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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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New post Updated on: 29 Nov 2013, 14:15
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A
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C
D
E

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Question Stats:

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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

Originally posted by udaymathapati on 02 Sep 2010, 13:15.
Last edited by Bunuel on 29 Nov 2013, 14:15, edited 1 time in total.
RENAMED THE TOPIC.
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Re: Teacher selection  [#permalink]

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New post 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: \(P=\frac{3}{9}*\frac{2}{8}=\frac{1}{12}\).

Answer: D.
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Re: Teacher selection  [#permalink]

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New post 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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Re: Teacher selection  [#permalink]

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New post 14 Sep 2010, 00:43
Hello Bunuel,

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

Thanks in advance,
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Re: Teacher selection  [#permalink]

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New post 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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Re: Teacher selection  [#permalink]

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

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

Thanks in advance,


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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Re: Membership of committee  [#permalink]

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New post 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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Re: The membership of a committee consists of 3 English teachers  [#permalink]

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New post 28 Dec 2017, 19:56
Hi all,

I was trying to solve this problem in reverse way and getting wrong answer. Im solving as “1 - probability of selecting both from maths and social” that is 1-6c2/9c2 = 7/12. Anyone explain where I went wrong .
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Re: The membership of a committee consists of 3 English teachers  [#permalink]

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New post 28 Dec 2017, 20:30
1
gvrk_77 wrote:
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

Hi all,

I was trying to solve this problem in reverse way and getting wrong answer. Im solving as “1 - probability of selecting both from maths and social” that is 1-6c2/9c2 = 7/12. Anyone explain where I went wrong .


The opposite event of both being English teachers is {both are Mathematics and Social teachers} + {one is English teacher and one is Mathematics or Social teacher}.

\(P=1 - (\frac{C^2_6}{C^2_9}+\frac{C^1_3*C^1_6}{C^2_9})=1-(\frac{15}{36}+\frac{18}{36})=\frac{1}{12}\)

Or: \(P=1-(\frac{6}{9}*\frac{5}{8}+2*\frac{3}{9}*\frac{6}{8})=\frac{1}{12}\) (we multiply 3/9*6/8 by 2 because English/Not English can occur in two ways: {English, Not English} and {Not English, English}).

Answer: D.

Hope it's clear.
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Re: The membership of a committee consists of 3 English teachers  [#permalink]

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New post 13 Oct 2019, 07:19
udaymathapati wrote:
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


total = 12
so P of English 3c2 ; 3
2/9*1/8*3 = 1/12
IMO D
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Re: The membership of a committee consists of 3 English teachers   [#permalink] 13 Oct 2019, 07:19
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