$A$ biased coin with probability $p$ $(0 < p < 1)$ of getting a head is tossed until a head appears for the first time. If the probability that the number of tosses required is even is $\frac{2}{5}$, then $p=$

If a discrete random variable $X$ is defined as follows:
$P(X=x) = \begin{cases} \frac{k(x+1)}{5^x}, & x=0, 1, 2, \ldots \\ 0, & \text{otherwise} \end{cases}$
then $k=$

The probability distribution of $X$ is
$\begin{array}{|c|c|c|c|c|} \hline X & 0 & 1 & 2 & 3 \\ \hline P(X) & 0.3 & k & 2k & 2k \\ \hline \end{array}$
Find the value of $k$.

If the probability distribution of a discrete random variable $X$ is given by $P(X=k) = \frac{2^{-k}(3k+1)}{2^c}$ for $k = 0, 1, 2, \ldots, \infty$,then find the value of $P(X \leq c)$. Note: The expression provided in the prompt is likely $P(X=k) = \frac{(3k+1)}{2^{k+c}}$. Given $\sum_{k=0}^{\infty} P(X=k) = 1$,we have $\frac{1}{2^c} \sum_{k=0}^{\infty} (3k+1) \left(\frac{1}{2}\right)^k = 1$.

$5$ persons entered a lift cabin on the ground floor of a $7$-floor house. Suppose that each of them independently and with equal probability can leave the cabin at any floor beginning with the first. The probability of all the $5$ persons leaving the cabin at different floors is

For the following probability distribution,find the $Var(X)$.

$X$	$-2$	$-1$	$0$	$1$	$2$	$3$
$P(X)$	$0.1$	$0.2$	$0.2$	$0.3$	$0.15$	$0.05$

(Given : $(0.25)^2 = 0.0625$,$(0.35)^2 = 0.1225$,$(0.45)^2 = 0.2025$)