Two cards are drawn simultaneously from a well shuffled pack of 52 cards. If X is the random variable of getting queens, then the value of $2 E(X)+3 E\left(X^2\right)$ for the number of queens is

A random variable $X$ has the following probability distribution

$$ \begin{array}{|l|c|c|c|c|c|} \hline \mathrm{X}: & 0 & 1 & 2 & 3 & 4 \\ \hline \mathrm{P}(\mathrm{X}): & \mathrm{k} & 2 \mathrm{k} & 4 \mathrm{k} & 2 \mathrm{k} & \mathrm{k} \\ \hline \end{array} $$

then the value of $\mathrm{P}(1 \leqslant \mathrm{X}<4 \mid \mathrm{X} \leqslant 2)=$

If two numbers $p$ and $q$ are chosen randomly from the set $\{1,2,3,4\}$, one by one, with replacement, then the probability of getting $\mathrm{p}^2 \geq 4 \mathrm{q}$ is

If $X \sim B(n, p)$ then $\frac{P(X=k)}{P(X=k-1)}=$