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Probablity

Expert replies
Source: — Data Sufficiency |

Whats the OA

by prabhu3645 » Mon Apr 14, 2008 6:39 pm
Is the OA B. My logic is as follows.
P(W)= probablity that the ball is white.
Number of even balls under any circumstance, irrespective of the balls being white, red or blue is 12 and hence P(E) = 12/25. (lets say there are 10 red balls, 14 white and 1 blue ball or any other combination, there would only be 12 even numbered balls)
P(W) - P (E) = P(W) - 12/25 = 0.20

Thus P(W) is 17/25 and there are 17 white balls.

We need to find the probability that the ball is either a white or an even number which would be P(W) + P(E) - P(W+E) = 17/25 + 12/25 - 8/25 = 21/25

Hence B only should be sufficient to prove this
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No data given on number distribution.

by rdadbhawala » Mon Apr 14, 2008 9:01 pm
My answer is E.

We need P(E), P(W) and P(E & W) to know P(E U W).

Statement 1 only says that P(E & W) is zero. We can not deduce other quantities with that data.

Statement 2 says P(E) - P(W) = 0.2. But this does not give us any data on any of the required entities.

If we combine the 2 statements, we can say that we only need P(E) and P(W) to get the answer, but that data is still not available. You can try to make a tabulation with the information available: Red, Blue, White as column headers and Odd, Even as Row Headers. Fill the information available, and we find that we are not able to compute any data.

Also, one of the responses says that the number of even numbers is 12. But there is no information given on the distribution of the numbers, and hence I wouldn't assume that. It also says that "P(W+E) = 8/25". I don't know where that came from.
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by prabhu3645 » Tue Apr 15, 2008 5:50 am
I think what ever be the distribution amongst the tree colors, the number of even number balls will still be 12, it would not vary.
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The question only says that the balls have been numbered from 1 to 10. It doesn't say that the numbering is linear, each number is used (i.e. none of the numbers are missed), etc.

Suppose, there are 9 White, 9 Blue and 7 Red balls numbered as 1 to 9, 1 to 9, and 1 to 7 respectively. Here, we have 14 Odd numbers v/s 11 even numbers. And what if I am not using the even numbers at all; and there's no information given on repetition of numbers either.

I would say that you are over assuming; that's something I wouldn't do on a GMAT question.
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by prabhu3645 » Tue Apr 15, 2008 7:12 am
I think youve got a point. cheers mate
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