Thread: [SOLVED] Two Conditional Probability Problems
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Old September 15th, 2008, 05:13 AM
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Originally Posted by akolman View Post
[snip]
2. Bowl I contains 6 red chips and 4 blue chips. Five of these 10 chips are selected at random and without replacement and put in bowl II, which was originally empty. One chip is then drawn at random from bowl II. Given that this chip is blue, find the conditional probability that 2 red chips and 3 blue chips are transferred from bowl I to bowl II.

Answer:\frac {5}{14}
\Pr(\text{3 B taken from bowl I} \, | \, \text{B chip drawn from bowl II}) = \frac{ \Pr(\text{3 B taken from bowl I and B chip drawn from bowl II})}{\Pr(\text{B chip drawn from bowl II})}.


\Pr(\text{3 B taken from bowl I and B chip drawn from bowl II}) = \frac{^6C_2 \, ^4C_3}{^{10}C_5} \times \frac{3}{5}.


\Pr(\text{B chip drawn from bowl II}) = \frac{^6C_4 \, ^4C_1}{^{10}C_5} \times \frac{1}{5} ~ + ~ \frac{^6C_3 \, ^4C_2}{^{10}C_5} \times \frac{2}{5} ~ + ~ \frac{^6C_2 \, ^4C_3}{^{10}C_5} \times \frac{3}{5} ~ + ~ \frac{^6C_1 \, ^4C_4}{^{10}C_5} \times \frac{4}{5}.


Do you see why?
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