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

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

Answers (1)

Given-

X= no. of twos seen

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

\begin{aligned} &\therefore P(X=0)=P_{n o t 2} \cdot P_{n o t2} \cdot P_{n o t 2}=\frac{5}{6} \times \frac{5}{6} \times \frac{5}{6}=\frac{125}{216}\\ &P(X=1)=\left(\frac{5}{6} \times \frac{5}{6} \times \frac{1}{6}\right)+\left(\frac{5}{6} \times \frac{1}{6} \times \frac{5}{6}\right)+\left(\frac{1}{6} \times \frac{5}{6} \times \frac{5}{6}\right)\\ &=\frac{25}{36} \times \frac{3}{6}\\ &=\frac{25}{72}\\\end{aligned}

\\P(X=2) =\left(\frac{5}{6} \times \frac{1}{6} \times \frac{1}{6}\right)+\left(\frac{1}{6} \times \frac{1}{6} \times \frac{5}{6}\right)+\left(\frac{1}{6} \times \frac{5}{6} \times \frac{1}{6}\right) \\ =3\left(\frac{5}{6} \times \frac{1}{6} \times \frac{1}{6}\right) \\ =\frac{5}{72} \\ P(X=3)=P_{2} \cdot P_{2} \cdot P_{2} \\ =\frac{1}{6} \times \frac{1}{6} \times \frac{1}{6} \\ =\frac{1}{216}

\begin{aligned} &\text { As it is known that, }\\ &\mathrm{E}(\mathrm{X})=\Sigma \mathrm{XP}(\mathrm{X})=0 \times \frac{125}{216}+1 \times \frac{25}{72}+2 \times \frac{15}{216}+3 \times \frac{1}{216}\\ &=\frac{75+30+3}{216}\\ &=\frac{108}{216}\\ &=\frac{1}{2} \end{aligned}

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