## Thursday, March 22, 2007

### Five Balls Are Removed

A hat contains a number N of blue balls and red balls. If five balls are removed randomly from the hat, the probability is precisely 1/2 that all five balls are blue. What is the smallest value of N for which this is possible.

1. For the smallest value of N possible, the numbers of blue balls and total balls are going to have to be relatively close together. In fact, of all the combinations that would work, the smallest N would have to be the case where those two values are closest. We'll take the most extreme case to start and assume they are only one apart. Then, you get

P=(N-1)/N*(N-2)/(N-1)*(N-3)/(N-2)*(N-4)/(N-3)*(N-5)/(N-4)=1/2

for the probability that all five balls are blue (again we're assuming there are N total balls and N-1 blue balls, and each time we draw a blue ball, we subtract one from each). You can obviously see that everything cancels out except (N-5)/N=1/2, so N=10.

10 total balls, 9 blue balls.

2. I couldn't have said it better myself. So... I won't. ;-)

Nice work Abe!

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