Probability problem with boxes and balls

Question

I've got to solve the following problem:

We've got n boxes. In each box there are 'b' white balls and 'r' red balls. We take a ball from the 1st box and we enter it in the 2nd box; then, we take a ball from the 2nd box and we enter it in the 3rd box... and like that successively. By last, we take a ball from the last box (the box number 'n'). Which probability has got that last ball to be white?

I've defined Z_i = "the ball we take from the box number i is white" and I've tried to calculate P(Z_n) using the total probability theorem and also the conditional probability but I haven't obtain any solid answer. I think that P(Z_n) depends on P(Z_n-1) but that one depends on P(Z_n-2) and like that successively so I don't know how to calculate each of those probabilities neither P(Z_n).

Answer 1

Let us make the $n$ variable and let $p_{n}$ denote the probability on a white ball under condition that $n$ boxes are used where $n$ ranges over $\mathbb{N}_{+}$.

Then:

• $p_{1}=\frac{b}{b+r}$,
• $p_{2}=\frac{b}{b+r}\frac{b+1}{b+r+1}+\frac{r}{b+r}\frac{b}{b+r+1}=\frac{b\left(b+r+1\right)}{\left(b+r\right)\left(b+r+1\right)}=\frac{b}{b+r}$

Now a bell starts ringing, doesn't it?...

With induction you can prove that: $$p_{n}=\frac{b}{b+r}$$ for every $n$.

Answer 2

Consider starting from Box 2 and calculate the probability to pick a white ball out of the box.

$$P(W) = P(W|InsertedW)P(InsertedW) + P(W|Inserted B)P(Inserted B)$$ $$P(W) = \frac{w+1}{w+b+1}\frac{w}{w+b} + \frac{w}{w+b+1}\frac{b}{w+b}$$ $$P(W) = \frac{w}{w+b}(\frac{w+1}{w+b+1} + \frac{b}{w+b+1}) = \frac{w}{w+b}$$

Can you repeat this n times ?

It also helps to think of what happens if n = 1 or n = 2.

Answer 3

Imagine the colors as $b$ different shades of white and $r$ different shades of red, one ball of each shade in each box. Then all $b+r$ shades are equally likely for the last ball, since by symmetry nothing favors one shade over another.

So the probability of a white ball on the last draw is (# white shades)/(total # shades) = $b/(b+r)$.