Easier way to solve this mixed conditional Probability problem?

Question

The following is a solved sample problem in a Persian book. The answer of the book is very confusing and used lots of events. I tried to find an easier approach to solve it, but seems like a dead end.

Suppose we have 2 boxes; A and B. Box A contains 3 white balls and 4 red balls. Box B has 3 red balls and 2 white balls. We randomly choose one of the boxes and pick a random ball out of it, then we put it into the other box. After that, we pick 2 balls out of the later box (in which we have just put a ball inside of it) randomly. What is the probability that we have transferred a red ball in the first move, given that the later 2 balls are white?

Here's the answer provided by the book:

$W_i$ is event of picking a white ball from box $i$ on the first move

$R_i$ is event of picking a red ball from box $i$ on the first move

$WW_i$ is event of picking two white balls from box $i$ on the second move

$A$ is event of picking from box A on first move

$B$ is event of picking from box B on first move

$D$ is the event that we are looking for

Then,

$P(D) = P(A∩D) + P(B∩D)=P(A)*P(D|A)+P(B)*P(D|B)$

where

$P(A)=P(B)= \frac{1}{2}$

and

$$P(D|A) = P(R_A|WW_B) = \frac{P(R_A)*P(WW_B|R_A)}{P(R_A)*P(WW_B|R_A)+P(W_A)*P(WW_B|W_A)}$$

$$= \frac{\frac{4}{7}*\frac{\binom{2}{2}}{\binom{6}{2}}}{\frac{4}{7}*\frac{\binom{2}{2}}{\binom{6}{2}}+\frac{3}{7}*\frac{\binom{3}{2}}{\binom{6}{2}}}=\frac{4}{13}$$

also $$P(D|B) = P(R_B|WW_A) = \frac{P(R_B)*P(WW_A|R_B)}{P(R_B)*P(WW_A|R_B)+P(W_B)*P(WW_A|W_B)}$$ $$= \frac{\frac{3}{5}*\frac{\binom{3}{2}}{\binom{8}{2}}}{\frac{3}{5}*\frac{\binom{3}{2}}{\binom{8}{2}}+\frac{2}{5}*\frac{\binom{4}{2}}{\binom{8}{2}}}=\frac{9}{21}$$

therfore

$P(D) = \frac{1}{2}*\frac{4}{13}+\frac{1}{2}*\frac{9}{21}=\frac{201}{546}=\frac{67}{182}$

Answer 1

Use Bayes' Theorem:

Where:

$R1$ is event of transferring red ball on first move

$2W$ is event of picking two balls on second move

$A$ is event of picking from box A on 1st move

what you are looking for is:

$$P(R1|2W) = \frac{P(2W|R1) * P(R1)}{P(2W)}$$

where:

$P(2W) = P(2W|R1)*P(R1) + P(2W|\neg R1)*P(\neg R1)$

Also:

$P(R1) = P(R1|A)*P(A) + P(R1|\neg A)*P(\neg A)$

where

$P(R1|A) = \frac{4}{7}$

$P(R1|\neg A) = \frac{3}{5}$

$P(A) = P(\neg A) = \frac{1}{2}$

and of course $P(\neg R1) = 1-P(R1)$

And finally:

$P(2W|R1) = P(2W|R1\land A)*P(A)+P(2W|R1\land \neg A)*P(\neg A)$

where

$P(2W|R1\land A) = \frac{2}{6} * \frac{1}{5}$

$P(2W|R1 \land \neg A) = \frac{3}{8} * \frac{2}{7}$

and

$P(2W|\neg R1) = P(2W|\neg R1\land A)*P(A)+P(2W|\neg R1\land \neg A)*P(\neg A)$

where

$P(2W|\neg R1\land A) = \frac{3}{6} * \frac{2}{5}$

$P(2W|\neg R1 \land \neg A) = \frac{4}{8} * \frac{3}{7}$

Plug this all in (and you get $\frac{2993}{8039}$) ...so ... yeah, not simple ... and I don't know how to make it simpler either.

Addition

OK, so my answer is not the book's answer, so (at least) one of the approaches is mistaken. Which one? (or both?)

Frankly, I find the $D$ 'event' that the book talks about a little confusing as it is defined as 'the event we are looking for' ... so is that 'the event of having-picked-a-red-ball-on-the-first-move-given-that-two-white-balls-are-picked-on-the-second-move'? ... I mean, is that really an event?

OK, so the book says:

$P(D) = $

$P(D|A)*P(A) + P(D|B)*P(B) =$

$P(R_A|WW_B)*P(A) + P(R_B|WW_A)*P(B)$

But if I were to follow that method, and given that in general that:

$P(A|B) = P(A|B \land C) * P(C|B) + P(A|B \land \neg C) * P(\neg C|B)$

I would say:

$ P(R1|2W) = P(R1|2W \land A)*P(A|2W) + P(R1|2W \land B)*P(B|2W) =$

$P(R_A|2W \land A)*P(A|2W) + P(R_B|2W \land B)*P(B|2W) =$

$P(R_A|WW_B)*P(A|2W) + P(R_B|WW_A)*P(B|2W)$

so ... the book made a mistake (the book has $P(A)$ and $P(B)$ instead of $P(A|2W)$ and $P(B|2W)$ in this last equation). Conceptually, the mistake is that it treated $R1|WW$ as an event $D$, and so when it went from $P(D|A)$ to $P(R_A|WW_B)$, it basically did something like this:

$P(D|A) = $

$P(R1|2W)|A) =$

$ P(R1|2W \land A) =$

$ P(R_A|WW_B)$

The third step makes sense, and even though the first two steps seem to make sense, they really don't. In fact, this is a good example why you don't want to treat something like $A|B$ as an event!