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Probability Problem

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by gmat009 » Thu Oct 02, 2008 8:25 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?
A. 10
B. 11
C. 12
D. 15
E. 18

This is the way I did:
Blue cards= 9 Red cards=x Total cards=9+x
9C2/(9+x)C2 =6/11
Calculating this I am getting x=20,-3.
I am not sure my method is wrong or I am making calculation mistake.
OA is C
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by mental » Thu Oct 02, 2008 9:00 am
Let total number of cards be x
9 are blue

probability of two cards = 9/x*8/x-1 = 72/x(x-1)
this is given as 6/11

72/ x(x-1) = 6/11

x^2 -x -132=0

solving we get, x=12, x=-11

Ans 12.
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by mental » Thu Oct 02, 2008 9:03 am
your equation was correct, probably some sign mistake

(9C2)/(9+x)c2 = 9*8/(9+x)(8+x) = 6/11

132 = 72+17x+x^2

or x^2 + 17x -60 = 0

gives factors as (x +20)(x-3)=0
x=3, -20

total cards = 3+9=12
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by gmat009 » Thu Oct 02, 2008 9:07 am
mental wrote:your equation was correct, probably some sign mistake

(9C2)/(9+x)c2 = 9*8/(9+x)(8+x) = 6/11

132 = 72+17x+x^2

or x^2 + 17x -60 = 0

gives factors as (x +20)(x-3)=0
x=3, -20

total cards = 3+9=12
Thanks.........
I got x=3,-20 but forgot to add 9 and 3
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