Saturday, June 8, 2013

Another probability problem

An interesting set of problems was posted on "Math, Math Education, Math Culture" LinkedIn group few weeks ago. I engaged myself in addressing few of them, so I will be posting my solutions in the next few posts. Just to clarify this from the beginning, interesting doesn't necessarily mean difficult. This particular post is dedicated to the probability problem and here is the original LinkedId discussion, if you happen to have a LinkedIn account. Otherwise, continue reading this post. Here is the problem ...

We have n urns, each of these urns contains A white balls and B black balls. We assume a ball from the first urn is randomly picked and then placed into the second urn, then another ball from the second urn is randomly picked and then placed into the third urn, and so on, until a ball from the last urn is finally randomly picked. If this last ball is white, what probability has this fact?

One way to attack the problem is to use total probability (Wikipedia is probably more descriptive on this topic).

Let's define the following two events:
- Wi - white ball is picked from i-th urn
- Bi - black ball is picked from i-th urn.
It is worth noting that $P\left ( W_{i} \right )+P\left ( B_{i} \right )=1$.

Now, applying total probability we have: $$P\left ( W_{n} \right ) = P\left ( W_{n} | W_{n-1} \right )\cdot P\left ( W_{n-1} \right )+P\left ( W_{n} | B_{n-1} \right )\cdot P\left ( B_{n-1} \right )$$ Where:
- $P\left ( W_{n} | W_{n-1} \right )=\frac{A+1}{A+B+1}$ - translated as "probability to pick a white ball from n-th urn, knowing that a white ball was picked from n-1-th urn"
- $P\left ( W_{n} | B_{n-1} \right )=\frac{A}{A+B+1}$ - translated as "probability to pick a white ball from n-th urn, knowing that a black ball was picked from n-1-th urn".

Putting all these together, we obtain: $$P\left ( W_{n} \right )= P\left ( W_{n-1} \right )\cdot \frac{1}{A+B+1}+\frac{A}{A+B+1}$$ Or, if we note $k= \frac{1}{A+B+1}$, then: $$P\left ( W_{n} \right )= P\left ( W_{n-1} \right )\cdot k+A\cdot k$$

Continuing doing this recursively, we obtain: $$P\left ( W_{n} \right )= P\left ( W_{1} \right )\cdot k^{n-1} + A\cdot \left ( k^{n-1}+k^{n-2}+...+k^{2}+k \right )$$ Where (because W1 relates to the very first urn): $$P\left ( W_{1} \right )=\frac{A}{A+B}=A\cdot \frac{k}{1-k}$$

Finally: $$P\left ( W_{n} \right )=A\cdot \frac{k^{n}}{1-k}+A\cdot \frac{k^{n}-k}{k-1}=\frac{A}{A+B}$$

Which means, the trick with picking randomly a ball and moving it to the next urn has no effect on the final probability. Interesting, isn't it?