41. Let $A_{1}, A_{2}, \ldots, A_{i}, \ldots, A_{n}$ be $n$ independent events in total. The probabilities of each event $A_{i}$ occurring are $p_{i}$ $(i=1,2, \ldots, n)$. Suppose that as a result of the trial, all events or some of them may occur, or none of them may occur. Find the probability of at least one of the events $A_{i}$ occurring, if the probabilities $p_{i}:$ a) are different; b) are equal to each other.
S o l u t i o n. In addition to the $\boldsymbol{n}$ independent events $A_{i}$, consider their opposites $\bar{A}_{i}$, as follows:

$A$ - the occurrence of at least one of the events $A_{i}$;

$\bar{A}$ - the non-occurrence of any of the events $A_{i}$.

Since $P\left(A_{i}\right)=p_{i}$, then $P\left(\bar{A}_{i}\right)=1-p_{i}=q_{i}$.

The event $\bar{A}$ consists in the simultaneous occurrence of the events $\bar{A}_{1}$, $\bar{A}_{2}, \ldots$, and $\bar{A}_{n}$, i.e., it is equal to the product of these events: $\bar{A}=\bar{A}_{1} \bar{A}_{2} \ldots \bar{A}_{n}$. The event $\bar{A}$ is the opposite of the event $A$, so $P(A)=1-P(\bar{A})$.

a) By the theorem of multiplication of probabilities of independent events, we find

$$
P(\bar{A})=q_{1} q_{2} \ldots q_{n}
$$

Thus,

$$
P(A)=1-q_{1} q_{2} \ldots q_{n}
$$

b) Since $P\left(A_{1}\right)=P\left(A_{2}\right)=\ldots=P\left(A_{n}\right)=p$, then

$$
P\left(\bar{A}_{1}\right)=P\left(\bar{A}_{2}\right)=\ldots P\left(\bar{A}_{n}\right)=1-p=q
$$

By the theorem of multiplication of probabilities of independent events, we find $P(\bar{A})=q^{n}$.

Thus,

$$
P(A)=1-q^{n} \text {. }
$$