A bag contains 6 balls. Two balls are drawn from it at random and both are found to be black. The probability that the bag contains at least 5 black balls is
Step-by-Step Solution
Key Concept: Use Bayes' theorem with hypotheses: bag has 5 or 6 black balls (only these allow drawing 2 black balls with reasonable probability given 6 total)
Let $H_k$ = bag has $k$ black balls. Prior: $P(H_k) = 1/5$ for $k=2,3,4,5,6$ (assuming equal priors). $P(BB|H_k) = \binom{k}{2}/\binom{6}{2} = k(k-1)/30$. $P(BB) = \frac{1}{5}\sum_{k=2}^{6}\frac{k(k-1)}{30} = \frac{1}{150}(2+6+12+20+30) = \frac{70}{150} = 7/15$. $P(H_5|BB)+P(H_6|BB) = \frac{1/5 \cdot (20+30)/30}{7/15} = \frac{50/150}{7/15} = \frac{1/3}{7/15} = 5/7$. Answer: (1)
Correct Answer: $\frac{5}{7}$