Question 998473: As a fundraiser, raffle tickets are printed that have three out of the 32 NFL teams on it each week. No two people have the exact same three teams in a given week. The person with the ticket of the highest combined score for their three teams in the week is the grand prize winner for that week. If you buy a ticket and all tickets are sold, what is the probability that you are the grand-prize winner for that week? Write your answer as a simplified fraction.
Answer by mathmate(429)   (Show Source):
You can put this solution on YOUR website!
Given:
Unique raffle tickets bearing 3 teams chosen from 32 NFL teams, and drawn once a week.
Need to find the probability of winning.
Solution:
First step is to find how many unique tickets there are.
We choose 3 teams out of 32, the number of ways is 32 choose 3, or
C(32,3) = 32!/(3!(32-3)!) = 4960
Therefore the probability of winning is 1/4960.
Note:
The number of unique teams can be worked out this way as well:
number of choices for first team=32
number of choices for second team=31
number of choices for third team=30
So total number of ways (multiplication rule)=32*31*30
If order does not count (e.g. ABC,ACB,BAC,BCA,CAB,CBA), we have over counted by 6 times, so number of ways to choose 3 teams (order does not count)
=32*31*30/6=4960
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