HELP with a Combination & Probabilty Problem

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hengirl03: Senior | Next Rank: 100 Posts; Posts: 86; Joined: Tue Aug 05, 2008 10:50 am; Thanked: 7 times

by hengirl03 » Sun Aug 17, 2008 8:25 am

Hi. I'm trying to figure out the probability that 2 U.S. Olympians will each win a medal at the Olympics.

The facts:
- There are 2 U.S. Olympians competing
- There are 12 athletes total
- There are 3 medals to be awarded

So far, I've figured out that there are 12C2 sets that could occur, or 33 sets. However, I don't know how do figure out the probability. Is it 33/12! ???? Because that seems really small.

Thanks!!!!

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pepeprepa: Legendary Member; Posts: 661; Joined: Tue Jul 08, 2008 12:58 pm; Location: France; Thanked: 48 times

by pepeprepa » Sun Aug 17, 2008 8:38 am

Can also wonder the probability of Jamaican women sprinters who are 1st, 2nd and 3rd given there were 3 Jamaican girls and 8 athletes.

Here, you have 12C3=12!/3!9!=220 total possibilities of podiums with 12 athletes.

You can figure out the different podiums you can have with 2 americans (A and B).
AB_ (in this one A is first, B is second and _ represent the place of one of the 10 other athletes)
A_B
BA_
B_A
_BA
_AB

For each of these configurations there are 10 different guys who can complete the podium, so 10 possibilities per configuration.
Given there are 6 configurations, the podiums with two Americans are 60.

60/220 is the probability or 3/11

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sudhir3127: Legendary Member; Posts: 829; Joined: Mon Jul 07, 2008 10:09 pmFollowed by: 3 members; Location: INDIA; Thanked: 84 times

by sudhir3127 » Sun Aug 17, 2008 9:36 am

Even i got 3/11

Almost the same approach...

chosing 3 athlets out of 12 in 12C3 ways = 220

we need to put 2 American on the podium out of 12 ( only 3 are eligible)

apart from these 2 Amercian we are left with 10 others competing for 1 place in 10 ways.

since the order is not important . we can arrange the 2 americans in 3 places in 3P2 ways which is 6.

thus the total probability is
(6*10)/220 = 3/11

hope its clear .. do let us know if u have any doubs..

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Ian Stewart: GMAT Instructor; Posts: 2623; Joined: Mon Jun 02, 2008 3:17 amFollowed by: 355 members; Location: Montreal; Thanked: 1090 times; GMAT Score:780

by Ian Stewart » Sun Aug 17, 2008 11:53 am

Guys, the probability that the first athlete wins a medal is 3/12, so there's no way the probability that both win a medal can be higher than 3/12. That is, 3/11 can't be the answer, unless the USA athletes are a whole lot better than the other athletes.

If we are to assume that everyone has an equal probability of winning, then athlete A has a 3/12 chance of winning a medal, and assuming A wins, B has a 2/11 chance of winning a medal. So the probability both win should be

(3/12)*(2/11) = 1/22

Or, you can say that there are 12C3 ways of choosing three medal winners. If both Americans must win, there are 2C2 (that is, one) way of choosing the two Americans, and 10C1 ways of choosing the third medal winner. So the probability should be:

(2C2 * 10C1)/(12C3) = 10/220 = 1/22

Notice that this answer is exactly 1/6 of 3/11, the answer given above. When you list the podiums involving the Americans (AB_, A_B, _AB, etc.) you're assuming order matters, and when you counted the total number of events (12C3), you assumed order did not matter. It doesn't matter whether you assume order matters here as long as you're consistent: you need to count the numerator and the denominator in the same way.

For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

ianstewartgmat.com

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pepeprepa: Legendary Member; Posts: 661; Joined: Tue Jul 08, 2008 12:58 pm; Location: France; Thanked: 48 times

by pepeprepa » Sun Aug 17, 2008 2:40 pm

Thanks a lot for pointing this error of using ordered and non ordered events

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