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A stack of cards

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A stack of cards

by alex.gellatly » Thu Jul 12, 2012 5:40 am
In a stack of cards, 9 cards are blue and the rest are red. If 2 cards are to be chosen at random from the stack without replacement, the probability that the cards chosen will both be blue is 6/11. What is the number of cards in the stack?

10
11
12
15
18

I solved this correctly by plugging in answer choices. I was lucky with my first choice. However, I do not like relying on plug and chug. Can anyone help me with an algebraic approach?
Thanks
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by Birottam Dutta » Thu Jul 12, 2012 6:03 am
Let total number of cards be X.

Probability that the first is blue is 9/X.

Probability that the second is blue is 8/(X-1).

Probability that the first 2 are blue is (9/X)* 8/(X-1) = 6/11 (as per the question)

So, on solving we get, x(X-1) = 132 => x^2 - x - 132 = 0 => (x-12) (x+11) = 0.

This gives two values of x 12 and -11. As x cannot be negative, answer is 12.

Hence, C!
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by GMATGuruNY » Thu Jul 12, 2012 7:27 am
alex.gellatly wrote:In a stack of cards, 9 cards are blue and the rest are red. If 2 cards are to be chosen at random from the stack without replacement, the probability that the cards chosen will both be blue is 6/11. What is the number of cards in the stack?

10
11
12
15
18

I solved this correctly by plugging in answer choices. I was lucky with my first choice. However, I do not like relying on plug and chug. Can anyone help me with an algebraic approach?
Thanks
Let T = the total number of cards.

Since 9 of the cards are blue, P(BB) = 9/T * 8/(T-1) = 72/(T(T-1)).
Thus:
6/11 = 72/(T(T-1))
T(T-1) = (72*11)/6 = 12*11.
Thus, T = 12.

The correct answer is C.
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