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gidimba
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GMat Prep 07 Jun 2006, 08:53
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Help - stumped!!!



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shobhitb
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GMat Prep 07 Jun 2006, 09:47
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gidimba
Help - stumped!!!
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E

P(White U Even) = P (White) + P(Even) - P (White and Even)

Statement 1 gives P(White and Even) = 0

But we do not know P(White), P(Even) hence insufficient

Statement 2 gives P(White)-P(Even) = 0.2
but no idea of P(White and Even) - Hence insufficient

Together

Still no idea of P(white) + P(Even)

Hence E
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jaynayak
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GMat Prep 09 Jun 2006, 03:36
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Will go with C.

P U B = P(A) + P(B) - P(A&B)

1) gives us P(a&b) = 0
2) lets say there are w white balls. And there are total 25 balls.
therfore

P(a) + P(b) = 2/10
W/25 + 5/25 = 2/10

w = 10

Hence P U B = 10/25 + 5/25 - 0

C is sufficient
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shobhitb
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GMat Prep 09 Jun 2006, 03:47
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jaynayak
Will go with C.

P U B = P(A) + P(B) - P(A&B)

1) gives us P(a&b) = 0
2) lets say there are w white balls. And there are total 25 balls.
therfore

P(a) + P(b) = 2/10
W/25 + 5/25 = 2/10 (the question said minus and where did the five come from?)

w = 10

Hence P U B = 10/25 + 5/25 - 0

C is sufficient
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Jaynayak - I am stumped!. Could you please explain in detail how u got the figures in bold. Thanks in advance
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mendiratta_1812
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GMat Prep 09 Jun 2006, 04:24
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Quote:
P(a) + P(b) = 2/10
W/25 + 5/25 = 2/10
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Pls explain this part.
In equation (2), it's the subtraction of probabilities not the sum. So how can you equate the sum to .2 ?
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haas_mba07
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GMat Prep 09 Jun 2006, 04:53
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Going with C for this,

2 gives P(W) - P(E) - 0.2 => # of white = 10

1 gives P(W&Even) = 0 => No white with even

P(W OR Even) = P(W) + P(E) - P(W & Even)

We know all, so can compute.
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GMat Prep 09 Jun 2006, 04:59
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An afterthought...

the trick may be computing the P(Even) which can be computed if the assumption is that the balls (which are 25) are assigned numbers 1 through 10, uniquely or duplicated i.e. more than one ball can have an even or odd number.

On second thoughts, it would seem likely that as the number of balls > 10, the numbers would have to repeat.

In which case the assignment could be in any manner.

If we assume that the numbers are unique i.e. non-repeating then all the balls would not have numbers (which is not the case as per the problem).

So, in summary, changing this to E.
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tl372
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GMat Prep 09 Jun 2006, 12:01
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shobhitb
gidimba
Help - stumped!!!

E

P(White U Even) = P (White) + P(Even) - P (White and Even)

Statement 1 gives P(White and Even) = 0

But we do not know P(White), P(Even) hence insufficient

Statement 2 gives P(White)-P(Even) = 0.2
but no idea of P(White and Even) - Hence insufficient

Together

Still no idea of P(white) + P(Even)

Hence E
Show more


Got E doing it the same way as shobhitb.

P(white) and P(Even) could be anything. Also, since every ball has a number painted on it, there must be repeats. However, we have no idea what numbers are repeated, odds or evens.



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Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
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