If $P(A)=0.25, P(B)=0.55$ and $P(A \cup B)=0.65$,then $P(B' \mid A) =$ . . . . . . .

Let $E_{1}$ and $E_{2}$ be two events such that the conditional probabilities $P(E_{1} \mid E_{2}) = \frac{1}{2}$,$P(E_{2} \mid E_{1}) = \frac{3}{4}$ and $P(E_{1} \cap E_{2}) = \frac{1}{8}$. Then:

$A$ 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:

Given $P(A)=0.5, P(B)=0.4, P(A \cap B)=0.3$,then $P(A^{\prime} / B^{\prime})$ is equal to

Let $A$ and $B$ be two events with $P(A) = \frac{1}{7}$,$P(A|B) = \frac{2}{3}$,and $P(B) = \frac{2}{7}$. Then,the value of $P(B|A)$ is

The probabilities of Virat becoming the Man of the Match in the $1^{st}$,$2^{nd}$,and $3^{rd}$ matches of India in the World Cup $2015$ are $\frac{3}{7}$,$\frac{2}{7}$,and $\frac{1}{7}$ respectively. If Virat got the Man of the Match in exactly one match,what is the probability that he got it in the $3^{rd}$ match?