Let X denotes the number of doublets obtained in 3 throws of a pair of dice.

Match List-I with List-II

Choose the correct answer from the options given below:

(A)-(IV), (B)-(III), (C)-(II), (D)-(I)

The correct answer is Option (3) → (A)-(IV), (B)-(III), (C)-(II), (D)-(I)

Probability of getting a doublet (same numbers on both dice) in one throw:

$p=\frac{6}{36}=\frac{1}{6}$

Probability of not getting a doublet:

$q=1-p=\frac{5}{6}$

Let $X$ = number of doublets in 3 throws.

$P(X=r)=\frac{3!}{r!(3-r)!}\,p^{r}\,q^{3-r}$

$P(X=0)=\frac{3!}{0!3!}\left(\frac{1}{6}\right)^{0}\left(\frac{5}{6}\right)^{3}=\frac{125}{216}$ → (IV)

$P(X=1)=\frac{3!}{1!2!}\left(\frac{1}{6}\right)\left(\frac{5}{6}\right)^{2}=\frac{3\times25}{216}=\frac{75}{216}$ → (III)

$P(X=2)=\frac{3!}{2!1!}\left(\frac{1}{6}\right)^{2}\left(\frac{5}{6}\right)=\frac{3\times5}{216}=\frac{15}{216}$ → (II)

$P(X=3)=\frac{3!}{3!0!}\left(\frac{1}{6}\right)^{3}\left(\frac{5}{6}\right)^{0}=\frac{1}{216}$ → (I)

Correct Matching:

(A) → (IV), (B) → (III), (C) → (II), (D) → (I)