Balls are randomly removed from a bag without replacement. If the probability that the first five balls withdrawn are all green is one-half, what is the fewest possible number of balls in the bag at the start?
I believe the answer is 10 balls because we can assume that there are 5 green balls to begin with. Does this logic hold?
Best Answer
A difference of $1$ ball between green balls and $n$ will yield the minimum $n$ providing it has a whole number solution, so .........
$$\frac{\binom{n-1}{5}}{\binom{n}{5}} = \frac{(\frac{(n-1)!}{(n-6)!5!})}{(\frac{n!}{(n-5)!5!})} = \frac{(n-1)!(n-5)!5!}{(n-6)!5!n!} = \frac{1}{2}$$
$$\frac{(n-1)!(n-5)!}{(n-6)!n!} = \frac{1}{2}$$
$$\frac{(n-5)}{n} = \frac{1}{2}$$
$$2n - 10 = n$$
$$n = 10$$