Let's determine the maximum and minimum of the following expression:

$$
P\left(A_{1} A_{2} A_{3} A_{4}\right)-P\left(A_{1}\right) \cdot P\left(A_{2}\right) \cdot P\left(A_{3}\right) \cdot P\left(A_{4}\right)
$$

where $A_{1}, A_{2}, A_{3}, A_{4}$ are arbitrary events, and $P(A)$ denotes the probability of event $A$.
Let the expression under investigation be denoted by $p\left(A_{1}, A_{2}, A_{3}, A_{4}\right)$, or simply $p$. If the events $A_{i}$ are independent, then $p=0$, and due to the maximum of $p$ being positive, we expect the minimum of $p$ to be a negative number. Since the probability of the events is a number between 0 and 1 (including the boundaries), the absolute value of both extreme values can be at most 1.

The maximum is obtained when the probability of the event $B=A_{1} A_{2} A_{3} A_{4}$ is large, and the product $S=P\left(A_{1}\right) \cdot P\left(A_{2}\right) \cdot P\left(A_{3}\right) \cdot P\left(A_{4}\right)$ is small. First, let's determine the minimum of the product $S$ for a fixed $B$. Since $B$ implies $A_{i}$, $P(B) \leqq P\left(A_{i}\right)(i=1,2,3,4)$, and thus $S \geqq[P(B)]^{4}$. Here, the lower bound can be achieved by the substitution $A_{i}=B(i=1,2,3,4)$. Therefore, it is sufficient to determine the maximum of the function

$$
f(x)=x-x^{4}
$$

over the interval $0 \leqq x \leqq 1$. The derivative of the function vanishes when:

$$
f^{\prime}(x)=1-4 x^{3}=0
$$

if $x^{3}=\frac{1}{4}$, then

$$
f(x)=x\left(1-x^{3}\right)=\frac{3}{4} \cdot \frac{1}{\sqrt[3]{4}}=0.4725
$$

The minimum is obtained when $P(B)$ is small, and $S$ is large. The first condition is sufficiently met when $B$ is the impossible event, in which case we expect $S$ to be the largest when each event in the sample space is in exactly three of the $A_{i}$ events, i.e., $P\left(A_{i}\right)=\frac{3}{4}(i=1,2,3,4)$, and in this case $p=-\left(\frac{3}{4}\right)^{4}=-0.3164$.

This can be achieved by partitioning the sample space into four equally probable parts, denoted by $C_{i}$, and defining the $A_{i}$ as follows:

$$
\begin{aligned}
& A_{1}=C_{1}+C_{2}+C_{3} \\
& A_{2}=C_{1}+C_{2}+C_{4} \\
& A_{3}=C_{1}+C_{3}+C_{4} \\
& A_{4}=C_{2}+C_{3}+C_{4}
\end{aligned}
$$

Let $P(B)=x \geqq 0$ in general, and determine the maximum of $S$. Then

$$
1-x=P(\bar{B})=P\left(\bar{A}_{1}+\bar{A}_{2}+\bar{A}_{3}+\bar{A}_{4}\right) \leqq P\left(\bar{A}_{2}\right)+P\left(\bar{A}_{2}\right)+P\left(\bar{A}_{3}\right)+P\left(\bar{A}_{4}\right)
$$

It is therefore convenient to express $S$ using the $\bar{A}_{i}$ events:

$$
S=\left[1-P\left(\bar{A}_{1}\right)\right]\left[1-P\left(\bar{A}_{2}\right)\right]\left[1-P\left(\bar{A}_{3}\right)\right]\left[1-P\left(\bar{A}_{4}\right)\right]
$$

By the arithmetic and geometric mean inequality,

$$
S \leqq\left(1-\frac{P\left(\bar{A}_{1}\right)+P\left(\bar{A}_{2}\right)+P\left(\bar{A}_{3}\right)+P\left(\bar{A}_{4}\right)}{4}\right)^{4}
$$

and we can further increase this by substituting $\frac{1-x}{4}$ for the second term:

$$
S \leqq\left(1-\frac{1-x}{4}\right)^{4}=\frac{(3+x)^{4}}{4^{4}}
$$

Therefore, if $P(B)=x$, then

$$
p \geqq x-\frac{(3+x)^{4}}{4^{4}}
$$

which is monotonically increasing on the interval $0 \leqq x \leqq 1$, so its minimum is at $x=0$. This proves that the value found initially is the minimum of $p$.