How many ball are there in the bag?

Question

A teacher gives a student a bag which contains white and black balls. The teacher asks the student to pick $5$ balls out of the bag. The students picks 5 white balls and the teacher says : "The probability of this event is $0.5$". Can you determine how many white and black balls are in the bag ?

I assume that the trials are independent and that the balls are picked from the bag without replacement (the problem does not mention this). Let $n_b$ (resp. $n_w$) denote the number of black (resp. white) balls in the bag. The probability of picking $5$ white balls from the bag (without replacement) is:

$$ p = \frac{1}{2} = \frac{n_w (n_w - 1) (n_w - 2) (n_w - 3) (n_w - 4)}{5!(n_w + n_b)^5} = \frac{n_w!}{(n_w - 5)!5!(n_w + n_b)^5}$$

[The probability of picking the $1$st ball from the bag is $n_w/(n_w+n_b)$. For the second, it is $(n_w - 1)/(n_w + n_b)$... and so on].

Is this the right way to go with this problem ?

Answer 1

That does not look right, since you keep dividing by $n_w + n_b$, which assumes that you pick with replacement.

With $B$ the number of black balls, and $W$ the number of white balls, you should simply get that:

$$P = \frac{W \choose 5}{{W + B} \choose 5}$$

Or, using your method:

$$ p = \frac{n_w (n_w - 1) (n_w - 2) (n_w - 3) (n_w - 4)}{(n_w + n_b)(n_w + n_b-1)(n_w + n_b-2)(n_w + n_b-3)(n_w + n_b-4)}$$

One solution is $B=1$ and $W=9$ ... I suspect there are no other solutions but don't know how to prove that ...

Answer 2

We can pick $5$ balls (without replacement) from a bowl with $w$ white balls and $b$ black balls as

$$\frac{{w \choose 5}}{{b+w} \choose 5} = \frac{1}{2}$$

So

$$2{w \choose 5} = {b+w \choose 5} $$

we can cancel the common $5!$ terms, and get

$$2w(w-1)(w-2)(w-3)(w-4) = (b+w)(b+w - 1)(b+w-2)(b+w-3)(b+w-4)$$

which is still pretty nasty (might need some number theory to find integer solutions with $b \ge 1, w \ge 5$ (Wolframalpha had trouble).

The identity $${10 \choose 5 } = {9 \choose 5} + {9 \choose 4}\implies {10 \choose 5} = 2 { 9 \choose 5}$$ shows that $9$ white balls and 1 black does give a solution. Wolframalpha promises 27 integer solutions with no restrictions on $b,w$, but gives fewer than that...