252. The random variable $X$ is given by the distribution function

$$
F(x)=\left\{\begin{array}{llr}
0 & \text { for } & x \leqslant-1 \\
(3 / 4) x+3 / 4 & \text { for } & -11 / 3
\end{array}\right.
$$

Find the probability that in the result of the trial, the variable $X$ will take a value in the interval $(0,1 / 3)$.
Solution. The probability that $X$ will take a value within the interval $(a, b)$ is equal to the increment of the distribution function over this interval: $P(a<X<b)=F(b)-F(a)$. Setting $a=0$, $b=1 / 3$, we get

$$
\begin{gathered}
P(0<X<1 / 3)=F(1 / 3)-F(0)= \\
=[(3 / 4) x+3 / 4]_{x=1 / 3}-[(3 / 4) x+3 / 4]_{x=0}=1 / 4
\end{gathered}
$$