MHB Occupancy Problem (Wackerly & Mendenhall & Schaeffer Problem 2.93)

[Ackbach]

Problem 2.93. Five identical bowls are labeled $1, 2, 3, 4,$ and $5$. Bowl $i$ contains $i$ white and $5-i$ black balls, with $i=1, 2, \dots, 5$. A bowl is randomly selected and two balls are randomly selected (without replacement) from the contents of the bowl.

• What is the probability that both balls selected are white?
• Given that both balls selected are white, what is the probability that bowl $3$ was selected?

Answer. Let $B_i$ be the event that bowl $i$ is selected. Let $W_1$ be the event that the first ball taken is white, and let $W_2$ be the event that the second ball taken is white. We are given the following data:
$$P(B_i)=1/5;\quad P(W_1|B_i)=i/5;\quad P(W_2|B_i\cap W_1)=(i-1)/4; \quad P(W_2|B_i\cap \overline{W_1})=i/4.$$

• We are asked to compute $P(W_2|W_1)$. If $W_1$ has occurred, then the probability of getting a second white ball from bowl $i$ is $(i-1)/4$. So, we find $P(W_2|W_1)$ as follows:
  \begin{align*}
  P(W_2|W_1)&=\sum_{i=1}^{5}P(B_i\cap W_1) \, P(W_2|B_i\cap W_1) \\
  &=\sum_{i=1}^{5}P(B_i) \, P(W_1|B_i) \, P(W_2|B_i\cap W_1) \\
  &=\sum_{i=1}^{5}\left(\frac15 \right) \left(\frac{i}{5} \right)\left(\frac{i-1}{4}\right) \\
  &=\frac{1}{100}\sum_{i=1}^{5}(i^2-i) \\
  &=\frac{1}{100}\left(\frac{5(6)(11)}{6}-15\right) \\
  &=\frac25.
  \end{align*}
• We are asked to compute $P(B_3|W_1\cap W_2)$. Note that
  $$P(W_1\cap W_2)\, P(B_3|W_1\cap W_2)=P(B_3) \, P(W_1\cap W_2|B_3) $$
  $$\implies P(B_3|W_1\cap W_2)=\frac{P(B_3) \, P(W_1\cap W_2|B_3)}{P(W_1\cap W_2)}
  =\frac{P(B_3) \, P(W_1\cap W_2|B_3)}{P(W_1) \, P(W_2|W_1)}.$$
  To assemble the ingredients of this formula, we note immediately that
  \begin{align*}
  P(B_3)&=\frac15 \\
  P(W_1\cap W_2|B_3)&=\frac35 \cdot \frac24 =\frac35 \cdot \frac12 = \frac{3}{10} \\
  P(W_2|W_1)&=\frac25 \\
  P(W_1)&=\sum_{i=1}^{5}P(B_i) \, P(W_1|B_i)=\frac15 \sum_{i=1}^{5}\frac{i}{5}=
  \frac{15}{25}=\frac35.
  \end{align*}
  It follows, then, that
  $$P(B_3|W_1\cap W_2)=\frac{\frac15 \cdot \frac{3}{10}}{\frac35 \cdot \frac25}
  =\frac{3}{50} \cdot \frac{25}{6}=\frac{1}{2}\cdot \frac{1}{2}=\frac{1}{4}.$$

The first part a is, I am fairly sure, correct. However, I think part b is incorrect. What am I doing wrong? Thank you for your time!

 

[Fermat1]

Problem 2.93. Five identical bowls are labeled $1, 2, 3, 4,$ and $5$. Bowl $i$ contains $i$ white and $5-i$ black balls, with $i=1, 2, \dots, 5$. A bowl is randomly selected and two balls are randomly selected (without replacement) from the contents of the bowl.

• What is the probability that both balls selected are white?
• Given that both balls selected are white, what is the probability that bowl $3$ was selected?

