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10 members of a society, including members A and B, attended a me

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Math Revolution GMAT Instructor
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10 members of a society, including members A and B, attended a me  [#permalink]

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New post 06 Mar 2018, 02:58
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A
B
C
D
E

Difficulty:

  15% (low)

Question Stats:

77% (01:25) correct 23% (01:40) wrong based on 84 sessions

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[GMAT math practice question]

10 members of a society, including members A and B, attended a meeting. 2 of these members were selected to form a committee. What is the probability that the 2 members selected were members A and B?

A. 1/10
B. 1/15
C. 1/20
D. 1/30
E. 1/45

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Re: 10 members of a society, including members A and B, attended a me  [#permalink]

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New post 06 Mar 2018, 07:06
Top Contributor
MathRevolution wrote:
[GMAT math practice question]

10 members of a society, including members A and B, attended a meeting. 2 of these members were selected to form a committee. What is the probability that the 2 members selected were members A and B?

A. 1/10
B. 1/15
C. 1/20
D. 1/30
E. 1/45


Let's apply some probability rules

P(A and B both selected) = P(1st selection is one member of the pair AND 2nd selection is the other member of the pair)
= P(1st selection is one one member of the pair) x P(2nd selection is the other member of the pair)
= 2/10 x 1/9
= 1/45

Answer: E

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Re: 10 members of a society, including members A and B, attended a me  [#permalink]

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New post 06 Mar 2018, 07:18
MathRevolution wrote:
[GMAT math practice question]

10 members of a society, including members A and B, attended a meeting. 2 of these members were selected to form a committee. What is the probability that the 2 members selected were members A and B?

A. 1/10
B. 1/15
C. 1/20
D. 1/30
E. 1/45


Favourable ways of choosing committee = 1 (A and B both)

Total ways of selecting a 2 member committee out of 10 members = 10C2 = 45

Probability = 1/45

answer: Option E
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Re: 10 members of a society, including members A and B, attended a me  [#permalink]

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New post 08 Mar 2018, 01:25
=>

The number of committees of 2 people that include members A and B is equal to the number of ways that 2 people can be selected from 2 people, which is 2C2 = 1.
The number different possible committees of 2 people, is equal to the number of ways that 2 people can be selected from 10 people, which is 10C2 = \(\frac{(10*9)}{(1*2)} = 45.\)
The probability that the committee contains members A and B is 2C2 / 10C2 = \(\frac{1}{45}\).

Therefore, the answer is E.

Answer: E
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Re: 10 members of a society, including members A and B, attended a me  [#permalink]

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New post 20 Mar 2018, 16:23
MathRevolution wrote:
[GMAT math practice question]

10 members of a society, including members A and B, attended a meeting. 2 of these members were selected to form a committee. What is the probability that the 2 members selected were members A and B?

A. 1/10
B. 1/15
C. 1/20
D. 1/30
E. 1/45


Probability of choosing both A and B = \(\frac{2}{10} * \frac{1}{9} = \frac{1}{45}\)

The first part of the equation is \(\frac{2}{10}\) because it can be either A or B, while the 2nd is \(\frac{1}{9}\) because it's the only remaining member
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Re: 10 members of a society, including members A and B, attended a me &nbs [#permalink] 20 Mar 2018, 16:23
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