*(Advanced Level Statistics - J. Crawshaw & J. Chambers)*

Page 244, Ex.4b Expectation, Question 10. **Book Answer: 2.75**

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.

I have

**2.875** as answer instead of

**2.75 **from the book.

Again more ambiguities.

Here my Tree Diagram and solution.

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

And their corresponding Probabilities,

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

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

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

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

$\displaystyle \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:

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

=

**2.875**
$\displaystyle \neq 2.75$

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