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Bowl A contains 6 red chips and 4 blue chips. Five chips are randomly chosen and transferred without replacement to Bowl B. One chip is drawn at random from Bowl B. Given that this chip is blue, find the conditional probability that 2 red chips and 3 blue chips are transferred from bowl A to bowl B.

Attempt:

$$P(A|B) = \frac{P(B|A)\cdot P(A)}{P(B)}$$

Let $B$ = chip is blue and $A$ = 2 red and 3 blue are chosen.

$$\begin{align} &P(A) = \frac {\binom 6 2 \cdot \binom 4 3}{\binom {10} 5}\\ &P(B|A) = \frac 3 5 \end{align}$$

By Bayes Rule, $P(A|B) = \left(\dfrac 3 5\right)\cdot \dfrac{ \binom 6 2 \binom 4 3}{\binom {10} 5\cdot \dfrac{4}{10}}$.

Is this correct?

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3 Answers 3

Yes, this is correct and a good way to solve the problem.

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There are $\frac{10!}{6!4!}$ (= 210) possible arrangements for the chips, and $\frac{5!}{2!3!}$ arrangements for the chips desired in bowl B. Any given arrangement of bowl B can occur for every corresponding arrangement in bowl A (also $\frac{5!}{2!3!}$ combinations)

The total number of possiblilities with the correct bowl B is therefore $\frac{5!}{2!3!}\dot{}\frac{5!}{2!3!}=100$

Substitute P(A) = 100/210 to get P(A|B) = (3/5)(100/210)/(4/10) = 5/7, or about 71%

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In your statement, I think, you should clarify what B exactly stand for. Here, I suppose you're taking B = blue chips been drawn from the bowl B. So, you have to calculate $P(B)$ in a more explicit way. According to Bayes Rule, we got: $$ P(A|B)=\frac{P(B|A_0)P(A_0)}{\sum_0^nP(B|A_n)P(A_n)}. $$ Where, $A_n$ stands for other situations when you transfer 5 chips from bowl A to bowl B. And thus $P(B)=\sum_0^nP(B|A_n)P(A_n)$. Hope that would be helpful to you.

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