If $m$ and ${\sigma ^2}$ are the mean and variance of a random variable $X$,whose distribution is given by:

$X=x$	$0$	$1$	$2$	$3$	$4$
$P(X=x)$	$\frac{1}{3}$	$\frac{1}{2}$	$0$	$\frac{1}{6}$	$0$

,then:

If a random variable $X$ has the following probability distribution,then its variance is nearly:

$X=x$	$-3$	$-2$	$-1$	$0$	$1$	$2$	$3$
$P(X=x)$	$0.05$	$0.1$	$2K$	$0$	$0.3$	$K$	$0.1$

If $X$ is a Poisson variate with $P(X=0)=0.8$,then the variance of $X$ is

India plays two matches each with West Indies and Australia. In any match,the probabilities of India getting $0, 1,$ and $2$ points are $0.45, 0.05,$ and $0.50$ respectively. Assuming that the outcomes are independent,the probability of India getting at least $7$ points is:

Given the probability density function: $f(x) = \begin{cases} 3(1 - 2x^2), & 0 < x < 1 \\ 0, & \text{otherwise} \end{cases}$ The probability $P\left(\frac{1}{4} < X < \frac{1}{3}\right)$ is given by: $P\left(\frac{1}{4} < X < \frac{1}{3}\right) = \int_{1/4}^{1/3} 3(1 - 2x^2) \, dx$

It is known that a box of $8$ batteries contains $3$ defective pieces and a person randomly selects two batteries from the box. If $X$ is the number of defective batteries selected,then $P(X \leq 1) = $