Compute $P(A | B),$ if $P(B)=0.5$ and $P(A \cap B)=0.32$.

Three numbers are chosen at random without replacement from $\{1, 2, 3, 4, 5, 6, 7, 8\}$. The probability that their minimum is $3$,given that their maximum is $6$,is:

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

$A$ biased die is tossed and the respective probabilities for various faces to turn up are given below:

$Face$	$1$	$2$	$3$	$4$	$5$	$6$
$P(F)$	$0.1$	$0.24$	$0.19$	$0.18$	$0.15$	$0.14$

If an even face has turned up,then the probability that it is face $2$ or face $4$ is:

Two events $E$ and $F$ are independent. If $P(E) = \frac{3}{5}$ and $P(F) = \frac{3}{10}$, then $P(E'/F) + P(F'/E) = \text{ . . . . . . }$

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