What is the difference between this question and number 31? I believe that 31 is: $ \frac{{3 \choose 1}}{100} $.
So would this one be $ \frac{{3 \choose 3}}{100} $? which would end up as 1/100? -Brandy
I am not entirely sure that this is the case. I believe you can look at them as three separate events. The event Kumar wins, the event Janice wins and the event Pedro wins. If you look at it this way E1= $ \frac{{3 \choose 1}}{100} $, E2=$ \frac{{2 \choose 1}}{99} $, and E3 = $ \frac{{1 \choose 1}}{98} $. I am not 100 percent sure, but that is the way I look at it instead of trying to pick three random people to match three random prizes, they each have a separate chance of winning a prize and it varies dependent on who is given the first prize. Please correct me if I am thinking incorrectly.
--JRHaynie--
I am little bit confused about this problem... I was thinking that each one of them should have C(100,1)chances because they are in the same events individually... so, can it be E1= $ \frac{{3 \choose 1}}{100} $, E2=$ \frac{{3 \choose 1}}{100} $, and E3 = $ \frac{{3 \choose 1}}{100} $ ???? --Lee 22:07, 18 February 2009 (UTC)
I don't think you can count the events individually, since if one wins, it lessens the chance of the other two to win. So I did the problem by counting the number of ways they could all win (3 * 2 * 1) over the total number of ways (100C3 * 3!).
- Karen
I agree. Here's a couple of ways to get to Karen's answer:
First, consider permutations of winners. There are 3! acceptable permutations that include all 3 of our chosen winners, divided by 100P3 possible permutations of winners.
Second (alternatively), consider just combinations of winners, and disregard order. There is 1 acceptable group of winners, divided by 100C3 groups of winners. In other words, cancel out the 3!'s, and the two answers are equivalent.
Hope this helps, -Zoe
