Bowl C contains 6 red and 4 blue chips. 5 of these chips were selected at random and without replacement and put in bowl D, which was originally empty.
a) What is the probability that 2 red chips and 3 blue chips were transferred from bowl C to bowl D?
b) One chip is then drawn at random from bowl D. Given that this chip is blue, find the conditional probability that 2 red chips and 3 blue chips were transferred from bowl C to D.
Actually, I managed to solve point a). If I am correct, P(2 red and 3 blue chips were transferred from bowl C to D)=5/21.
Probability from b) can be transformed by the Bayes rule. P(draw a blue chip from bowl D| 2 red and 3 blue chips were transferred from bowl C to D)=3/5. It is left to find probability of drawing a blue chip from bowl D. Should I use the total probability formula and calculate conditional probabilities for all possible amounts of blue chips in bowl D (4 out of 5, 3 out of 5, etc.) or is there another way?
P.S. one of my groupmates suggested that probability of drawing a blue chip would be 4/10, which, to my mind, is a complete nonsense.
If the event of transferring $2$ red and $3$ blue chips to the bowl $D$ is $A$ and the event of drawing a blue chip from bowl $D$ is $B$,
$P(A|B) = \cfrac{P(B|A) \cdot P(A)}{P(B)}$
$P(B)$ is simply $\cfrac{2}{5}$ as there are $4$ blue chips out of $10$ to start with in bowl $C$. The approach you wrote in your question is not incorrect but it is unnecessary. It does not change the unconditional probability of drawing a blue chip just because we first randomly moved a few chips from bowl $C$ to $D$ and then picked from $D$.
As you wrote, $P(B|A)$ is $\cfrac{3}{5}$ as it is the probability of drawing a blue chip given we have $2$ red and $3$ blue chips in bowl $D$.
$P(A)$ is $\cfrac{5}{21}$, what you already calculated correctly in $(a)$.
So you now have everything you need to find $P(A|B)$.