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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

a) The value of Z is computed for each X and Y possibility here 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

As each of the above outcome is equally likely and have a probability of 1/16 = 0.0625 occurrence, therefore the PDF for Z here is computed 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

The expected value of Z now is computed here as:

E(Z) = \sum zP(z)

E(Z) = 2*0.0625 + 3*0.125 + .... + 7*0.125 + 8*0.0625 = 5

Therefore 5 is the required expected value of Z here.

b) The probability here is computed using Bayes theorem as:
P(Z = 7 | Z > 5)

= P(Z = 7) / P(Z = 6, 7 or 8)

= 0.125 / (0.125 + 0.1875 + 0.0625) = 1/3

Therefore 0.3333 is the required probability here.

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

= Probability of not getting Z = 7 in the first 4 games

= (1 - 0.125)4 = 0.5862

Therefore 0.5862 is the required probability here.

d) Probability that we get Z = 4, five times in the 10 roll is computed using binomial probability function as:

(9) 10 5 10.187571 -0.1875) = 0.0207

Therefore 0.0207 is the required probability here.

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