367. The duration of trouble-free operation of an element has an exponential distribution $F(t)=1$ - $\mathrm{e}^{-0.01 t}(t>0)$. Find the probability that during a time period of $t=50$ hours: a) the element will fail; b) the element will not fail.
Solution. a) Since the distribution function $F(t)=1-\mathrm{e}^{-0.01 t}$ defines the probability of the element failing within a time duration $t$, substituting $t=50$ into the distribution function, we get the probability of failure:

$$
F(50)=1-\mathrm{e}^{-0.01 \cdot 50}=1-\mathrm{e}^{-0.3}=1-0.606=0.394 ;
$$

b) the events "the element will fail" and "the element will not fail" are complementary, so the probability that the element will not fail is

$$
P=1-0.394=0.606
$$

This same result can be obtained directly using the reliability function $R(t)=\mathrm{e}^{-\lambda \cdot t}$, which defines the probability of the element operating without failure for a time duration $t$:

$$
R(50)=\mathrm{e}^{-0.01 \cdot 50}=\mathrm{e}^{-0.3}=0.606
$$