Example 20. From an urn containing 3 blue and 2 red balls, balls are drawn sequentially according to a random selection scheme without replacement. Find the probability $P_{k}$ that a red ball will appear for the first time on the $k$-th trial ( $k=1,2,3,4$ ).
Solution. We introduce the following notations for events: $A_{k}$ - "a red ball appears at the $k$-th trial", $B_{k}$ - "the first red ball appears at the $k$-th trial" ($k=1,2,3,4$). The events $B_{k}$ can be expressed through $A_{i}$ and $\bar{A}_{i}$: $B_{1}=A_{1}, B_{2}=\bar{A}_{1} A_{2}, B_{3}=\bar{A}_{1} \bar{A}_{2} A_{3}, B_{4}=\bar{A}_{1} \bar{A}_{2} \bar{A}_{3} A_{4}$.

According to formula (1.8.11), we obtain

$$
\begin{gathered}
P\left(B_{1}\right)=P\left(A_{1}\right), P\left(B_{2}\right)=P\left(\bar{A}_{1}\right) P\left(A_{2} / \bar{A}_{1}\right), \\
P\left(B_{3}\right)=P\left(\bar{A}_{1}\right) P\left(\bar{A}_{2} / \bar{A}_{1}\right) P\left(A_{3} / \bar{A}_{1} \bar{A}_{2}\right)
\end{gathered}
$$

$$
P\left(B_{4}\right)=P\left(\bar{A}_{1}\right) P\left(\bar{A}_{2} / \bar{A}_{1}\right) P\left(\bar{A}_{3} / \bar{A}_{1} \bar{A}_{2}\right) P\left(A_{4} / \bar{A}_{1} \bar{A}_{2} \bar{A}_{3}\right)
$$

Using formula (1.2.1), we find:

$$
\begin{gathered}
P\left(A_{1}\right)=\frac{2}{5}, P\left(\bar{A}_{1}\right)=\frac{3}{5}, P\left(\bar{A}_{i+1} / \bar{A}_{1} \ldots \bar{A}_{i}\right)=\frac{3-i}{5-i}, \\
P\left(A_{i+1} / \bar{A}_{1} \bar{A}_{2} \ldots \bar{A}_{i}\right)=\frac{2}{5-i} \quad i=1,2,3 .
\end{gathered}
$$

Thus, the required probabilities are respectively:

$$
\begin{gathered}
p_{1}=P\left(B_{1}\right)=\frac{2}{5}=0.4 ; p_{2}=P\left(B_{2}\right)=\frac{3}{5} \cdot \frac{2}{4}=0.3 \\
p_{3}=P\left(B_{3}\right)=\frac{3}{5} \cdot \frac{2}{4} \cdot \frac{2}{3}=0.2 ; p_{4}=P\left(B_{4}\right)=\frac{3}{5} \cdot \frac{2}{4} \cdot \frac{1}{3} \cdot \frac{1}{2}=0.1
\end{gathered}
$$