| Line 11: | Line 11: | ||
--JRHaynie-- | --JRHaynie-- | ||
| + | |||
| + | I am little bit confused about this problem... I was thinking that each one of them should have C(100,1)changes because they are in the same events individually... so, can it be E1= <math>\frac{{3 \choose 1}}{100}</math>, E2=<math>\frac{{3 \choose 1}}{100}</math>, and E3 = <math>\frac{{3 \choose 1}}{100}</math> ???? --[[User:Lee|Lee]] 22:07, 18 February 2009 (UTC) | ||
Revision as of 18:07, 18 February 2009
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)changes 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)
