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(D) We are given that two integers $r$ and $s$ are drawn without replacement from the set $\{1, 2, \ldots, n\}$.We need to find the conditional probab

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$\frac{k-1}{n-1}$

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(D) We are given that two integers $r$ and $s$ are drawn without replacement from the set $\{1, 2, \ldots, n\}$.We need to find the conditional probability $P(r \leq k \mid s \leq k)$.By the definition of conditional probability, $P(r \leq k \mid s \leq k) = \frac{P(r \leq k \cap s \leq k)}{P(s \leq k)}$.First, the probability that $s \leq k$ is $P(s \leq k) = \frac{k}{n}$.Next, the probability that both $r \leq k$ and $s \leq k$ is the probability of choosing two distinct integers from the set $\{1, 2, \ldots, k\}$ out of the total ways to choose two distinct integers from the set $\{1, 2, \ldots, n\}$.This is given by $P(r \leq k \cap s \leq k) = \frac{k(k-1)}{n(n-1)}$.Therefore, the conditional probability is $P(r \leq k \mid s \leq k) = \frac{\frac{k(k-1)}{n(n-1)}}{\frac{k}{n}} = \frac{k(k-1)}{n(n-1)} 	imes \frac{n}{k} = \frac{k-1}{n-1}$.

List-$I$	List-$II$
$A$. $P(E_2)$	$(i)$ $\frac{1}{2}$
$B$. $P(\frac{E_1}{E_2})$	$(ii)$ $\frac{5}{6}$
$C$. $P(\frac{\bar{E}_2}{E_1})$	$(iii)$ $\frac{1}{3}$
$D$. $P(\bar{E}_1 \cup \bar{E}_2)$	$(iv)$ $\frac{1}{6}$
	$(v)$ $\frac{2}{3}$

Two integers $r$ and $s$ are drawn one at a time without replacement from the set $\{1, 2, \ldots, n\}$. Then $P(r \leq k \mid s \leq k) =$

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