A pair of dice is thrown twice in succession. | vedclass

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(C) When a pair of dice is thrown, the total number of outcomes is $n(S) = 6 	imes 6 = 36$.Prime numbers on a die are $\{2, 3, 5\}$. The outcomes whe

Vedclass · Accepted Answer

$\frac{1}{36}$

Vedclass · Answer

(C) When a pair of dice is thrown, the total number of outcomes is $n(S) = 6 	imes 6 = 36$.Prime numbers on a die are $\{2, 3, 5\}$. The outcomes where both dice show prime numbers are $\{(2,2), (2,3), (2,5), (3,2), (3,3), (3,5), (5,2), (5,3), (5,5)\}$. There are $9$ such outcomes.Probability of getting prime numbers on both dice in the first throw is $P(A) = \frac{9}{36} = \frac{1}{4}$.Composite numbers on a die are $\{4, 6\}$ (Note: $1$ is neither prime nor composite). The outcomes where both dice show composite numbers are $\{(4,4), (4,6), (6,4), (6,6)\}$. There are $4$ such outcomes.Probability of getting composite numbers on both dice in the second throw is $P(B) = \frac{4}{36} = \frac{1}{9}$.Since the two throws are independent, the required probability is $P(A) 	imes P(B) = \frac{1}{4} 	imes \frac{1}{9} = \frac{1}{36}$.

$A$ pair of dice is thrown twice in succession. The probability of getting prime numbers on both the dice in the first throw and composite numbers on both the dice in the second throw is

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