Let $A$ and $B$ be two events such that $P(A) = \frac{3}{8}$, $P(B) = \frac{5}{8}$ and $P(A \cup B) = \frac{3}{4}$. Then $P(A'|B) - P(A|B) =$ . . . . . .

For two events $A$ and $B$,$P(B) \neq 0$ and $P(A \mid B) = 1$,then . . . . . . .

$A$ die is thrown two times and the sum of the scores appearing on the die is observed to be a multiple of $4$. Then the conditional probability that the score $4$ has appeared at least once is

Two cards are drawn one by one from a pack of cards. The probability of getting the first card as an ace and the second as a coloured card is (without replacement).

Two events $A$ and $B$ are such that $P(A)=\frac{1}{4}$, $P(A|B)=\frac{1}{4}$ and $P(B|A)=\frac{1}{2}$. Consider the following statements:
$(I) P(\bar{A}|\bar{B})=\frac{3}{4}$
$(II) A$ and $B$ are mutually exclusive
$(III) P(A|B)+P(A|\bar{B})=1$
Then,

$A$ fair die is tossed until a six is obtained. Let $X$ be the number of required tosses,then the conditional probability $P(X \geq 5 \mid X > 2)$ is: