$A$ box contains $10$ mangoes,out of which $4$ are rotten. $2$ mangoes are taken out together. If one of them is found to be good,what is the probability that the other is also good?

One ticket is selected at random from $100$ tickets numbered $00, 01, 02, \dots, 98, 99$. If $X$ and $Y$ denote the sum and the product of the digits on the tickets,then $P(X = 9 | Y = 0)$ equals

If $A$ and $B$ are two events such that $P(A) \neq 0$ and $P(B) \neq 1$,then $P\left( \frac{\overline{A}}{\overline{B}} \right) = $

$A$ bag contains $N$ balls out of which $3$ are white,$6$ are green,and the remaining $(N-9)$ balls are blue. Three balls are drawn randomly one after the other without replacement. Let $W_i, G_i$,and $B_i$ denote the events that the ball drawn in the $i^{\text{th}}$ draw is white,green,and blue,respectively. If $P(W_1 \cap G_2 \cap B_3) = \frac{2}{5N}$ and $P(B_3 \mid W_1 \cap G_2) = \frac{2}{9}$,then $N$ equals:

Assume that each born child is equally likely to be a boy or a girl. If two families have two children each,then the conditional probability that all children are girls given that at least two are girls is

Suppose that $E_1$ and $E_2$ are two events of a random experiment such that $P(E_1) = \frac{1}{4}$,$P(E_2 / E_1) = \frac{1}{2}$ and $P(E_1 / E_2) = \frac{1}{4}$. Observe the lists given below. The correct matching of List-$I$ with List-$II$ is:

List-$I$	List-$II$
$(A)$ $P(E_2)$	$(i)$ $1/4$
$(B)$ $P(E_1 \cup E_2)$	$(ii)$ $5/8$
$(C)$ $P(\bar{E}_1 / \bar{E}_2)$	$(iii)$ $1/8$
$(D)$ $P(E_1 / \bar{E}_2)$	$(iv)$ $1/2$
	$(v)$ $3/8$
	$(vi)$ $3/4$