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something blue balls and 9 red balls

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something blue balls and 9 red balls

by jsl » Mon Oct 27, 2008 3:00 pm
A box contains some Blue balls and 9 Red balls. If the probability that two balls randomly removed without replacement from the box are Red is 6/11, what is the probability that the third ball randomly removed without replacement of the balls drawn is a blue ball?

3/11
9/55
3/7
2/9
Cannot be Determined
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by rohangupta83 » Mon Oct 27, 2008 3:11 pm
there are 9 red balls
number of blue balls = x

therefore total number of balls = 9+x

probability that the first ball drawn is a red ball = 9/(9+x)

probability that the second ball is also red = 8/(8+x)

probability that the first two balls drawn are red = 6/11 (given)

Therefore, [9/(9+x)]*[8/(8+x)] = 6/11
OR
11*9*8=6*(9+x)(8+x)
11*9*8/6 = 72 +17x + x^2
OR
x^2 + 17x +72 - 132 = 0
x^2 + 17x - 60 = 0
x^2 +20x - 3x - 60 = 0
x(x+20) - 3(x+20) = 0
(x-3)(x+20) = 0

x = 3 or -20

x can't be negative therefore x = 3
i.e. number of blue balls = 3

Hence, the probability that the third ball is a blue ball is 3/(8+3)
or
3/11 (i think if my math was correct)

Hope this helps !!
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by Ian Stewart » Mon Oct 27, 2008 3:51 pm
The question, or at least the OA, really doesn't make any sense (and the same is true of some other questions from the same source). It and other questions of the same provenance were discussed at length here:

www.beatthegmat.com/need-help-t20237.html
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

ianstewartgmat.com
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by jsl » Tue Oct 28, 2008 4:38 am
OA is 9/55
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