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(D) Let $S$ be the sum of the numbers on the two dice. The possible values of $S$ range from $2$ to $12$.The set of prime sums is $P = \{2, 3, 5, 7, 1

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(D) Let $S$ be the sum of the numbers on the two dice. The possible values of $S$ range from $2$ to $12$.The set of prime sums is $P = \{2, 3, 5, 7, 11\}$. The number of outcomes for these sums are: $P(2)=1, P(3)=2, P(5)=4, P(7)=6, P(11)=2$. Total outcomes for prime = $1+2+4+6+2 = 15$. So,$P(	ext{Prime}) = \frac{15}{36}$.The set of perfect square sums is $Q = \{4, 9\}$. The number of outcomes for these sums are: $P(4)=3, P(9)=4$. Total outcomes for perfect square = $3+4 = 7$. So,$P(	ext{Perfect Square}) = \frac{7}{36}$.The probability of getting neither a prime nor a perfect square is $1 - (\frac{15}{36} + \frac{7}{36}) = 1 - \frac{22}{36} = \frac{14}{36}$.Let $E$ be the event that a perfect square occurs before a prime. The probability $P(E) = \frac{7/36}{1 - 14/36} = \frac{7}{22}$.We are given that a perfect square occurs before a prime. We want the probability that this perfect square is odd. The odd perfect square in our set is $9$,which has $4$ outcomes. The even perfect square is $4$,which has $3$ outcomes.Given the condition,the probability that the perfect square is $9$ (odd) is $\frac{P(9)}{P(4) + P(9)} = \frac{4}{3+4} = \frac{4}{7}$.Thus,$p = \frac{4}{7}$.Therefore,$14p = 14 	imes \frac{4}{7} = 8$.

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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