Answer. Let $B_i$ be the event that bowl $i$ is selected. Let $W_1$ be the event that the first ball taken is white, and let $W_2$ be the event that the second ball taken is white. We are given the following data:
$$P(B_i)=1/5;\quad P(W_1|B_i)=i/5;\quad P(W_2|B_i\cap W_1)=(i-1)/4; \quad P(W_2|B_i\cap \overline{W_1})=i/4.$$

• We are asked to compute $P(W_2|W_1)$. If $W_1$ has occurred, then the probability of getting a second white ball from bowl $i$ is $(i-1)/4$. So, we find $P(W_2|W_1)$ as follows:
  \begin{align*}
  P(W_2|W_1)&=\sum_{i=1}^{5}P(B_i\cap W_1) \, P(W_2|B_i\cap W_1) \\
  &=\sum_{i=1}^{5}P(B_i) \, P(W_1|B_i) \, P(W_2|B_i\cap W_1) \\
  &=\sum_{i=1}^{5}\left(\frac15 \right) \left(\frac{i}{5} \right)\left(\frac{i-1}{4}\right) \\
  &=\frac{1}{100}\sum_{i=1}^{5}(i^2-i) \\
  &=\frac{1}{100}\left(\frac{5(6)(11)}{6}-15\right) \\
  &=\frac25.
  \end{align*}
• We are asked to compute $P(B_3|W_1\cap W_2)$. Note that
  $$P(W_1\cap W_2)\, P(B_3|W_1\cap W_2)=P(B_3) \, P(W_1\cap W_2|B_3) $$
  $$\implies P(B_3|W_1\cap W_2)=\frac{P(B_3) \, P(W_1\cap W_2|B_3)}{P(W_1\cap W_2)}
  =\frac{P(B_3) \, P(W_1\cap W_2|B_3)}{P(W_1) \, P(W_2|W_1)}.$$
  To assemble the ingredients of this formula, we note immediately that
  \begin{align*}
  P(B_3)&=\frac15 \\
  P(W_1\cap W_2|B_3)&=\frac35 \cdot \frac24 =\frac35 \cdot \frac12 = \frac{3}{10} \\
  P(W_2|W_1)&=\frac25 \\
  P(W_1)&=\sum_{i=1}^{5}P(B_i) \, P(W_1|B_i)=\frac15 \sum_{i=1}^{5}\frac{i}{5}=
  \frac{15}{25}=\frac35.
  \end{align*}
  It follows, then, that
  $$P(B_3|W_1\cap W_2)=\frac{\frac15 \cdot \frac{3}{10}}{\frac35 \cdot \frac25}
  =\frac{3}{50} \cdot \frac{25}{6}=\frac{1}{2}\cdot \frac{1}{2}=\frac{1}{4}.$$

The first part a is, I am fairly sure, correct. However, I think part b is incorrect. What am I doing wrong? Thank you for your time!

Shouldn't the numerator be $P(B_{3}{\cap}W_{1}{\cap}W_{2})=\frac{1}{5}.\frac{3}{5}.\frac{1}{2}=\frac{3}{50}$? EDIT: sorry, that's what you've got. Why do you think the answer's wrong?

 

[Fermat1]

I've found your problem. In a) you should be finding $P(W_{1}{\cap}W_{2})$, not $P(W_{2}|W_{1})$. Then for b) you get $\frac{3}{50}.\frac{5}{2}=\frac{3}{20}$

 

[Ackbach]

Ok, so let's do this:

• We are asked to compute $P(W_1\cap W_2)$. Now the entire sample space
  $\displaystyle S=\bigcup_{i=1}^{5}B_i$, with $B_i\cap B_j=\varnothing$ whenever $i\not=j$. That is, the $B_{i}$ partition $S$. It follows from the Law of Total Probability that
  $$P(W_1\cap W_2)=\sum_{i=1}^{5}P(B_i) \, P(W_1\cap W_2|B_i).$$
  Now
  $$P(W_1\cap W_2|B_i)=\frac{i}{5} \cdot \frac{i-1}{4}=\frac{i^2-i}{20}.$$
  Hence,
  $$P(W_1\cap W_2)=\frac{1}{100}\sum_{i=1}^{5}(i^2-i)=\frac{2}{5}.$$
• Then, following your lead, we have
  $$P(B_3|W_1\cap W_2)=\frac{P(B_3) \, P(W_1\cap W_2|B_3)}{P(W_1\cap W_2)}
  =\frac{(1/5)(3/5)(2/4)}{2/5}=\frac{3}{20}.$$

Is this reasoning correct? The numerical answers agree with the back of the book, and it certainly seems correct to me.

 

[Fermat1]

Ok, so let's do this:

• Is this reasoning correct? The numerical answers agree with the back of the book, and it certainly seems correct to me.

• 
  
  Yes.