If $A$ and $B$ are two independent events such that $P(A^{\prime}) = 0.75$,$P(A \cup B) = 0.65$ and $P(B) = p$,then the value of $p$ is

$A$ cubical die with faces marked $1, 2, 3, ..., 6$ is tossed such that the probability of throwing the number $t$ is proportional to $t^2$. The probability that the number $5$ has appeared,given that the number turned up is not even,is:

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

$Face$	$1$	$2$	$3$	$4$	$5$	$6$
$Probability$	$0.1$	$0.32$	$0.21$	$0.15$	$0.05$	$0.17$

The die is tossed and you are told that either face $1$ or $2$ has turned up. Then the probability that it is face $1$ is:

If $A$ and $B$ are events such that $P(A | B) = P(B | A)$,then

Assume that each born child is equally likely to be a boy or a girl. If a family has two children,what is the conditional probability that both are girls given that at least one is a girl?

$A$ fair die is rolled. Consider events $E=\{1,3,5\}, F=\{2,3\}$ and $G=\{2,3,4,5\}$. Find $P(E | F)$ and $P(F | E)$.