In a game, two dice are thrown simultaneously | vedclass

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(B) Step $1$: Calculate the probability that person $A$ wins. The sum of two dice ranges from $2$ to $12$. The prime sums are ${2, 3, 5, 7, 11}$. The

Vedclass · Accepted Answer

$\frac{40}{663}$

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(B) Step $1$: Calculate the probability that person $A$ wins. The sum of two dice ranges from $2$ to $12$. The prime sums are ${2, 3, 5, 7, 11}$. The number of outcomes for these sums are: $2(1), 3(2), 5(4), 7(6), 11(2)$. Total favorable outcomes $= 1 + 2 + 4 + 6 + 2 = 15$. So, $P(A) = \frac{15}{36} = \frac{5}{12}$.Step $2$: Calculate the probability that person $B$ wins. There are $12$ face cards (Jack, Queen, King of each suit) and $16$ prime-numbered cards (Ace is not prime, so $2, 3, 5, 7$ of each suit, $4 	imes 4 = 16$). Total ways to draw $2$ cards from $52$ is $^{52}C_2 = \frac{52 	imes 51}{2} = 1326$. Favorable ways $= ^{12}C_1 	imes ^{16}C_1 = 12 	imes 16 = 192$. So, $P(B) = \frac{192}{1326} = \frac{32}{221}$.Step $3$: Since $A$ and $B$ are independent, $P(A \cap B) = P(A) 	imes P(B) = \frac{5}{12} 	imes \frac{32}{221} = \frac{5 	imes 8}{3 	imes 221} = \frac{40}{663}$.

List-$I$	List-$II$
$(I)$ Probability of $(X_2 \geq Y_2)$ is	$(P)$ $\frac{3}{8}$
$(II)$ Probability of $(X_2 > Y_2)$ is	$(Q)$ $\frac{11}{16}$
$(III)$ Probability of $(X_3 = Y_3)$ is	$(R)$ $\frac{5}{16}$
$(IV)$ Probability of $(X_3 > Y_3)$ is	$(S)$ $\frac{355}{864}$
	$(T)$ $\frac{77}{432}$

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