Find the probability distribution of the number of sixes in three throws of a die. Also find the mean and variance of the distribution.

Mean: $\frac{1}{2}$, Variance: $\frac{5}{12}$

The correct answer is Option (2) → Mean: $\frac{1}{2}$, Variance: $\frac{5}{12}$

When a die is thrown, probability of getting a six = $p = \frac{1}{6}$,

so $q = 1-p=1-\frac{1}{6}=\frac{5}{6}$. Here $n = 3$.

Thus, we have a binomial distribution with $p =\frac{1}{6},q=\frac{5}{6}$ and $n = 3$.

If X denotes the number of sixes obtained, then X can take values 0, 1, 2, 3.

$P(0)= {^3C}_0 q^3 = (\frac{5}{6})^2 = \frac{125}{216}$,

$P(1)={^3C}_1pq^2 = 3.\frac{1}{6}.(\frac{5}{6})^2=\frac{75}{216}$

$P(2) = {^3C}_2 p^2q = 3.(\frac{1}{6})^2.\frac{5}{6}=\frac{15}{216}$ and

$P(3) = {^3C}_3p^3 = (\frac{1}{6})^3 = \frac{1}{216}$

∴ Required probability distribution is $\begin{pmatrix}0&1&2&3\\\frac{125}{216}&\frac{75}{216}&\frac{15}{216}&\frac{1}{216}\end{pmatrix}$

∴ Mean = $μ = Σp_ix_i =\frac{125}{216}×0+\frac{75}{216}×1+\frac{15}{216}×2+\frac{1}{216}×3=\frac{108}{216}=\frac{1}{2}$

$Σp_i{x_i}^2 =\frac{125}{216}×0^2+\frac{75}{216}×1^2+\frac{15}{216}×2^2+\frac{1}{216}×3^2$

$=\frac{75}{216}+\frac{60}{216}+\frac{9}{216}=\frac{144}{216}=\frac{2}{3}$

∴ Variance = $Σp_i{x_i}^2- μ^2=\frac{2}{3}-(\frac{1}{2})^2=\frac{2}{3}-\frac{1}{4}=\frac{5}{12}$