$A$ and $B$ are two events such that $P(A)=0.54$,$P(B)=0.69$ and $P(A \cap B)=0.35$. Find $P(A' \cap B')$.

If the probability of $X$ to fail in the examination is $0.3$ and that for $Y$ is $0.2$,then the probability that either $X$ or $Y$ fail in the examination is

There are three events $A, B, C$,one of which must and only one can happen. The odds are $8:3$ against $A$,$5:2$ against $B$,and the odds against $C$ are $43:17k$. Then the value of $k$ is:

From the employees of a company,$5$ persons are selected to represent them in the managing committee of the company. Particulars of five persons are as follows :

$S.No.$	Name	Sex	Age in years
$1.$	Harish	$M$	$30$
$2.$	Rohan	$M$	$33$
$3.$	Sheetal	$F$	$46$
$4.$	Alis	$F$	$28$
$5.$	Salim	$M$	$41$

$A$ person is selected at random from this group to act as a spokesperson. What is the probability that the spokesperson will be either male or over $35$ years?

$P(A \cup B) = P(A \cap B)$ if and only if the relation between $P(A)$ and $P(B)$ is

$A$ random variable $X$ has the probability distribution:

$X$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$
$P(X)$	$0.15$	$0.23$	$0.12$	$0.10$	$0.20$	$0.08$	$0.07$	$0.05$

For the events $E = \{X \text{ is a prime number}\}$ and $F = \{X < 4\}$,then $P(E \cup F)$ is