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Hello,

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.

The relevant relations are the Law of Total Probability:

[tex] P(B) = \sum_{i=1}^n P(A_i)P(B|A_i) [/tex]

and the Bayes Theorem:

[tex] P(A_i|B) = \dfrac{P(A_i)P(B|A_i)}{\sum_{i=1}^n P(A_i)P(B|A_i)} [/tex]

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:

[tex] P(B) = P(A1)P(B|A1) + P(A2)P(B|A2) + P(A3)P(B|A3) = [/tex]

[itex]

\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}

[/itex]

b) [tex] 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 [/tex]

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:

[tex] P(B) = P(A1)P(B|A1) + P(A2)P(B|A2) + P(A3)P(B|A3) = [/tex]

[itex]

\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}

[/itex]

Also, for (b), it computes:

[tex] 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}} [/tex]

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

Is the textbook wrong or am I missing something?

Thank you.

## 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:

[tex] P(B) = \sum_{i=1}^n P(A_i)P(B|A_i) [/tex]

and the Bayes Theorem:

[tex] P(A_i|B) = \dfrac{P(A_i)P(B|A_i)}{\sum_{i=1}^n P(A_i)P(B|A_i)} [/tex]

## 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:

[tex] P(B) = P(A1)P(B|A1) + P(A2)P(B|A2) + P(A3)P(B|A3) = [/tex]

[itex]

\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}

[/itex]

b) [tex] 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 [/tex]

**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:

[tex] P(B) = P(A1)P(B|A1) + P(A2)P(B|A2) + P(A3)P(B|A3) = [/tex]

[itex]

\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}

[/itex]

Also, for (b), it computes:

[tex] 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}} [/tex]

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

Is the textbook wrong or am I missing something?

Thank you.

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