In an urn, there are 5 white and 4 blue balls, and in another, 2 white and 8 blue. We draw one ball from each urn and, without looking at them, put them into a third, empty urn. What is the probability that drawing one ball from the third - containing two balls - urn, it will be white?
We draw a white ball from the third urn if any of the following three mutually exclusive events occur:

1. We draw a white ball from the first urn, a blue ball from the second urn, and then a white ball from the third urn.
2. We draw a blue ball from the first urn, a white ball from the second urn, and then a white ball from the third urn.
3. We draw a white ball from both the first and second urns.

The probability of the first event is denoted by \( v_{1} \), the second event by \( v_{2} \), and the third event by \( v_{3} \).

According to the multiplication rule,

\[
v_{1}=\frac{5}{9} \cdot \frac{8}{10} \cdot \frac{1}{2}=\frac{2}{9}
\]

similarly,

\[
v_{2}=\frac{4}{9} \cdot \frac{2}{10} \cdot \frac{1}{2}=\frac{2}{45}
\]

and

\[
v_{3}=\frac{5}{9} \cdot \frac{2}{10}=\frac{1}{9}
\]

The desired probability, since all three events are mutually exclusive, is the sum of the probabilities of the three events, that is,

\[
v=v_{1}+v_{2}+v_{3}=\frac{2}{9}+\frac{2}{45}+\frac{1}{9}=\frac{17}{45} \approx 0.378
\]

László Kovács (Debrecen, Ref. g. IV. o. t.)