1. **Problem statement:** Find the probability that exactly two dark chocolates are selected from a sequence of three chocolates drawn without replacement. 2. **Understanding the problem:** We have a total of 25 chocolates: 10 dark (D) and 15 white (W). We select 3 chocolates one after another without replacement. 3. **Event definition:** Let’s define the event "exactly two dark chocolates are selected". 4. **Approach:** We want the probability of selecting exactly two dark chocolates in three draws. This can happen in three ways: - D, D, W - D, W, D - W, D, D 5. **Calculate each sequence probability:** - For D, D, W: $$P(D_1) = \frac{10}{25}$$ $$P(D_2|D_1) = \frac{9}{24}$$ $$P(W_3|D_1,D_2) = \frac{15}{23}$$ So, $$P(D,D,W) = \frac{10}{25} \times \frac{9}{24} \times \frac{15}{23}$$ - For D, W, D: $$P(D_1) = \frac{10}{25}$$ $$P(W_2|D_1) = \frac{15}{24}$$ $$P(D_3|D_1,W_2) = \frac{9}{23}$$ So, $$P(D,W,D) = \frac{10}{25} \times \frac{15}{24} \times \frac{9}{23}$$ - For W, D, D: $$P(W_1) = \frac{15}{25}$$ $$P(D_2|W_1) = \frac{10}{24}$$ $$P(D_3|W_1,D_2) = \frac{9}{23}$$ So, $$P(W,D,D) = \frac{15}{25} \times \frac{10}{24} \times \frac{9}{23}$$ 6. **Sum the probabilities:** $$P(\text{exactly two dark}) = P(D,D,W) + P(D,W,D) + P(W,D,D)$$ Calculate each term: $$P(D,D,W) = \frac{10 \times 9 \times 15}{25 \times 24 \times 23} = \frac{1350}{13800} = \frac{9}{92}$$ $$P(D,W,D) = \frac{10 \times 15 \times 9}{25 \times 24 \times 23} = \frac{1350}{13800} = \frac{9}{92}$$ $$P(W,D,D) = \frac{15 \times 10 \times 9}{25 \times 24 \times 23} = \frac{1350}{13800} = \frac{9}{92}$$ Sum: $$P = \frac{9}{92} + \frac{9}{92} + \frac{9}{92} = \frac{27}{92}$$ **Final answer:** $$\boxed{\frac{27}{92}}$$ --- 1. **Problem statement:** Given event A: first chocolate selected is white. Given event B: exactly two dark chocolates are selected. Find the conditional probability $$P(A|B)$$. 2. **Recall conditional probability formula:** $$P(A|B) = \frac{P(A \cap B)}{P(B)}$$ 3. **Calculate $$P(B)$$:** From previous calculation, $$P(B) = \frac{27}{92}$$ 4. **Calculate $$P(A \cap B)$$:** Event A and B means first chocolate is white and exactly two dark chocolates are selected. From the sequences for exactly two dark chocolates, only the sequence W, D, D satisfies A. So, $$P(A \cap B) = P(W,D,D) = \frac{15}{25} \times \frac{10}{24} \times \frac{9}{23} = \frac{1350}{13800} = \frac{9}{92}$$ 5. **Calculate $$P(A|B)$$:** $$P(A|B) = \frac{\frac{9}{92}}{\frac{27}{92}} = \frac{9}{92} \times \frac{92}{27} = \frac{9}{27} = \frac{1}{3}$$ **Final answer:** $$\boxed{\frac{1}{3}}$$