There are three bugs 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 the probability that it came from the first group is ______.
Step-by-Step Solution
Key Concept: Apply Bayes' theorem by computing the weighted conditional probabilities for each group, then normalize.
With 3 bags from group 1 and 2 from group 2: $p_1 = \frac{3}{5}$, $p_2 = \frac{2}{5}$. The probability of drawing a black ball is $\frac{2}{3}$ from group 1 and $\frac{4}{5}$ from group 2. Thus $p = \frac{3}{5} \cdot \frac{2}{3} + \frac{2}{5} \cdot \frac{4}{5} = \frac{6}{15} + \frac{8}{25}$. Using Bayes' theorem, the probability the black ball came from group 1 is $\frac{\frac{6}{35}}{\frac{6}{35} + \frac{8}{25}} = \frac{15}{43}$.
Correct Answer: 0.3488