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Marbles

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by crackgmat007 » Sun Jul 19, 2009 7:32 pm
A jar contains 8 red marbles and y white marbles. If Joan takes 2 random marbles from the jar, is it more likely that she will have 2 red marbles than that she will have one marble of each color?

(1) y ≤ 8
(2) y ≥ 4

OA - B Pls explain
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Source: — Data Sufficiency |
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Re: Marbles

by aj5105 » Sun Jul 19, 2009 10:33 pm
crackgmat007 wrote:A jar contains 8 red marbles and y white marbles. If Joan takes 2 random marbles from the jar, is it more likely that she will have 2 red marbles than that she will have one marble of each color?

(1) y ≤ 8
(2) y ≥ 4

OA - B Pls explain
pick 2 marbles out of 8. 8C2 = 28.

pick 1 red, 1 white. 8C1 * y C1 (where y is the total white marbles)

you want to find out which of the two is big.

(B) gives the answer. even you take y = 4, 8*4 = 32, which is greater than 28. therefore any value greater than 4 gives a number greater than 32.
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by crackgmat007 » Mon Jul 20, 2009 9:48 am
I thought this problem is about probability because it says 'which one is more likely'. Let me know what you think.
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by Domnu » Mon Jul 20, 2009 12:48 pm
This is a probability problem. Here's how I did it:

The probability of picking 2 red marbles is

56/[(8+y)(7+y)]

Here's how:

There are nCr(8, 2) ways to pick 2 red marbles, and nCr(8+y, 2) ways to choose marbles in general. Now, nCr(8,2) = 8!/(2!6!) = 56, and nCr(8+y,2) = (7+y)(8+y)/2, so we get the above.

Similarly, for choosing 1 red and 1 white marble, we have 8 ways of choosing a red marble and y ways of choosing a white marble, so our probability is

8y/nCr(8+y,2) = 16y/(8+y)(7+y)

We need to be able to definitely compare

16y/(8+y)(7+y) and 56/(8+y)(7+y)

Multiplying both out by denominators, we can just compare 16y and 56. If y <= 8, this doesn't give us anything... y could be 4, in which case 16y > 56, or y could be 2 in which case 16y < 56. So, cross off A. However, looking at B, if y >= 4, 16y > 56 always. So, our answer is B.
Have you wondered how you could have found such a treasure? -T
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by raghavsarathy » Mon Jul 20, 2009 5:05 pm
IMO - E

1. y <=8

The probability will change based on the value of y. If y=8 then the probability of having bith read marbles will be less and this probability will increase as the vlaue of y reduces.

However we cannot determine the value of this probability since y is not known

2. y>=4
Again here , the probability of choosing both red will be maximum when y=4 and will keep reducing as we increase the vlaue of y

Since value of y is not known we cannot determiine


Even using both statements y can have values from 4 to 8. Not sufficient

For me a sufficient soln would be something like

y<=7

Since we know the maximum value of the number of white marbles.

I am still not able to comprehend the OA !
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