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