Given that $A$ and $B$ are two events such that $P(B) = \frac{3}{5}$,$P\left(\frac{A}{B}\right) = \frac{1}{2}$,and $P(A \cup B) = \frac{4}{5}$,then $P(A)$ is equal to:

$A$ college student has to appear for two examinations $A$ and $B$. The probabilities that the student passes in $A$ and $B$ are $\frac{2}{3}$ and $\frac{3}{4}$ respectively. If it is known that the student passes at least one among the two examinations,then the probability that the student will pass both the examinations 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

If $E_1$ and $E_2$ are two events of the sample space such that $P(E_1) = \frac{1}{4}$, $P(E_1 | E_2) = \frac{1}{2}$ and $P(E_2 | E_1) = \frac{1}{3}$, then $P(E_1 | \bar{E}_2) = $

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

Let $A$ and $B$ be two events such that $P(A) = \frac{5}{11}$, $P(B) = \frac{2}{11}$, and $P(A \cup B) = \frac{3}{11}$, then $P(A'|B')$ is . . . . . . .