The mean of the numbers obtained on throwing a die having written 1 on three faces, 2 on two faces and 5 on one face is
The mean of the numbers obtained on throwing a die having written 1 on three faces, 2 on two faces and 5 on one face is
(A) 1
(B) 2
(C) 5
(D) $\frac{8}{3}$
Let X be the random variable representing a number on the die.
The total number of observations is six.
$\therefore P(X=1)=\frac{3}{6}=\frac{1}{2}$
$\mathrm{P}(\mathrm{X}=2)=\frac{2}{6}=\frac{1}{3}$
$\mathrm{P}(\mathrm{X}=5)=\frac{1}{6}$
Therefore, the probability distribution is as follows.
Mean $=\mathrm{E}(\mathrm{X})=\sum p_{i} x_{i}$
$=\frac{1}{2} \times 1+\frac{1}{3} \times 2+\frac{1}{6} \cdot 5$
$=\frac{1}{2}+\frac{2}{3}+\frac{5}{6}$
$=\frac{3+4+5}{6}$
$=\frac{12}{6}$
$=2$
The correct answer is B.
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