Calculating Probability for Transferred and Selected Balls

[Thunderbird88]

Hello,

Homework Statement

Suppose we have two boxes, numbered 1 and 2.
Box 1 contains 10 white and 6 numbered red balls, while Box 2 contains 8 white and 12 numbered red balls.
We take out 2 balls from Box 1 and are transferred in Box 2. Then, we choose 1 ball from Box 2.

a) Find the probability to take out one red ball from Box 2 and
b) Find the probability that we transferred one red and one white ball from Box 1 in Box 2, given that we took out a red ball from Box 2.

Homework Equations

The relevant relations are the Law of Total Probability:
$$P(B) = \sum_{i=1}^n P(A_i)P(B|A_i)$$

and the Bayes Theorem:
$$P(A_i|B) = \dfrac{P(A_i)P(B|A_i)}{\sum_{i=1}^n P(A_i)P(B|A_i)}$$

The Attempt at a Solution

My solution is:

Suppose we have the following events:

A1: we transferred two white balls from Box 1 to Box 2
A2: we transferred one white and one red ball from Box 1 to Box 2
A3: we transferred two red balls from Box 1 to Box 2
B: we took out one red ball from Box 2

a) Applying the Law of total probability, I have:

$$P(B) = P(A1)P(B|A1) + P(A2)P(B|A2) + P(A3)P(B|A3) =$$
$\begin{align} = \dfrac{\binom{10}{2}}{\binom{16}{2}} \cdot \dfrac{\binom{12}{1}}{\binom{22}{1}} + \dfrac{\binom{10}{1} \cdot \binom{6}{1}}{\binom{16}{2}} \cdot \dfrac{\binom{13}{1}}{\binom{22}{1}} + \dfrac{\binom{6}{2}}{\binom{16}{2}} \cdot \dfrac{\binom{14}{1}}{\binom{22}{1}} =\\\\ = \dfrac{45}{120} \cdot \dfrac{12}{22} + \dfrac{10 \cdot 6}{120} \cdot \dfrac{13}{22} + \dfrac{15 \cdot 14}{120 \cdot 22} = 0.5795 \end{align}$

b) $$P(A2|B) = \dfrac{\text{P(A2)} \cdot \text{P(B|A2)}}{\text{P(B)}} = \dfrac{\dfrac{\binom{10}{1} \cdot \binom{6}{1}}{\binom{16}{2}} \cdot \dfrac{\binom{13}{1}}{\binom{22}{1}}}{0.5795} = 0.5099$$

4. Textbook's Solution

The textbook confuses me, because it defines the event B as "we take out a white ball from Box 2" and then for question (a), it computes:

$$P(B) = P(A1)P(B|A1) + P(A2)P(B|A2) + P(A3)P(B|A3) =$$

$\begin{align} = \dfrac{\binom{10}{2}}{\binom{16}{2}} \cdot \dfrac{\binom{10}{1}}{\binom{22}{1}} + \dfrac{\binom{10}{1} \cdot \binom{6}{1}}{\binom{16}{2}} \cdot \dfrac{\binom{9}{1}}{\binom{22}{1}} + \dfrac{\binom{6}{2}}{\binom{16}{2}} \cdot \dfrac{\binom{8}{1}}{\binom{22}{1}} \text{It stops here.} \end{align}$

Also, for (b), it computes:

$$P(A2|B) = \dfrac{\text{P(A2)} \cdot \text{P(B|A2)}}{\text{P(B)}} = \dfrac{\binom{10}{1} \cdot \binom{6}{1}}{\binom{16}{2}} \cdot \dfrac{\binom{9}{1}}{\binom{22}{1}}$$

Shouldn't this be divided by P(B), as well?

Is the textbook wrong or am I missing something?

Thank you.

 

[mfb]

I agree that this should get divided by P(B).