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Question
A die is thrown three times. Let X be ‘the number of twos seen’. Find the expectation of X.
Solution
Here, we have X = 0, 1, 2, 3 ......[∵ Die is thrown 3 times]
And p = `1/6`, q= `5/6`
∴ P(X = 0) = P(not 2) . P(not 2) . P(not 2)
= `5/6 * 5/6 * 5/6`
= `125/216`
P(X = 1) = P(2) . P(not 2) . P(not 2) + P(not 2) . P(2) . P(not 2) + P(not 2) . P(not 2) . P(2)
= `1/6 * 5/6 * 5/6 + 5/6 * 1/6 * 5/6 * 5/6 * 1/6`
= `25/216 + 25/216 + 25/216`
= `75/216`
P(X = 2) = P(2) . P(2) . P(not 2) + P(2) . P(not 2) . P(2) + P(not 2) . P(2) . P(2)
= `1/6 * 1/6 * 5/6 + 1/6 * 5/6 * 1/6 + 5/6 * 1/6 * 1/6`
= `5/216 + 5/216 + 5/216`
= `15/216`
P(X = 3) = P(2) . P(2) . P(2)
= `1/6 * 1/6 * 1/6`
= `1/216`
Now E(X) = `sum_("i" = 1)^"n" "p"_"i"x_"i"`
= `0 xx 125/216 + 1 xx 75/216 + 2 xx 15/216 + 3 xx 1/216`
= `0 + 75/216 + 30/216 + 3/216`
= `(75 + 30 + 3)/216`
= `108/216`
= `1/2`
Hence, the required expectation is `1/2`.
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