Probability of selecting balls from Bag B

Question

There are two bags A and B. Bag A contains $3$ white and $4$ red balls whereas bag B contains $4$ white and $3$ red balls. Three balls are drawn at random (without replacement) from one of the bags and are found to be two white and one red. Find the probability that these were drawn from bag B?

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MY WORK SO FAR:

I found that $P(A) = P(B) = 0.5$. I am unable to find $P(A_2 \mid B)$, $P(B \mid A_2)$, $P(B \mid A_1)$, $P(B \mid A_2)$. I have taken all these values as $3/14$ since $3$ balls are selected from a total of $14$ ($7+7$) and I got ans as $0.5$. but not sure...

Answer 1

Bag $A$ contains $3$ white and $4$ red balls.

The conditional probability for the observed Event of extracting 2 white balls among 3 balls extracted, given this bag was choosen is:

$$\mathsf P(E\mid A) ~=~ \dfrac{\dbinom{3}{2}\dbinom{4}{1}}{\dbinom{7}{3}}~=~\dfrac{12}{35}$$

You should now be able to find $\mathsf P(E\mid B)$ and subsequently $\mathsf P(B\mid E)$

Answer 2

Since the two boxes have equal probabilities of being chosen, we can simply find the odds in favor of $B$ and convert it to a $Pr$. I leave it to you to shroud it in the expected terminology.

Odds in favor of $B = \dfrac{\dbinom42\dbinom31}{\dbinom73} \div \dfrac{\dbinom32\dbinom41}{\dbinom73} = 18:12 = 3:2$

P(drawn from $B) = \dfrac35$