I'm posting one puzzle, riddle, math, or statistical problem a day. Try to answer each one and post your answers in the comments section. I'll post the answer the next day. Even if you have the same answer as someone else, feel free to put up your answer, too!
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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
ReplyDeleteP=(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.
I couldn't have said it better myself. So... I won't. ;-)
ReplyDeleteNice work Abe!