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# A committee of 3 has to be formed randomly from a group of 6

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Intern
Joined: 14 Jul 2006
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A committee of 3 has to be formed randomly from a group of 6 [#permalink]

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23 Aug 2007, 19:48
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

A committee of 3 has to be formed randomly from a group of 6 people. Tom and Mary are in this group of 6. What is the probability that Tom will be selected into the committee but Mary will not?

1/10
1/5
3/10
2/5
1/2

This question is a lot easier than I made it out to be. What do you see in the question that tips you off to the easiest method to solve? I keep confusing something in these types of probability questions. Thanks for the help!

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Manager
Joined: 03 Sep 2006
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23 Aug 2007, 20:47
Brrr - I got 1/10

1st person - Tom: 1/6
2nd person - not Mary 4/5
3rd person - not Mary - 3/4

P = 1/6 * 4/5 * 3/4 = 1/10

P.S> If Tom is 2nd or 3rd - there are no such answers for this probability in the answer choices

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Manager
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23 Aug 2007, 22:39
total committees : 6c3 = 20

committees with only tom = 6

ans : 6/20 = 3/10

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Manager
Joined: 08 Oct 2006
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23 Aug 2007, 23:00
3/10 it is.
Total committes posssible = 6C3
Committee with Tom= 4C2=6

Thus 6/20= 3/10

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Manager
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23 Aug 2007, 23:11
empty_spaces or excelgmat,
could you please explain why Committee with Tom = 6?

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Manager
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23 Aug 2007, 23:28
Whatever wrote:
empty_spaces or excelgmat,
could you please explain why Committee with Tom = 6?

you want tom and 2 other people
2 others from a pool of 4 can be chosen in 4c2 ways = 6

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Manager
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23 Aug 2007, 23:35
empty_spaces wrote:
you want tom and 2 other people
2 others from a pool of 4 can be chosen in 4c2 ways = 6

Ic, thanks. I forgot to exclude Mary to get 4 people =)

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23 Aug 2007, 23:35
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# A committee of 3 has to be formed randomly from a group of 6

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