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by ketkoag » Wed Jun 10, 2009 9:27 am
What is the total number of some green balls and yellow balls?
1) The probability that a ball randomly selected is green is 1/10
2) One green ball picked out at fist, the probability that another ball randomly selected is green is 3/8.

[spoiler]please confirm the answer, IMO : E[/spoiler]
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by Musiq » Wed Jun 10, 2009 9:32 am
Ketkoag,

May I ask where you got this question from....it's not very GMAT-like.

The answer is E, because there could be other colors of balls present...no where are we told that the only two kinds of balls are Green and Yellow.
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by ankit1383 » Mon Aug 10, 2009 6:06 am
IMO...C
can someone confirm....(If we assume there are only two types of ball)
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by aspiregmat » Mon Aug 10, 2009 6:39 am
If no other information given about other color balls its definately E.

Lets assume that there are only green and yellow balls:

1) Probability of picking a green ball = 1/10 so only thing we know is green:yellow = 1:9

2) Probability of picking a green ball in second try = 3/8 and 1 green is already out so green:yellow = 4:5 or 7:9 and so on...

combining both == again only ratios which are also not matching...

So IMO E even if we assume only green and yellow balls.
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by ankit1383 » Mon Aug 10, 2009 6:55 am
Let the ratio be x:9x from statement 1

From second statement
(x/10x)* (x-1)/(10x-1)=3/8 (since both times green ball selected)

we can solve this...

where m i getting ths wrong
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by aspiregmat » Mon Aug 10, 2009 7:16 am
2) One green ball picked out at fist, the probability that another ball randomly selected is green is 3/8.

It does not say that the probability of picking two green balls is 3/8. It gives the probability of only the second ball.

Even if we take that.... the green and yellow balls will come in fraction which doesn't looks logical. IMO though not sure...
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