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:

$A$ and $B$ are two events such that $P(A) = 0.8$,$P(B) = 0.6$,and $P(A \cap B) = 0.5$. Then the value of $P(A/B)$ is:

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

$A$ 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:

In a hostel,$60 \%$ of the students read Hindi newspaper,$40 \%$ read English newspaper and $20 \%$ read both Hindi and English newspapers. $A$ student is selected at random. If she reads English newspaper,find the probability that she reads Hindi newspaper.

Let $X$ be a random variable which takes values $1, 2, 3, 4$ such that $P(X=r) = K r^3$ where $r = 1, 2, 3, 4$. Then: