12.1.2. In an urn, there are 3 white balls and 5 black balls. Three balls are drawn at random one by one, and after each draw, the ball is returned to the urn. Find the probability that among the drawn balls there will be:

a) exactly two white balls, b) at least two white balls.
Solution. We have a scheme with replacement, i.e., the composition of balls does not change each time:

a) when drawing three balls, two of them should be white, and one should be black. In this case, the black ball can be either the first, the second, or the third. Applying the theorems of addition and multiplication of probabilities together, we have:

$$
\begin{gathered}
P=P_{1 \sigma} \cdot P_{2 \sigma} \cdot P_{34}+P_{1 \sigma} \cdot P_{24} \cdot P_{3 \sigma}+P_{14} \cdot P_{2 \sigma} \cdot P_{3 \sigma}= \\
=\frac{3}{8} \cdot \frac{3}{8} \cdot \frac{5}{8}+\frac{3}{8} \cdot \frac{5}{8} \cdot \frac{3}{8}+\frac{5}{8} \cdot \frac{3}{8} \cdot \frac{3}{8}=\frac{135}{512}
\end{gathered}
$$

b) drawing not fewer than two white balls means that there should be either two or three white balls:

$$
\begin{gathered}
P(\text { not fewer })=P(2 \text { white })+P(3 \text { white })=\frac{135}{512}+P_{16} \cdot P_{2 \sigma} \cdot P_{36}= \\
=\frac{135}{512}+\frac{3}{8} \cdot \frac{3}{8} \cdot \frac{3}{8}=\frac{162}{512} .
\end{gathered}
$$