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# MGMAT Probability

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Senior Manager
Joined: 18 Jun 2010
Posts: 303
Schools: Chicago Booth Class of 2013
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Kudos [?]: 122 [0], given: 194

MGMAT Probability [#permalink]  17 Aug 2010, 10:27
00:00

Difficulty:

5% (low)

Question Stats:

80% (02:16) correct 20% (02:57) wrong based on 10 sessions
Gents,

Can we solve the task using "1-X" method?

A basketball coach will select the members of a five-player team from among 9 players, including John and Peter. If the five players are chosen at random, what is the probability that the coach chooses a team that includes both John and Peter?

1/9
1/6
2/9
5/18
1/3
[Reveal] Spoiler: OA
Senior Manager
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Kudos [?]: 18 [0], given: 3

Re: MGMAT Probability [#permalink]  17 Aug 2010, 13:11
you can but that will require more cases and will be time consuming.

straight forward way is to group J and P as always present on the team and as order does not matter so we just need to find the total number of ways to select rest three players = 7c3

total number of ways of selecting 5 players out of 9 = 9c5

probability = 7c3/9c5 = 5/18
Senior Manager
Joined: 18 Jun 2010
Posts: 303
Schools: Chicago Booth Class of 2013
Followers: 20

Kudos [?]: 122 [0], given: 194

Re: MGMAT Probability [#permalink]  17 Aug 2010, 22:22
ingoditrust wrote:
you can but that will require more cases and will be time consuming.

probability = 7c3/9c5 = 5/18

Look at my approach - it's easy, nice and not at all time consuming. But I'm somehow getting the wrong answer
I decided to find the probability of not getting these two guys in the choosen group and then substruct it from 1.
What's wrong with my approach?

1-(7/9)*(6/8)*(5/7)*(4/6)*(3/5)= 15/18
Intern
Joined: 15 Aug 2010
Posts: 23
Location: Mumbai
Schools: Class of 2008, IIM Ahmedabad
Followers: 4

Kudos [?]: 11 [0], given: 0

Re: MGMAT Probability [#permalink]  17 Aug 2010, 23:30
Financier wrote:
I decided to find the probability of not getting these two guys in the choosen group and then substruct it from 1.
What's wrong with my approach?

Prob of choosing both J and P = 1 - ( Prob of choosing neither J nor P + Prob of choosing J and not P + Prob of choosing P and not J)

Hope this helps. This was highlighted in one of the other probability threads earlier this day only.

Thanks.
_________________

Naveenan Ramachandran
4GMAT - Mumbai

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Re: MGMAT Probability [#permalink]  18 Aug 2010, 14:24
+1 D
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Re: MGMAT Probability   [#permalink] 18 Aug 2010, 14:24
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