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If a die is thrown at random, then the expectation of the number on it is
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3.5
(D)
When a die is thrown, probability of getting any number from 1 to 6 is $\frac{1}{6}$. Expectation of a number on it to occur
$\begin{array}{l}
=\left(1 \times \frac{1}{6}\right)+\left(2 \times \frac{1}{6}\right)+\left(3 \times \frac{1}{6}\right)+\left(4 \times \frac{1}{6}\right)+\left(5 \times \frac{1}{6}\right)+\left(6 \times \frac{1}{6}\right) \\
=\frac{1}{6}(1+2+3+4+5+6)=\frac{21}{6}=3.5
\end{array}$
When a die is thrown, probability of getting any number from 1 to 6 is $\frac{1}{6}$. Expectation of a number on it to occur
$\begin{array}{l}
=\left(1 \times \frac{1}{6}\right)+\left(2 \times \frac{1}{6}\right)+\left(3 \times \frac{1}{6}\right)+\left(4 \times \frac{1}{6}\right)+\left(5 \times \frac{1}{6}\right)+\left(6 \times \frac{1}{6}\right) \\
=\frac{1}{6}(1+2+3+4+5+6)=\frac{21}{6}=3.5
\end{array}$
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