If $X$ is a Poisson variate such that $P(X=1) = 2P(X=2)$,then $P(X=3)$ is equal to:

Let the mean and standard deviation of the probability distribution given by the table below be $\mu$ and $\sigma$ respectively. If $\sigma - \mu = 2$,then find the value of $\sigma$.

$X=x$	$-3$	$0$	$1$	$\alpha$
$P(X=x)$	$\frac{1}{4}$	$K$	$\frac{1}{4}$	$\frac{1}{3}$

Let $X$ denote the number of hours you study during a randomly selected school day. The probability that $X$ can take the values $x$ has the following form,where $k$ is some unknown constant.
$P(X=x) = \begin{cases} 0.1, & \text{if } x=0 \\ kx, & \text{if } x=1 \text{ or } 2 \\ k(5-x), & \text{if } x=3 \text{ or } 4 \\ 0, & \text{otherwise} \end{cases}$
What is the probability that you study at least two hours? Exactly two hours? At most two hours?

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

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

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

If the probability density function of a continuous random variable $X$ is $f(x) = \frac{x^3}{3}$ for $-1 < x < 2$,and $f(x) = 0$ otherwise,then the cumulative distribution function $F(x)$ for $-1 < x < 2$ is:

In a meeting,$60 \%$ of the members favour and $40 \%$ oppose a certain proposal. $A$ member is selected at random. We define a random variable $X$ such that $X=0$ if the member opposes the proposal and $X=1$ if the member is in favour. Then,$\text{Var}(X) = $