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 step by step, we will use the concept of conditional probability.

### Step 1: Define the Groups and Events
- **Group 1 (G1)**: 3 bags, each containing 5 white balls and 2 black balls.
- **Group 2 (G2)**: 2 bags, each containing 1 white ball and 4 black balls.
- **Event B**: A black ball is drawn.

### Step 2: Calculate the Probability of Drawing a Black Ball from Group 1
To find the probability of drawing a black ball from Group 1:
- The probability of choosing one of the 3 bags from Group 1 is \( P(G1) = \frac{3}{5} \) (since there are 5 bags in total).
- The probability of drawing a black ball from a bag in Group 1 is \( P(B|G1) = \frac{2}{7} \) (2 black balls out of 7 total balls).

Thus, the joint probability of choosing Group 1 and drawing a black ball is:
\[ P(G1 \cap B) = P(G1) \times P(B|G1) = \frac{3}{5} \times \frac{2}{7} = \frac{6}{35} \]

### Step 3: Calculate the Probability of Drawing a Black Ball from Group 2
To find the probability of drawing a black ball from Group 2:
- The probability of choosing one of the 2 bags from Group 2 is \( P(G2) = \frac{2}{5} \).
- The probability of drawing a black ball from a bag in Group 2 is \( P(B|G2) = \frac{4}{5} \) (4 black balls out of 5 total balls).

Thus, the joint probability of choosing Group 2 and drawing a black ball is:
\[ P(G2 \cap B) = P(G2) \times P(B|G2) = \frac{2}{5} \times \frac{4}{5} = \frac{8}{25} \]

### Step 4: Calculate the Total Probability of Drawing a Black Ball
Now, we can find the total probability of drawing a black ball, \( P(B) \):
\[ P(B) = P(G1 \cap B) + P(G2 \cap B) = \frac{6}{35} + \frac{8}{25} \]

To add these fractions, we need a common denominator. The least common multiple of 35 and 25 is 175.
- Convert \( \frac{6}{35} \) to have a denominator of 175:
\[ \frac{6}{35} = \frac{6 \times 5}{35 \times 5} = \frac{30}{175} \]
- Convert \( \frac{8}{25} \) to have a denominator of 175:
\[ \frac{8}{25} = \frac{8 \times 7}{25 \times 7} = \frac{56}{175} \]

Now, add the two fractions:
\[ P(B) = \frac{30}{175} + \frac{56}{175} = \frac{86}{175} \]

### Step 5: Calculate the Conditional Probability
Finally, we can find the conditional probability that the black ball came from Group 1, \( P(G1|B) \):
\[ P(G1|B) = \frac{P(G1 \cap B)}{P(B)} = \frac{\frac{6}{35}}{\frac{86}{175}} \]

To simplify this, multiply by the reciprocal:
\[ P(G1|B) = \frac{6}{35} \times \frac{175}{86} = \frac{6 \times 5}{86} = \frac{30}{86} = \frac{15}{43} \]

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