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:
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:
Therefore 0.0207 is the required probability here.
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