If a Poisson variate $X$ satisfies $P(X=2) = P(X=3)$, then $P(X=5) =$

Let $F(x)$ be the cumulative distribution function (c.d.f.) of a continuous random variable $X$. If $F(b) = 0.7$ and $P(X > a) = 0.4$, then the value of $P(a < X < b)$ is ...

The p.m.f. of a random variable $X$ is given by $P(X=x) = \begin{cases} \frac{\binom{5}{x}}{2^5}, & x = 0, 1, 2, 3, 4, 5 \\ 0, & \text{otherwise} \end{cases}$. Then which of the following is not correct?

Rajesh has just bought a $VCR$ from Maharashtra Electronics and the shop offers an after-sales service contract for Rs. $1000$ for the next five years. Considering the experience of $VCR$ users,the following distribution of maintenance expenses for the next five years is formed:

Expenses	$0$	$500$	$1000$	$1500$	$2000$	$2500$	$3000$
Probability	$0.35$	$0.25$	$0.15$	$0.10$	$0.08$	$0.05$	$0.02$

The expected value of the maintenance cost is:

In a city,$10$ accidents take place in a span of $50$ days. Assuming that the number of accidents follows the Poisson distribution,the probability that three or more accidents occur in a day is:

If the p.m.f. of a random variable $X$ is given by the following table,then the standard deviation of $X$ is (given $p+q=1$):

$x$	$0$	$1$	$2$
$P(X=x)$	$q^2$	$2pq$	$p^2$