$A$ random variable $X$ has the following probability distribution. Then,$P(2 \leq X < 5) = $ . . . . . .

$X = x$	$1$	$2$	$3$	$4$	$5$	$6$
$P(X = x)$	$K$	$3K$	$5K$	$7K$	$8K$	$K$

$A$ coin is tossed until one head appears or a tail appears $4$ times in succession. The probability distribution of the number of tosses $X$ is:

If the mean of the following probability distribution of a random variable $X$ is $\frac{46}{9}$,then the variance of the distribution is:

$X$	$0$	$2$	$4$	$6$	$8$
$P(X)$	$a$	$2a$	$a+b$	$2b$	$3b$

If random variable $X$ is the waiting time in minutes for a bus and the probability density function of $X$ is given by $f(x) = \begin{cases} \frac{1}{5}, & 0 \leq x \leq 5 \\ 0, & \text{otherwise} \end{cases}$,then the probability of the waiting time being not more than $4$ minutes is = . . . . . . .

If three dice are thrown, then the mean of the sum of the numbers appearing on them is

Given the probability density function (p.d.f.) of the random variable $X$ as $f(x) = \frac{1}{2a}$ for $0 < x < 2a$ and $f(x) = 0$ otherwise, where $a > 0$, which of the following is correct?