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18 Aug 2006, 10:19
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Sorry messed up S1, I was going off of memory. Its fixed now.
There are a total of 25 balls and each ball can only be either red, white or blue. Also, each ball is numbered from 1-10. What is the probability of choosing a white ball OR an even ball?
S1) Probability of choosing a white ball that is even is 0.
S2) Probability of choosing a white ball minus the probability of choosing an even ball is 0.2.
There are a total of 25 balls and each ball can only be either red, white or blue. Also, each ball is numbered from 1-10. What is the probability of choosing a white ball OR an even ball?
S1) Probability of choosing a white ball that is even is 0.
S2) Probability of choosing a white ball minus the probability of choosing an even ball is 0.2.
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18 Aug 2006, 10:45
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Given T= 25, and balls are {R, W, B}
Success = Ball is white or ball is even
P(W or Even) = P(W) + P(E) - P(W and E)
S1: Given P(W & E) = 0
Not sufficient.
S2: P(W) - P(E) = 0.2
Still don't know P(W & E)
S1 & S2 :
P(W or E) = P(W) + P(E)
We know P(W) -P(E) = 0.2
Not sufficient to calculate P(W or E)
Answer: E
Success = Ball is white or ball is even
P(W or Even) = P(W) + P(E) - P(W and E)
S1: Given P(W & E) = 0
Not sufficient.
S2: P(W) - P(E) = 0.2
Still don't know P(W & E)
S1 & S2 :
P(W or E) = P(W) + P(E)
We know P(W) -P(E) = 0.2
Not sufficient to calculate P(W or E)
Answer: E
23 Aug 2006, 13:20
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agsfaltex
St I : is insufficent
St II :
let there be 'x' whte balls
then P(choosing white ball): x/25
further P(even balls)=12/25
so,
x/25 - 12/25 =.2
therefore : x=17
i feel st II i sufficient
Hence B)
please let me know if the logic is incrrect..
