A die is thrown three times. Let X be the 'number of twos seen', find the expectation of X.

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Solution

We have, X = number of twos seen

So, on throwing a die three times, we will have X = 0, 1, 2,3.

$∴ P (X = 0) = P_{(n o t 2)} . P_{(n o t 2)} . P_{(n o t 2)} = \frac{5}{6} . \frac{5}{6} . \frac{5}{6} = \frac{125}{216}$

$P (X = 1) = P_{{(n o t 2)} . P_{(n o t 2}} . P_{(2)} + P_{(n o t 2)} . P_{(2)} . P_{(n o t 2)} . P_{(n o t 2)} = \frac{5}{6} . \frac{5}{6} \frac{1}{6} + \frac{5}{6} . \frac{1}{6} . \frac{5}{6} + \frac{1}{6} . \frac{5}{6} . \frac{5}{6} = \frac{25}{36} . \frac{3}{6} = \frac{25}{72} P (X = 2) = P_{(n o t 2)} . P_{(2)} . P_{(2)} + P_{(2)} . P_{(2)} . P_{(n o t 2)} + P_{(2)} . P_{(n o t 2)} + P_{(2)} = \frac{5}{6} . \frac{1}{6} . \frac{1}{6} + \frac{1}{6} . \frac{1}{6} . \frac{5}{6} + \frac{1}{6} . \frac{5}{6} . \frac{1}{6} = \frac{1}{36} . [\frac{15}{6}] = \frac{15}{216} P (X = 3) = P_{(2)} . P_{(2)} . P_{(2)} = \frac{1}{6} . \frac{1}{6} . \frac{1}{6} = \frac{1}{216}$
We know that, $E (X) = \sum X P (X) = 0. \frac{125}{216} + 1. \frac{25}{72} + 2 . \frac{15}{216} + 3. \frac{1}{216} = \frac{75 + 30 + 3}{216} = \frac{108}{216} = \frac{1}{2}$

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