Find the probability distribution of the number of doublets in four t…

The correct answer is Option (3) → Mean: $\frac{2}{3}$, Variance: $\frac{5}{9}$

When a pair of dice is thrown, the sample space has 36 equally likely outcomes, out of which six outcomes are doublets-(1, 1), (2, 2), (3, 3), (4, 4), (5, 5) and (6, 6).

If $p$ = probability of getting a doublet, then $p =\frac{6}{36}=\frac{1}{6}$,

so $q = 1-\frac{1}{6}=\frac{5}{6}$. Here $n = 4$.

Thus, we have a binomial distribution with $p=\frac{1}{6},q=\frac{5}{6}$ and $n = 4$.

If X denotes the number of doublets obtained, then X takes the values 0, 1, 2, 3, 4.

$P(0) = {^4C}_0 q^2 = (\frac{5}{6})^4=\frac{325}{1296}$,

$P(1) = {^4C}_1pq^3=4.\frac{1}{6}.(\frac{5}{6})^3=\frac{500}{1296}$,

$P(2) = {^4C}_2p^2q^2=6(\frac{1}{6})^2(\frac{5}{6})^2=\frac{150}{1296}$,

$P(3) = {^4C}_3 p^3q = 4 (\frac{1}{6})^3\frac{5}{6}=\frac{20}{1296}$ and

$P(4) = {^4C}_4p^4 = (\frac{1}{6})^4=\frac{1}{1296}$.

∴ The required probability distribution is $\begin{pmatrix}0&1&2&3&4\\\frac{625}{1296}&\frac{500}{1296}&\frac{150}{1296}&\frac{20}{1296}&\frac{1}{1296}\end{pmatrix}$

Mean = $μ = ∑p_ix_i=\frac{1}{1296}(625 × 0 + 500 × 1 + 150 × 2 + 20 × 3+1 × 4)$

$=\frac{1}{1296}×864=\frac{2}{3}$

Now $Σp_i{x_i}^2=\frac{1}{1296}(625 × 0^2 + 500 × 1^2 + 150 × 2^2 + 20 × 3^2 + 1 × 4^2)$

$=\frac{1}{1296}(500+ 600+ 180 + 16) =\frac{1296}{1296}=1$.

∴ Variance = $Σp_i{x_i}^2 – μ^2 = 1 –(\frac{2}{3})^2=1-\frac{4}{9}=\frac{5}{9}$