Question

Suppose you roll two 4 sided dice. Let the probabilities of the first die be represented by random variable X and those of th

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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

E(Z ) = 5

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.

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