Question

A die is thrown three times. Let X be ‘the number of twos seen’. Find the expectation of X.

Answer 1

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`.