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

If $E_1$ and $E_2$ are two events of a sample space such that $P(E_1) = \frac{1}{4}$,$P(E_2 \mid E_1) = \frac{1}{2}$,and $P(E_1 \mid E_2) = \frac{1}{4}$,then $P(\bar{E}_1 \mid E_2) = $

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?

Suppose $E$ and $F$ are two events of a random experiment. If the probability of occurrence of $E$ is $1/5$ and the probability of occurrence of $F$ given $E$ is $1/10$, then the probability of non-occurrence of at least one of the events $E$ and $F$ is

Let $A$ and $B$ be two events such that $P(B \mid A) = \frac{2}{5}$,$P(A \mid B) = \frac{1}{7}$ and $P(A \cap B) = \frac{1}{9}$. Consider:
$(S1) P(A' \cup B) = \frac{5}{6}$
$(S2) P(A' \cap B') = \frac{1}{18}$.
Then:

Two dice are thrown. If it is known that the sum of numbers on the dice was less than $6$,the probability of getting a sum as $3$ is: