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

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Max@Math Revolution: Elite Legendary Member; Posts: 3991; Joined: Fri Jul 24, 2015 2:28 amFollowed by: 37 members; Location: Las Vegas, USA; Thanked: 19 times

by Max@Math Revolution » Tue Mar 06, 2018 2:54 am

[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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Vincen: Legendary Member; Posts: 2898; Joined: Thu Sep 07, 2017 2:49 pmFollowed by: 5 members; Thanked: 6 times

by Vincen » Tue Mar 06, 2018 3:22 am

I am confused here.

The probability that A was selected is 1/10, and the probability that B was selected is 1 /9. Hence, the answer should be $$\frac{1}{10}\cdot\frac{1}{9}=\frac{1}{90}.\ $$ Am I wrong?

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regor60: Master | Next Rank: 500 Posts; Posts: 416; Joined: Thu Oct 15, 2009 11:52 am; Thanked: 27 times

by regor60 » Tue Mar 06, 2018 6:37 am

Vincen wrote:I am confused here.

The probability that A was selected is 1/10, and the probability that B was selected is 1 /9. Hence, the answer should be $$\frac{1}{10}\cdot\frac{1}{9}=\frac{1}{90}.\ $$ Am I wrong?

Yes...and the probability is the same for B being selected first and A second

Order doesn't matter, just interested in A and B being selected as a pair, not which one is selected first or second

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Tue Mar 06, 2018 6:46 am

Max@Math Revolution 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

P(A or B is the first person selected) = 2/10. (Of the 10 people, 2 are A or B.)
P(A or B is the second person selected) = 1/9. (Of the 9 remaining people, 1 is A or B).
Since we want both events to happen, we MULTIPLY the probabilities:
(2/10)(1/9) = 1/45.

The correct answer is E.

Alternate approach:

From the 10 people, the number of ways to select a pair = 10C2 = (10*9)/(2*1) = 45.
Of these 45 pairs, only one -- AB -- constitutes a favorable outcome.
Thus:
P(AB is the selected pair) = 1/45.

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » Tue Mar 06, 2018 7:04 am

Max@Math Revolution 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

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

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Max@Math Revolution: Elite Legendary Member; Posts: 3991; Joined: Fri Jul 24, 2015 2:28 amFollowed by: 37 members; Location: Las Vegas, USA; Thanked: 19 times

by Max@Math Revolution » Thu Mar 08, 2018 1:18 am

=>

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 = (10*9) / (1*2) = 45.
The probability that the committee contains members A and B is 2C2 / 10C2 = 1/45.

Therefore, the answer is E.

Answer: E

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