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by arjunshn » Wed Apr 13, 2011 4:08 am
A bag contains 12 balls some of which are red and rest are blue. How many red balls are there in the bag?
1) The probability of randomly selecting two red balls from the bag is 1/11.
2) When two balls are selected randomly from the bag the probability of selecting one red and one blue ball is 9/22
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Source: — Data Sufficiency |
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by HSPA » Wed Apr 13, 2011 4:16 am
using 1 : 4/12 * 3/11 will do => 4 red

x, 12-x
(x/12 + 12-x-1/11) + 8/12 * 4/11.... got to go havent calculated this..

I guess D
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by arjunshn » Wed Apr 13, 2011 4:22 am
but hte answer is a, even i selected d
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by bubbliiiiiiii » Wed Apr 13, 2011 4:22 am
IMO D.

Let the number of red balls be x => blue are 12-x. we have to find x.

Option 1:
xc2/12c2=1/11 : Solve for X SUFFICIENT

Option 2:
(xc1*(12-x)c1)/12c2=9/22 : Solve for X SUFFICIENT
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by Testluv » Thu Apr 14, 2011 11:42 pm
bubbliiiiiiii wrote:IMO D.

Let the number of red balls be x => blue are 12-x. we have to find x.

Option 1:
xc2/12c2=1/11 : Solve for X SUFFICIENT

Option 2:
(xc1*(12-x)c1)/12c2=9/22 : Solve for X SUFFICIENT
Careful! Let's think about statement 2 again. Logically, the "x" in your equation could just as easily refer to the blue balls rather than red. Although we would have two integers whose sum was 12, we wouldn't know which integer referred to which color ball.

You can also see this algebraically. We know that xC1 is always x. So, if you sub in "x" for "xc1" and "12-x" for "(12-x)c1", you will end up with a quadratic equation, which of course will have two solutions.

The correct answer is A.
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by bubbliiiiiiii » Sat Apr 16, 2011 5:25 am
Testluv wrote:
bubbliiiiiiii wrote:IMO D.

Let the number of red balls be x => blue are 12-x. we have to find x.

Option 1:
xc2/12c2=1/11 : Solve for X SUFFICIENT

Option 2:
(xc1*(12-x)c1)/12c2=9/22 : Solve for X SUFFICIENT
Careful! Let's think about statement 2 again. Logically, the "x" in your equation could just as easily refer to the blue balls rather than red. Although we would have two integers whose sum was 12, we wouldn't know which integer referred to which color ball.

You can also see this algebraically. We know that xC1 is always x. So, if you sub in "x" for "xc1" and "12-x" for "(12-x)c1", you will end up with a quadratic equation, which of course will have two solutions.

The correct answer is A.
Thanks testluv a great concept shared. :)
Regards,

Pranay
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