Suppose for independent events $A_{2}, A_{3}, \ldots, A_{n}$,

$$
P\left(A_{i}\right)=\frac{1}{2 i^{2}}
$$

What is the probability that an odd number of the events $A_{2}, A_{3}, \ldots, A_{n}$ occur?
I. solution. Let $p_{n}$ be the probability asked for in the problem. Among the events $A_{2}, A_{3}, \ldots, A_{n+1}$, an odd number will occur either if an even number of the events $A_{2}, A_{3}, \ldots, A_{n}$ occur and $A_{n+1}$ also occurs, or if an odd number of the former occur and $A_{n+1}$ does not. Since the events $A_{i}$ are independent, the following recursive equation can be written:

$$
p_{n+1}=P\left(A_{n+1}\right)\left(1-p_{n}\right)+\left(1-P\left(A_{n+1}\right)\right) p_{n}
$$

According to the problem, $P\left(A_{2}\right)=\frac{1}{8}$ and $P\left(A_{3}\right)=\frac{1}{18}$. Clearly, $p_{2}=\frac{1}{8}$ and thus

$$
p_{3}=\frac{1}{18} \cdot \frac{7}{8}+\frac{17}{18} \cdot \frac{1}{8}=\frac{2}{12}
$$

It can be conjectured that the general formula is: $p_{n}=\frac{n-1}{4 n}$.

We prove the correctness of the formula by induction. The formula is correct for $n=2$ and $n=3$. Assume it is true for $n=k$. Write it for $n=k+1$:

$$
\begin{aligned}
p_{k+1} & =P\left(A_{k+1}\right)\left(1-p_{k}\right)+\left(1-P\left(A_{k+1}\right)\right) p_{k}= \\
& =\frac{1}{2(k+1)^{2}} \cdot \frac{3 k+1}{4 k}+\frac{2(k+1)^{2}-1}{2(k+1)^{2}} \cdot \frac{k-1}{4 k}= \\
& =\frac{(3 k+1)+(k-1)\left(2 k^{2}+4 k+1\right)}{8 k(k+1)^{2}}= \\
& =\frac{2 k^{3}+2 k^{2}}{8 k(k+1)^{2}}=\frac{2 k^{2}(k+1)}{8 k(k+1)^{2}}=\frac{k}{4(k+1)}
\end{aligned}
$$

Thus, the formula holds for $n=k+1$ as well, completing the induction.

The probability that an odd number of the events $A_{2}, A_{3}, \ldots, A_{n}$ occur is $\frac{n-1}{4 n}$.

II. solution. Let $p$ be the probability that an odd number of the events $A_{2}, A_{3}, \ldots, A_{n}$ occur. Define the random variables $X_{2}, \ldots, X_{n}$ as follows: let $X_{i}=-1$ if $A_{i}$ occurs, and $X_{i}=+1$ if $A_{i}$ does not occur.

The product $X_{2} X_{3} \ldots X_{n}$ is -1 precisely when an odd number of the events $A_{2}, A_{3}, \ldots, A_{n}$ occur, and +1 otherwise. Therefore, the expected value of the product is

$$
E\left(X_{2} X_{3} \ldots X_{n}\right)=p \cdot(-1)+(1-p) \cdot(+1)=1-2 p
$$

The expected value of the product of independent variables can also be calculated factor by factor, so

$$
\begin{aligned}
& E\left(X_{2} X_{3} \ldots X_{n}\right)=\prod_{i=2}^{n} E\left(X_{i}\right)=\prod_{i=2}^{n}\left(P\left(A_{i}\right) \cdot(-1)+\left(1-P\left(A_{i}\right)\right) \cdot(+1)\right)= \\
= & \prod_{i=2}^{n}\left(1-2 P\left(A_{i}\right)\right)=\prod_{i=2}^{n}\left(1-\frac{1}{i^{2}}\right)=\prod_{i=2}^{n} \frac{(i-1)(i+1)}{i^{2}}= \\
= & \frac{(1 \cdot 3) \cdot(2 \cdot 4) \cdot(3 \cdot 5) \ldots((n-3)(n-1))((n-2) \cdot n)((n-1)(n+1))}{2^{2} \cdot 3^{2} \ldots n^{2}}= \\
= & \frac{2 n(n+1)}{2^{2} \cdot n^{2}}=\frac{n+1}{2 n} .
\end{aligned}
$$

(For smaller $n$, the middle factors "cancel out".) From comparing (1) and (2), $1-2 p=\frac{n+1}{2 n}$, from which

$$
p=\frac{n-1}{4 n}
$$