$A$ box contains $100$ tickets numbered $1, 2, \dots, 100$. Two tickets are chosen at random. It is given that the maximum number on the two chosen tickets is not more than $10$. What is the probability that the minimum number on them is $5$?

$A$ candidate takes three tests in succession and the probability of passing the first test is $p$. The probability of passing each succeeding test is $p$ if he passes the preceding one, and $\frac{p}{2}$ if he fails the preceding one. The candidate is selected if he passes at least two tests. The probability that the candidate is selected is:

Let $X$ and $Y$ be two events of a sample space such that $P(X)=\frac{1}{3}$, $P(X|Y)=\frac{1}{2}$ and $P(Y|X)=\frac{2}{5}$, then which of the following is true?

Let $X$ and $Y$ be two events such that $P(X)=\frac{1}{3}$,$P(X \mid Y)=\frac{1}{2}$ and $P(Y \mid X)=\frac{2}{5}$. Then:
$A) P(X^{\prime} \mid Y)=\frac{1}{2}$
$B) P(X \cap Y)=\frac{1}{5}$
$C) P(X \cup Y)=\frac{2}{5}$
$D) P(Y)=\frac{4}{15}$

For any two events $A$ and $B$,if $P(A) + P(B) - P(A \cap B) = P(A)$,then . . . . . . .

It is given that $A$ and $B$ are such that $P(A) = \frac{1}{4}$,$P(A|B) = \frac{1}{2}$,and $P(B|A) = \frac{2}{3}$. Then $P(B) = $?