If $A$ and $B$ are two events such that $P(A)=\frac{1}{3}$,$P(B)=\frac{1}{2}$ and $P(A \cap B)=\frac{1}{6}$,then $P(A^{\prime} | B)$ is

Let $X$ be a random variable which takes values $1, 2, 3, 4$ such that $P(X=r) = K r^3$ where $r = 1, 2, 3, 4$. Then:

If $A$ and $B$ are two events such that $P(A) = \frac{3}{8}$,$P(B) = \frac{5}{8}$ and $P(A \cup B) = \frac{3}{4}$,then $P(A|B) = $

Two dice $A$ and $B$ are rolled. If it is known that the number on $B$ is $5$,then the probability that the sum of the numbers on the two dice will be greater than $9$ is

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

Two cards are drawn from a pack of $52$ playing cards one after the other. If $p_1$ is the probability of getting a queen in the first draw and a diamond card in the second draw when the first card drawn is replaced,and $p_2$ is the probability of the same event when the first card drawn is not replaced,then $\frac{p_1}{p_2} = $