Two dice are thrown 4 times simultaneously. W | vedclass

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(C) The probability of getting the same numbers on both dice in a single throw is $p = \frac{6}{36} = \frac{1}{6}$.Here,we use the binomial distributi

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$25/216$

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(C) The probability of getting the same numbers on both dice in a single throw is $p = \frac{6}{36} = \frac{1}{6}$.Here,we use the binomial distribution formula $P(X = r) = ^nC_r p^r q^{n-r}$,where $n = 4$,$r = 2$,$p = \frac{1}{6}$,and $q = 1 - p = \frac{5}{6}$.The required probability is $P(X = 2) = ^4C_2 \left(\frac{1}{6}ight)^2 \left(\frac{5}{6}ight)^{4-2}$.$P(X = 2) = 6 	imes \left(\frac{1}{36}ight) 	imes \left(\frac{25}{36}ight) = \frac{150}{1296} = \frac{25}{216}$.

Two dice are thrown $4$ times simultaneously. What is the probability that both dice show the same numbers exactly twice?

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