hi
can someone help me to develop this DS question so I can go on on my prep test. this is my weak point.
thank you
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We don't know how many balls are white or how many or even
Case A:
=====
Both events or mutually exlusive - Means none of the White balls and Even - That is why probability is 0
P(W intersect E) = 0
No enough info to get result
Case B:
========
P(W) - P(E) = 0.2
P(W) + P(E) != 1 because we have other colors
P(E) can be calculated only if you know how many even balls or how many whites - No info
P (W or E) = P(W) + P(E) - P(W and E)
A & B together does not give enough info
and hence E
Case A:
=====
Both events or mutually exlusive - Means none of the White balls and Even - That is why probability is 0
P(W intersect E) = 0
No enough info to get result
Case B:
========
P(W) - P(E) = 0.2
P(W) + P(E) != 1 because we have other colors
P(E) can be calculated only if you know how many even balls or how many whites - No info
P (W or E) = P(W) + P(E) - P(W and E)
A & B together does not give enough info
and hence E
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gmat009
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Just confused with small thing.gace780 wrote:We don't know how many balls are white or how many or even
Case A:
=====
Both events or mutually exlusive - Means none of the White balls and Even - That is why probability is 0
P(W intersect E) = 0
No enough info to get result
Case B:
========
P(W) - P(E) = 0.2
P(W) + P(E) != 1 because we have other colors
P(E) can be calculated only if you know how many even balls or how many whites - No info
P (W or E) = P(W) + P(E) - P(W and E)
A & B together does not give enough info
and hence E
Why are we saying we cannot calculate even number of balls.
P(Even balls)= 5/10 [2,4,6,8,10].
Plz. tell me where I am wrong
Please le3t me know if i am wrong.
from the second statement .
p(w) - p(E) = 0.2
and from the 1st stement p(w)p(E)=0.
(p(W) + p(E))^2 =0.04
this gives p(W)^2 + p(E)^2 ..
(p(W)+ p(E))^2 = p(W)^2 + p(E)^2..
SO IMO C.
from the second statement .
p(w) - p(E) = 0.2
and from the 1st stement p(w)p(E)=0.
(p(W) + p(E))^2 =0.04
this gives p(W)^2 + p(E)^2 ..
(p(W)+ p(E))^2 = p(W)^2 + p(E)^2..
SO IMO C.
the answer is E....I chose C when I took the test.
Just for reference, can someone please explain what the opposite is for the probability of getting a white ball or an even number as the question asks?
Isn't it P(not white) + Prob (not even)???
Also, I thought the prob of A or B is just PA + PB...if not, when is this the case?
Thanks!
Just for reference, can someone please explain what the opposite is for the probability of getting a white ball or an even number as the question asks?
Isn't it P(not white) + Prob (not even)???
Also, I thought the prob of A or B is just PA + PB...if not, when is this the case?
Thanks!
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EricLien9122
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- Posts: 106
- Joined: Thu May 22, 2008 10:50 am
- Thanked: 3 times
Q: P(w)+P(E)=?
1. P(w)*P(E)=0
Insufficient, b/c we can't manipulate the equation into P(W)+P(E)
2. P(W)-P(E)=0.2
Insufficient, b/c we can't manipulate the question in to P(W)+P(E)
1+2:
Still insufficient, b/c we can't manipulate the equation into P(E)+P(E).
1. P(w)*P(E)=0
Insufficient, b/c we can't manipulate the equation into P(W)+P(E)
2. P(W)-P(E)=0.2
Insufficient, b/c we can't manipulate the question in to P(W)+P(E)
1+2:
Still insufficient, b/c we can't manipulate the equation into P(E)+P(E).
