There are $3$ bags each containing $5$ white balls and $2$ black balls and $2$ bags each containing $1$ white balls and $4$ black balls, a black ball having been drawn, find the chance that it came from the first group.
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$\displaystyle \frac{28}{43}$
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$\displaystyle \frac{15}{43}$
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$\displaystyle \frac{15}{28}$
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$\displaystyle \frac{13}{28}$
B
Correct answer
Explanation
Let E be the event of drawing a black ball. P(E) = P(E|Group1)P(Group1) + P(E|Group2)P(Group2). Group 1: 3 bags, 5W 2B. P(E|G1) = 2/7. P(G1) = 3/5. Group 2: 2 bags, 1W 4B. P(E|G2) = 4/5. P(G2) = 2/5. P(E) = (2/7 * 3/5) + (4/5 * 2/5) = 6/35 + 8/25 = (30+56)/175 = 86/175. P(G1|E) = (6/35) / (86/175) = (6/35) * (175/86) = 30/86 = 15/43.