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25 Jul 2006, 13:16
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Who can explain me this q?
Thanks!
Is ok if a use this concept:
P(W or E)=P(W)+P(E)-P(W and E)????
Thanks!
Is ok if a use this concept:
P(W or E)=P(W)+P(E)-P(W and E)????
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Data Sufficiency (DS) Forum
Still interested in this question? Check out the "Best Topics" block below for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
25 Jul 2006, 13:24
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E
P(W OR E) = P(W) + P(E) - P(W AND E)
we have to find P(W OR E).
St1: P(W AND E) = 0 then P(W OR E) = P(W) + P(E): INSUFF
St2: P(W)-P(E) = 0.2 i.e P(W)= P(E) + 0.2: Clearly INSUFF
Combined:
We come up with
P(W OR E) = 2P(E) + 0.2: INSUFF
P(W OR E) = P(W) + P(E) - P(W AND E)
we have to find P(W OR E).
St1: P(W AND E) = 0 then P(W OR E) = P(W) + P(E): INSUFF
St2: P(W)-P(E) = 0.2 i.e P(W)= P(E) + 0.2: Clearly INSUFF
Combined:
We come up with
P(W OR E) = 2P(E) + 0.2: INSUFF
25 Jul 2006, 13:25
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Given Total number of balls = 25
Some of them are Red/Blue/White
The balls are numbered 1 to 10. Note that it is not clear what the numbering is, but only thing that you can deduce is that the numbers are replicated. There is no way of knowing if the numbering is replicated evenly across all colors or a specific color has all the duplicates.
S1: P(White & Even) = 0
This means no White ball has even numbers.
However, it doesn't give us any info about P(White or Even) = P(White) + P(Even) - P(White & Even)
= P(White) + P(Even)
Not sufficient.
S2: Given P(White) - P(Even) = 0.2
With just this information nothing can be deduced.
Not sufficient.
S1 & S2:
P(White or Even) = P(White) + P(Even)
0.2 = P(White) - P(Even)
P(White or Even) +0.2 = 2P(White)
Can't do much with this either.
Not sufficient.
Answer: E
Some of them are Red/Blue/White
The balls are numbered 1 to 10. Note that it is not clear what the numbering is, but only thing that you can deduce is that the numbers are replicated. There is no way of knowing if the numbering is replicated evenly across all colors or a specific color has all the duplicates.
S1: P(White & Even) = 0
This means no White ball has even numbers.
However, it doesn't give us any info about P(White or Even) = P(White) + P(Even) - P(White & Even)
= P(White) + P(Even)
Not sufficient.
S2: Given P(White) - P(Even) = 0.2
With just this information nothing can be deduced.
Not sufficient.
S1 & S2:
P(White or Even) = P(White) + P(Even)
0.2 = P(White) - P(Even)
P(White or Even) +0.2 = 2P(White)
Can't do much with this either.
Not sufficient.
Answer: E
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Data Sufficiency (DS) Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
