Let $A$ and $B$ be independent events with $P(A)=0.3$ and $P(B)=0.4$. Find $P(B | A)$.

If $2$ cards drawn at random from a well-shuffled pack of $52$ playing cards are from the same suit,then the probability of getting a face card and a card having a prime number is

Suppose that $E_1$ and $E_2$ are two events of a random experiment such that $P(E_1) = \frac{1}{4}$,$P(E_2 / E_1) = \frac{1}{2}$ and $P(E_1 / E_2) = \frac{1}{4}$. Observe the lists given below. The correct matching of List-$I$ with List-$II$ is:

List-$I$	List-$II$
$(A)$ $P(E_2)$	$(i)$ $1/4$
$(B)$ $P(E_1 \cup E_2)$	$(ii)$ $5/8$
$(C)$ $P(\bar{E}_1 / \bar{E}_2)$	$(iii)$ $1/8$
$(D)$ $P(E_1 / \bar{E}_2)$	$(iv)$ $1/2$
	$(v)$ $3/8$
	$(vi)$ $3/4$

For a biased die,the probabilities for different faces to turn up are

$Face$	$1$	$2$	$3$	$4$	$5$	$6$
$P(F)$	$0.2$	$0.22$	$0.11$	$0.25$	$0.05$	$0.17$

The die is tossed and you are told that either face $4$ or face $5$ has turned up. The probability that it is face $4$ is

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

If $A$ and $B$ are two events such that $P(A | B) = 0.6$,$P(B | A) = 0.3$,and $P(A) = 0.1$,then $P(\bar{A} \cap \bar{B})$ equals: