The probability distribution of number of doublets in three throws of a pair of dice is 

no. of possibilities of two dices = 6 × 6 = 36

no. of cases of doublets = 6

P(D) = P(doublet) = $\frac{6}{36} = \frac{1}{6}$ in one throw

$P(\bar{D})$ = P(doublet) = $\frac{5}{6}$ in one throw

X → no. of doublets in 3 throws

$P(X = 0) = { }^3 C_0 \times\left(\frac{5}{6}\right)^3=\frac{125}{216}$

$P(X = 1) = { }^3 C_1 \times\left(\frac{1}{6}\right) \left(\frac{5}{6}\right)^2=\frac{25×3}{216} = \frac{75}{216}$

$P(X=2)={ }^3 C_2 \times\left(\frac{1}{6}\right)^2 \times \frac{5}{6}=\frac{15}{216}$

$P(X=2)=3 C_3 \times\left(\frac{5}{6}\right)^0 \times\left(\frac{1}{6}\right)^3=\frac{1}{216}$

Option: 1