$A$ random variable $X$ takes the values $0, 1, 2$. Its mean is $1.2$. If $P(X=0)=0.3$,then $P(X=1)=$

The p.d.f. of a random variable $X$ is given by $f(x) = \frac{k}{\sqrt{x}}$ for $0 \leq x \leq 4$ and $f(x) = 0$ otherwise. Then $P(1 < X < 4) = $

Let $X$ denote the number of hours you study during a randomly selected school day. The probability that $X$ can take the values $x$ has the following form,where $k$ is some unknown constant.
$P(X=x) = \begin{cases} 0.1, & \text{if } x=0 \\ kx, & \text{if } x=1 \text{ or } 2 \\ k(5-x), & \text{if } x=3 \text{ or } 4 \\ 0, & \text{otherwise} \end{cases}$
What is the probability that you study at least two hours? Exactly two hours? At most two hours?

At a telephone enquiry system,the number of phone calls regarding relevant enquiries follows a Poisson distribution with an average of $5$ phone calls during $10$-minute time intervals. The probability that there is at most one phone call during a $10$-minute time period is:

$15$ coupons are numbered from $1$ to $15$. Seven coupons are selected at random with replacement. What is the probability that the maximum number on the selected coupons is $9$?

Let $F(x)$ be the cumulative distribution function (c.d.f.) of a continuous random variable $X$. If $F(b) = 0.7$ and $P(X > a) = 0.4$, then the value of $P(a < X < b)$ is ...