(Advanced Level Statistics - J. Crawshaw & J. Chambers)
Page 244, Ex.4b Expectation, Question 10.

A bag contains 3 Red balls and 1 Blue ball.
A Second Bag contains 1 Red ball and 1 Blue ball.
A ball is picked out of each Bag and is then placed in the other Bag.
What is the expected number of Red balls in the First Bag?

I remember having worked a similar question in other book "Statistics 1 - Steve Dobbs & Jane Miller, Exercise 4B Misc. Question 1" , which was a simple challenging probability question about transferring balls from a Bag A to Bag B and back to Bag A. And was easily solved with Tree Diagram.

But, the question quoted above, despite having some similarities, it seems that some more works (E(x) Expectation number, correct Tree Diagram for the transfer etc..) need to be done over there.

Again more ambiguities.

Here my Tree Diagram and solution.

The Random Discrete Variables is concluded to be $x=2,3,4$

And their corresponding Probabilities,

$P(X=2)=\left (\frac{1}{2}\times \frac{3}{4} \right )=\frac{3}{8}$

$P(X=3)=\left (\frac{1}{2}\times \frac{1}{4} \right )+\left ( \frac{1}{2}\times \frac{1}{2} \right )=\frac{3}{8}$

$P(X=4)=\left (\frac{1}{2}\times \frac{1}{2} \right )=\frac{1}{4}$

Now, summarizing these Variables $x$ (Numbers of Red ball in the 1st Bag), and their corresponding Probabilities $P(X=x)$:

$\begin{bmatrix} &x &2 &3 &4 \\ &P(X=x) &\frac{3}{8} &\frac{3}{8} &\frac{1}{4} \end{bmatrix}$

Finally, the Expected Means number:
$E(X)=\left (2\times \frac{3}{8} \right )+\left (3\times \frac{3}{8} \right )+\left (4\times \frac{1}{4} \right )$

=2.875
$\neq 2.75$

So what do you think, Feel welcomed to hand over your Tree Diagram, Solutions, Suggestions. Thank you !!!

2. ## Re: Mismatching Book Answer - Expectation (Advanced Statistics - Chambers)

Originally Posted by zikcau25
(Advanced Level Statistics - J. Crawshaw & J. Chambers)
Page 244, Ex.4b Expectation, Question 10.

I remembered having worked a similar question in other book "Statistics 1 - Steve Dobbs & Jane Miller, Exercise 4B Misc. Question 1" , which was a simple challenging probability question about transferring balls from a Bag A to Bag B and back to Bag A. And was easily solved with Tree Diagram.

But, the question quoted above, despite having some similarities, it seems that some more works (E(x) Expectation number, correct Tree Diagram for the transfer etc..) need to be done over there.

Again more ambiguities.

Here my Tree Diagram and solution.

The Random Discrete Variables is concluded to be $x=2,3,4$

And their corresponding Probabilities,

$P(X=2)=\left (\frac{1}{2}\times \frac{3}{4} \right )=\frac{3}{8}$

$P(X=3)=\left (\frac{1}{2}\times \frac{1}{4} \right )+\left ( \frac{1}{2}\times \frac{1}{2} \right )=\frac{3}{8}$

$P(X=4)=\left (\frac{1}{2}\times \frac{1}{2} \right )=\frac{1}{4}$

Now, summarizing these Variables $x$ (Numbers of Red ball in the 1st Bag), and their corresponding Probabilities $P(X=x)$:

$\begin{bmatrix} &x &2 &3 &4 \\ &P(X=x) &\frac{3}{8} &\frac{3}{8} &\frac{1}{4} \end{bmatrix}$

Finally, the Expected Means number:
$E(X)=\left (2\times \frac{3}{8} \right )+\left (3\times \frac{3}{8} \right )+\left (4\times \frac{1}{4} \right )$

=2.875
$\neq 2.75$

So what do you think, Feel welcomed to hand over your Tree Diagram, Solutions, Suggestions. Thank you !!!
the book is right

Pr[x=4] = 1/8 not 1/4 ;pr[blue from bag1]*pr[red from bag 2] = 1/4 * 1/2 = 1/8

Pr[x=3] = 3/8 + 1/8 = 1/2 not 3/8 ;pr[red from 1]*pr[red from 2] + pr[blue from 1]*pr[blue from 2] = 3/4*1/2 + 1/4*1/2 = 1/2

3. ## Re: Mismatching Book Answer - Expectation (Advanced Statistics - Chambers)

Thank you for paying attention. But I am still not fully convinced about the given answer, as it may not consider all the possibilities implied by the unrestricted (if not ambiguous) question itself. Your solution make believe as if the question tells you to start from First Bag only and ignore the location and effect when the ball get transferred and you consider only it's subtraction from the Red Balls in the First Bag only and not it's addition into the other Bag before the second transfer takes place.

I can't clearly read these conditions in that short question above.
Therefore all possibilities must be considered as follows:

(1)
It is equally likely to choose the first between the two Bags (First Bag and Second Bag) in order to start the first transfer = 1/2
(2)
The first transfer of 1 ball can starts either from First Bag first or from the Second Bag first.
(3)
The 1 ball that is being transferred affects the number of balls and it's probability in the next bag to which it is transferred.

This Tree Diagram below summarizes all these:

So,

$P(X=2)=\left (\frac{1}{2}\times \frac{3}{4}\times \frac{1}{3} \right )+\left (\frac{1}{2}\times \frac{1}{2}\times \frac{3}{5} \right )=\frac{11}{40}$

$P(X=3)=\left (\frac{1}{2}\times \frac{3}{4}\times \frac{2}{3} \right )+\left (\frac{1}{2}\times \frac{1}{4}\times \frac{2}{3} \right )+\left (\frac{1}{2}\times \frac{1}{2}\times \frac{4}{5} \right )+\left (\frac{1}{2}\times \frac{1}{2}\times \frac{2}{5} \right )$ = $\frac{19}{30}$

$P(X=4)=\left (\frac{1}{2}\times \frac{1}{4}\times \frac{1}{3} \right )+\left (\frac{1}{2}\times \frac{1}{2}\times \frac{1}{5} \right )=\frac{11}{120}$

Summarizing into table,

$\begin{bmatrix} &x &2 &3 &4 \\ &P(X=x) &\frac{11}{40} &\frac{19}{30} &\frac{11}{120} \end{bmatrix}$

Therefore,

$E(X)=\left ( 2\times \frac{11}{40} \right )+\left ( 3\times \frac{19}{30} \right )+\left ( 4\times \frac{11}{120} \right )$ = 2.81667

I want more convincing and detailed proof if ever I am wrong, you are welcomed !!!

4. ## Re: Mismatching Book Answer - Expectation (Advanced Statistics - Chambers)

sorry I'm not going to get into analyzing every aspect of the syntax of the question.

My answer assumes that a ball is picked from each bag, and once both balls have been picked they are transferred to the other bag. If you want to put a different interpretation on the question fine and you seem to be able to solve for your various interpretations.

5. ## Re: Mismatching Book Answer - Expectation (Advanced Statistics - Chambers)

Thanks again.

But I think a straight solution to that problem is a diagram/drawing (Tree Diagram) to clearly viewing the process in action.