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

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

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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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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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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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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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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

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=>

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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