Problem 714

Solution. If the third ball must be white, then the first two balls can be white, or white and black, or black and white, or black, i.e., there are four groups of non-
mutually exclusive events. Applying the multiplication theorem of probabilities to them, we get:

$$\begin{aligned} P & =P_{16} \cdot P_{2 \sigma} \cdot P_{3 \sigma}+\left(P_{1 \sigma} \cdot P_{2 u} \cdot P_{3 \sigma}+P_{14} \cdot P_{2 \sigma} \cdot P_{3 \sigma}\right)+P_{14} \cdot P_{2 u} \cdot P_{3 \sigma}= \\ & =\frac{6}{11} \cdot \frac{5}{10} \cdot \frac{4}{9}+\frac{6}{11} \cdot \frac{5}{10} \cdot \frac{5}{9}+\frac{5}{11} \cdot \frac{6}{10} \cdot \frac{5}{9}+\frac{5}{11} \cdot \frac{4}{10} \cdot \frac{6}{9}=\frac{540}{990}=\frac{6}{11} \end{aligned}$$

## 12.2. Random Variables