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Question
Find the probability distribution of the maximum of the two scores obtained when a die is thrown twice. Determine also the mean of the distribution.
Solution
Let X be the random variable scores when a die is thrown twice.
X = 1, 2, 3, 4, 5, 6
And S = {(1, 1), (1, 2), (2, 1), (2, 2), (1, 3), (2, 3), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), ..., (6, 6)}
So, P(X = 1) = `1/6*1/6 = 1/36`
P(X = 2) = `1/6*1/6 + 1/6*1/6 + 1/6*1/6 = 3/36`
P(X = 3) = `1/6*1/6 + 1/6*1/6 + 1/6*1/6 + 1/6*1/6 + 1/6*1/6 = 5/36`
Similarly P(X = 4) = `7/36`
P(X = 5) = `9/36`
And P(X = 6) = `11/36`
So, the required distribution is
| X | 1 | 2 | 3 | 4 | 5 | 6 |
| P(X) | `1/36` | `3/36` | `5/36` | `7/36` | `9/36` | `11/36` |
Now, the mean E(X) = `sum_("i" = 1)^"n" x_"i""p"_"i"`
= `1 xx 1/36 + 2 xx 3/36 + 3 xx 5/36 + 4 xx 7/36 + 5 xx 9/36 + 6 xx 11/36`
= `1/36 + 6/36 + 15/36 + 28/36 + 45/36 + 66/36`
= `161/36`
Hence, the required mean = `161/36`.
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