There are 3 bags each containing 5 white balls and 2 black balls, and 2 bags each containing 1 white ball and 4 black balls, a black ball having been drawn, find the chance that it came from the first group.

Answer

To solve the problem, we will use Bayes' theorem to find the probability that a black ball drawn came from the first group of bags.

### Step-by-step Solution:

1. **Define the Events**:
- Let \( A \) be the event that the black ball is drawn from the first group of bags (Group A).
- Let \( B \) be the event that the black ball is drawn from the second group of bags (Group B).
- Let \( C \) be the event that a black ball is drawn.

2. **Determine the Total Number of Bags**:
- Group A has 3 bags, and Group B has 2 bags. Thus, the total number of bags is:
\[ \text{Total bags} = 3 + 2 = 5 \]

3. **Calculate the Probability of Selecting Each Group**:
- The probability of selecting a bag from Group A:
\[ P(A) = \frac{3}{5} \]
- The probability of selecting a bag from Group B:
\[ P(B) = \frac{2}{5} \]

4. **Calculate the Probability of Drawing a Black Ball from Each Group**:
- In Group A, each bag contains 5 white balls and 2 black balls. Therefore, the probability of drawing a black ball from a bag in Group A is:
\[ P(C|A) = \frac{2}{7} \]
- In Group B, each bag contains 1 white ball and 4 black balls. Therefore, the probability of drawing a black ball from a bag in Group B is:
\[ P(C|B) = \frac{4}{5} \]

5. **Calculate the Total Probability of Drawing a Black Ball**:
- Using the law of total probability:
\[ P(C) = P(C|A) \cdot P(A) + P(C|B) \cdot P(B) \]
- Substituting the values:
\[ P(C) = \left(\frac{2}{7} \cdot \frac{3}{5}\right) + \left(\frac{4}{5} \cdot \frac{2}{5}\right) \]
- Calculating each term:
\[ P(C) = \frac{6}{35} + \frac{8}{25} \]
- Finding a common denominator (which is 175):
\[ P(C) = \frac{6 \cdot 5}{175} + \frac{8 \cdot 7}{175} = \frac{30 + 56}{175} = \frac{86}{175} \]

6. **Use Bayes' Theorem to Find \( P(A|C) \)**:
- Bayes' theorem states:
\[ P(A|C) = \frac{P(C|A) \cdot P(A)}{P(C)} \]
- Substituting the known values:
\[ P(A|C) = \frac{\left(\frac{2}{7}\right) \cdot \left(\frac{3}{5}\right)}{\frac{86}{175}} \]
- Simplifying:
\[ P(A|C) = \frac{\frac{6}{35}}{\frac{86}{175}} = \frac{6 \cdot 175}{35 \cdot 86} = \frac{30}{86} = \frac{15}{43} \]

### Final Answer:
The probability that the black ball drawn came from the first group is:
\[ \boxed{\frac{15}{43}} \]