefore 5 is the expected value here.

b) Using bayes theorem, probability here is computed as:
P(Z = 7 | Z > 5) = P(Z = 7) / P(Z = 6, 7 or 8)

= 0.125 / (0.1875 + 0.125 + 0.0625)

= 0.35

therefore 0.35 is the required probability here.

c) The probability that we will have to play 4 games before we roll Z = 7 is computed here as:

= [1 - P(Z = 7) ]⁴ = (1 - 0.125)⁴ = 0.5862

Therefore 0.5862 is the required probability here.

d) The probability that in 10 games played, we get Z = 4 five times is computed here using binomial probability function as:

$= \binom{10}{5}0.1875^5(1 - 0.1875)^5 = 0.0207$

Therefore 0.0207 is the required probability here.