The logically equivalent statement of $(\sim \mathrm{p} \wedge \mathrm{q}) \vee(\sim \mathrm{p} \wedge \sim \mathrm{q}) \vee(\mathrm{p} \wedge \sim \mathrm{q})$ is

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

The area of the region bounded by $\frac{x^2}{9}+\frac{y^2}{4}=1$ and the line $\frac{x}{3}+\frac{y}{2}=1$ is