tag:blogger.com,1999:blog-15628310.post5330562213040653512..comments2020-02-12T10:46:25.655-05:00Comments on Question of the day: Five Balls Are RemovedAnonymoushttp://www.blogger.com/profile/18153935609499338685noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-15628310.post-21619791489348842532007-03-23T08:45:00.000-04:002007-03-23T08:45:00.000-04:00I couldn't have said it better myself. So... I wo...I couldn't have said it better myself. So... I won't. ;-) <BR/><BR/>Nice work Abe!Anonymoushttps://www.blogger.com/profile/18153935609499338685noreply@blogger.comtag:blogger.com,1999:blog-15628310.post-72077672600322197932007-03-22T12:06:00.000-04:002007-03-22T12:06:00.000-04:00For the smallest value of N possible, the numbers ...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<BR/><BR/>P=(N-1)/N*(N-2)/(N-1)*(N-3)/(N-2)*(N-4)/(N-3)*(N-5)/(N-4)=1/2 <BR/><BR/>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. <BR/><BR/>10 total balls, 9 blue balls.Abehttps://www.blogger.com/profile/04424868492071587450noreply@blogger.com