1. **Problem statement:** A meeting has 4 Americans and 2 Germans. Three consuls are selected at random. We want to find the probability distribution of the random variable $G$, which represents the number of Germans selected. 2. **Formula and rules:** The total number of ways to select 3 consuls from 6 people is given by the combination formula: $$\text{Total ways} = \binom{6}{3}$$ The random variable $G$ can take values 0, 1, or 2 (since there are only 2 Germans). 3. **Calculate probabilities:** - For $G=0$ (no Germans selected), select all 3 from 4 Americans: $$P(G=0) = \frac{\binom{4}{3} \binom{2}{0}}{\binom{6}{3}} = \frac{4 \times 1}{20} = \frac{4}{20} = 0.2$$ - For $G=1$ (one German selected), select 1 German and 2 Americans: $$P(G=1) = \frac{\binom{2}{1} \binom{4}{2}}{\binom{6}{3}} = \frac{2 \times 6}{20} = \frac{12}{20} = 0.6$$ - For $G=2$ (two Germans selected), select 2 Germans and 1 American: $$P(G=2) = \frac{\binom{2}{2} \binom{4}{1}}{\binom{6}{3}} = \frac{1 \times 4}{20} = \frac{4}{20} = 0.2$$ 4. **Check probabilities sum:** $$0.2 + 0.6 + 0.2 = 1$$ This confirms the distribution is valid. 5. **Interpretation:** The probability distribution of $G$ is: - $P(G=0) = 0.2$ - $P(G=1) = 0.6$ - $P(G=2) = 0.2$ This means it is most likely to select exactly one German when choosing three consuls at random from the group.