Two cards are drawn successively with replacement from fair playing 52 cards. let X denote number of kings obtained when two cards are drawn, then $\mathrm{E}\left(\mathrm{X}^2\right)=$

If $\mathrm{X} \sim \mathrm{B}(35, \mathrm{p})$ such that $7 \mathrm{P}(\mathrm{X}=0)=\mathrm{P}(\mathrm{X}=1)$ then the value of $\frac{\mathrm{P}(\mathrm{X}=15)}{\mathrm{P}(\mathrm{X}=20)}$ is equal to

A student studies for X number of hours during a randomly selected school day. The probability that X can take the values, has the following form, where k is some constant.

$$ \mathrm{P}(\mathrm{X}=x)= \begin{cases}0.2, & \text { if } x=0 \\ \mathrm{k} x, & \text { if } x=1 \text { or } 2 \\ \mathrm{k}(6-x), & \text { if } x=3 \text { or } 4 \\ 0, & \text { otherwise }\end{cases} $$

The probability that the student studies for at most two hours is

The probability that a non leap year selected at random will contain 52 Saturdays or 53 Sundays is