Example 12. What is the minimum number of trials that need to be conducted so that with a probability of at least $\alpha(0<\alpha<1)$, one can expect the event $A$ to occur at least once, if the probability of event $A$ in one trial is $p$?
Solution. We require that the probability of event $A$ occurring at least once in $n$ trials (see formula (3.1.7)) is not less than $\alpha$:

$$
1-q^{n} \geq \alpha, \text { or }\left(1-(1-p)^{n}\right) \geq \alpha \text {. }
$$

Solving this inequality for $n$, we obtain the inequality

$$
n \geq \frac{\lg (1-\alpha)}{\lg (1-p)}
$$

From this, we conclude that the minimum number of trials $n_{0}$, satisfying the condition of the example, is determined by the formula

$$
n_{0}=\left[\frac{\lg (1-\alpha)}{\lg (1-p)}\right]+1
$$

where the square brackets denote the integer part of the number.

In particular, if $p=0.02$ and $\alpha=0.98$, then by the last formula we get $n_{0}=80$.