1. **Problem Statement:** Construct the probability distribution of the random variable $G$ representing the number of Germans selected when 3 consuls are chosen at random from 4 Americans and 2 Germans. 2. **Understanding the problem:** We have 6 consuls total: 4 Americans (A) and 2 Germans (G). We select 3 consuls randomly. 3. **Random variable $G$:** $G$ = number of Germans selected in the group of 3. Possible values of $G$ are 0, 1, or 2 because we cannot select more Germans than available. 4. **Total number of ways to select 3 consuls from 6:** $$\text{Total ways} = \binom{6}{3} = 20$$ 5. **Calculate probabilities for each value of $G$:** - $P(G=0)$: Select 0 Germans and 3 Americans. $$\binom{2}{0} \times \binom{4}{3} = 1 \times 4 = 4$$ - $P(G=1)$: Select 1 German and 2 Americans. $$\binom{2}{1} \times \binom{4}{2} = 2 \times 6 = 12$$ - $P(G=2)$: Select 2 Germans and 1 American. $$\binom{2}{2} \times \binom{4}{1} = 1 \times 4 = 4$$ 6. **Calculate probabilities:** $$P(G=k) = \frac{\text{Number of favorable ways}}{\text{Total ways}}$$ - $$P(G=0) = \frac{4}{20} = 0.2$$ - $$P(G=1) = \frac{12}{20} = 0.6$$ - $$P(G=2) = \frac{4}{20} = 0.2$$ 7. **Probability distribution table:** | $G$ | 0 | 1 | 2 | |-----|---|---|---| | $P(G)$ | 0.2 | 0.6 | 0.2 | **Final answer:** The probability distribution of $G$ is: $$P(G=0) = 0.2, \quad P(G=1) = 0.6, \quad P(G=2) = 0.2$$