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(D) Let $p$ be the probability of getting a doublet in a single throw of a pair of dice. $A$ pair of dice has $36$ total outcomes,and there are $6$ po

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$X$$0$$1$$2$$3$$4$$P(X)$$\frac{625}{1296}$$\frac{125}{324}$$\frac{25}{216}$$\frac{5}{324}$$\frac{1}{1296}$

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(D) Let $p$ be the probability of getting a doublet in a single throw of a pair of dice. $A$ pair of dice has $36$ total outcomes,and there are $6$ possible doublets: $(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)$.Thus,$p = \frac{6}{36} = \frac{1}{6}$ and $q = 1 - p = 1 - \frac{1}{6} = \frac{5}{6}$.The dice are thrown $n = 4$ times. Let $X$ be the random variable representing the number of doublets obtained. This follows a binomial distribution $B(n, p) = B(4, \frac{1}{6})$.The probability $P(X=k)$ is given by $\binom{n}{k} p^k q^{n-k}$.For $X=0: P(0) = \binom{4}{0} (\frac{1}{6})^0 (\frac{5}{6})^4 = 1 	imes 1 	imes \frac{625}{1296} = \frac{625}{1296}$.For $X=1: P(1) = \binom{4}{1} (\frac{1}{6})^1 (\frac{5}{6})^3 = 4 	imes \frac{1}{6} 	imes \frac{125}{216} = \frac{500}{1296} = \frac{125}{324}$.For $X=2: P(2) = \binom{4}{2} (\frac{1}{6})^2 (\frac{5}{6})^2 = 6 	imes \frac{1}{36} 	imes \frac{25}{36} = \frac{150}{1296} = \frac{25}{216}$.For $X=3: P(3) = \binom{4}{3} (\frac{1}{6})^3 (\frac{5}{6})^1 = 4 	imes \frac{1}{216} 	imes \frac{5}{6} = \frac{20}{1296} = \frac{5}{324}$.For $X=4: P(4) = \binom{4}{4} (\frac{1}{6})^4 (\frac{5}{6})^0 = 1 	imes \frac{1}{1296} 	imes 1 = \frac{1}{1296}$.

The probability distribution of the number of doublets in four throws of a pair of dice is given by

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$X$	$0$	$1$	$2$	$3$	$4$
$P(X)$	$\frac{1}{5}$	$\frac{1}{5}$	$\frac{1}{5}$	$\frac{1}{5}$	$\frac{1}{5}$

$X$	$0$	$1$	$2$	$3$	$4$
$P(X)$	$\frac{1}{4}$	$\frac{1}{4}$	$\frac{1}{4}$	$\frac{1}{4}$	$\frac{1}{4}$

$X$	$0$	$1$	$2$	$3$	$4$
$P(X)$	$\frac{625}{1296}$	$\frac{125}{324}$	$\frac{25}{216}$	$\frac{5}{324}$	$\frac{1}{1296}$