350. A continuous random variable $X$ is distributed according to the exponential law, given by the probability density function $f(x)=3 \mathrm{e}^{-3 x}$ for $x \geqslant 0$; for $x<0$, $f(x)=0$. Find the probability that in the experiment, $X$ falls into the interval $(0.13, 0.7)$.
Solution. We use the formula

$$
P(a<X<b)=\mathrm{e}^{-\lambda a}-\mathrm{e}^{-\lambda b} .
$$

Considering that, according to the condition, $a=0.13, b=0.7, \lambda=3$, and using the table of values of the function $\mathrm{e}^{-x}$, we get

$$
\begin{gathered}
P(0.13<X<0.7)=\mathrm{e}^{-3 \cdot 0.13}-\mathrm{e}^{-3 \cdot 0.7}=\mathrm{e}^{-0.39}-\mathrm{e}^{-3.1}= \\
=0.677-0.122=0.555 .
\end{gathered}
$$