$A$ die is loaded in such a way that each odd number is twice as likely to occur as each even number. If $E$ is the event that a number greater than or equal to $4$ occurs on a single toss of the die,then $P(E)$ is equal to:

In a city,it is found that $10$ accidents took place in a span of $50$ days. Assuming that the number of accidents follows the Poisson distribution,the probability that there will be $3$ or more accidents in a day in that city is

The following table represents the probability distribution of a random variable $X$ for some $k \in Q$. Find the mean of $X$.
$\begin{array}{|c|c|c|c|c|c|c|} \hline X=x & -2 & -1 & 0 & 1 & 2 & 3 \\ \hline P(X=x) & 0.1 & k & 0.2 & 2k & 0.3 & k \\ \hline \end{array}$

Two cards are drawn simultaneously from a well-shuffled pack of $52$ cards. If $X$ is the random variable representing the number of queens obtained,then the value of $2 E(X) + 3 E(X^2)$ is:

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 $\frac{1}{2^c} \sum_{k=0}^{\infty} (3k+1) \left(\frac{1}{2}\right)^k = 1$.

The following table shows the probability distribution of smart phones sold in a shop per day:

Number of smart phones $(x)$	$0$	$1$	$2$	$3$	$4$	$5$
Probability $(P(x))$	$k$	$0.3$	$0.15$	$0.15$	$0.1$	$2k$

Then $E(x) = ?$