Polya's Urn Scheme Using Bayes Rule

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youngbuddhist

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An urn contains w white balls and b black balls. A ball is drawn at random from the urn and then replaced, along with d more balls of its color. Then a ball is drawn at random from the w+b+d balls that are now in the urn. Let B1 be the event that the first ball drawn is black, and let B2 be the event that the second ball drawn is black.

a) Find P(B1). Find P(B2).
b) Find P(B2 given B1). Find P(B1 given B2).

I am very confused as to how to solve this problem. Can someone clearly explain to me in detail how to solve for all the probabilities (and work them out)? Thanks.
 
I hope that you understand that \(\displaystyle \L
P\left( {B_1 } \right) = \frac{b}{{w + b}}\;\& \;P\left( {W_1 } \right) = \frac{w}{{w + b}}.\)

\(\displaystyle \L \begin{array}{rcl}
P\left( {B_2 } \right) & = & P\left( {B_2 B_1 \cup B_2 W_1 } \right) \\
& = & P\left( {B_2 B_1 } \right) + P\left( {B_2 W_1 } \right) \\
& = & P\left( {B_2 |B_1 } \right)P\left( {B_1 } \right) + P\left( {B_2 |W_1 } \right)P\left( {W_1 } \right) \\
& = &\frac{{b + d}}{{w + b + d}}\frac{b}{{w + b}} + \frac{b}{{w + b + d}}\frac{w}{{w + b}} \\
\end{array}.\)

In that is the answer to all your questions.
 
Actually, the P(B1) and the P(B2) are the same, b/(w+b), according to my professor. I don't understand how she got P(B1) or P(B2)
 
Yes in fact they are equal!
Did you even bother to add up the last line that I gave you?
After all you do have a common denominator.
Why don’t you do that? See what you get. (You better know algebra!)
 
youngbuddhist said:
i still don't get how the P(B1) works out.
Click to expand...
Why is that? Take a simple example: four whites and six blacks.
Because 4+6=10 there are ten in the urn; thus \(\displaystyle P\left( {B_1 } \right) = \frac{6}{{10}}.\)

So with w whites and b blacks there are w+b in the urn and \(\displaystyle P\left( {B_1 } \right) = \frac{b}{{w+b}}\)
 
An eye-opener it is - just like Monty Hall problem.
 
Hehehe.. it's part of conditional probability eh. Sirs, I have seen the link and the answers are all the same for all Bj but still, it just took the P(B2) only. How to solve P(B3), and the general solution for P(Bj)?
 
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