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# The membership of a committee consists of 3 English teachers

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The membership of a committee consists of 3 English teachers [#permalink]

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02 Sep 2010, 13:15
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5% (low)

Question Stats:

79% (00:45) correct 21% (01:09) wrong based on 295 sessions

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The membership of a committee consists of 3 English teachers, 4 Mathematics teachers, and 2 Social Studies teachers. If 2 committee members are to be selected at random to write the committee’s report, what is the probability that the two members selected will both be English teachers?

A. 2/3
B. 1/3
C. 2/9
D. 1/12
E. 1/24
[Reveal] Spoiler: OA

Last edited by Bunuel on 29 Nov 2013, 14:15, edited 1 time in total.
RENAMED THE TOPIC.

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02 Sep 2010, 13:53
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udaymathapati wrote:
The membership of a committee consists of 3 English teachers, 4 Mathematics teachers, and 2 Social Studies teachers. If 2committee members are to be selected at random to write the committee’s report, what is the probability that the two members selected will both be English teachers?
A. 2/3
B. 1/3
C. 2/9
D. 1/12
E. 1/24

There are total 3+4+2=9 teachers out of which 3 teach English.

$$P=\frac{C^2_3}{C^2_9}=\frac{1}{12}$$

Or: $$P=\frac{3}{9}*\frac{2}{8}=\frac{1}{12}$$.

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14 Sep 2010, 00:43
Hello Bunuel,

Can you please explain why you have used combinations instead of permutations to calculate the probability?

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14 Sep 2010, 04:54
Hi sarafbhushan,

The combinatioan has to be used due to the below reason.

Let us say if we have 3 people A, B and C. If we are asked to select a couple. we can select AB/BC/CA: Please mnote that selecting AB is same as selecting BA as my couple is same in both the cases. Same is the case with selection teachers for a committee, as is the case in the question.

3C2 = 3C1 (not selecting 1) = 3 (AB, BC, CA)
3P2 = 6 (If Ab is not = BA)

Hope it is clear

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14 Sep 2010, 06:16
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sarafbhushan wrote:
Hello Bunuel,

Can you please explain why you have used combinations instead of permutations to calculate the probability?

As explained above, order of the members in the committee is not important. For example, $$C^2_9$$ gives the # of all committees of 2 we can choose out of 9 teachers (a, b, c, d, e, f, g, h, i): (a, b), (a,c), (a,d), ... Now, committee (a,b) is the same committee as (b,a), so that's why we should use C (which counts (a,b) only once) instead of P (which counts (a,b) as well (b,a)).

Check Probability and Combinations chapters of Math Book for more on this issues (link in my signature).

Hope it's clear.
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14 Sep 2010, 07:02
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Probability of first member an English teacher = 3/9
Probability of second member an English teacher = 2/8
Probability of both being english teacher = 3/9 x 2/8 =1/12 (D)
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25 Oct 2010, 05:38
monirjewel wrote:
The membership of committee consists of 3 English teachers, 4 mathematics teacher, and 2 social studies teachers. If 2 committee members are to be selected at random to write the committee's report, what is the probability that the two members selected will be both be English teachers?
A) 2/3
B) 1/3
C) 2/9
D) 1/12
E 1/24

My approach would be-
3C2/9C2 = 1/12

I am not sure why the OA is (C).

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29 Nov 2013, 13:28
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Re: The membership of a committee consists of 3 English teachers [#permalink]

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03 Dec 2014, 05:31
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Re: The membership of a committee consists of 3 English teachers [#permalink]

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16 Jan 2016, 11:48
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Re: The membership of a committee consists of 3 English teachers [#permalink]

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09 Jul 2017, 09:04
Hello from the GMAT Club BumpBot!

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Re: The membership of a committee consists of 3 English teachers   [#permalink] 09 Jul 2017, 09:04
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