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# Each of the 25 balls in a certain box is either red , blue

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Manager
Joined: 14 Dec 2008
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Each of the 25 balls in a certain box is either red , blue [#permalink]

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03 Sep 2009, 10:07
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Question Stats:

100% (02:23) correct 0% (00:00) wrong based on 1 sessions

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Each of the 25 balls in a certain box is either red , blue or white and has a number from 1 to 10 painted on it. if one ball is to be selected at randon from the box , what is the probalility that the ball selected will either be white or have an even number painted on it.

1. the probability that the ball will both be white and have an even number painted on it is 0.
2. thet probability that the ball will both be white minus that the ball will have an even number painted on it is 0.2.

Kudos [?]: 30 [0], given: 39

Current Student
Joined: 12 Jun 2009
Posts: 1837

Kudos [?]: 270 [0], given: 52

Location: United States (NC)
Concentration: Strategy, Finance
Schools: UNC (Kenan-Flagler) - Class of 2013
GMAT 1: 720 Q49 V39
WE: Programming (Computer Software)

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03 Sep 2009, 11:26
manojgmat wrote:
Each of the 25 balls in a certain box is either red , blue or white and has a number from 1 to 10 painted on it. if one ball is to be selected at randon from the box , what is the probalility that the ball selected will either be white or have an even number painted on it.

1. the probability that the ball will both be white and have an even number painted on it is 0.
2. thet probability that the ball will both be white minus that the ball will have an even number painted on it is 0.2.

So we need to know the probability of a white ball drawn and the probability of an even number drawn.

1. so that's telling us there are NO EVEN WHITE BALLS. INSUFF since we dont know the even balls or white balls numbers
2. ProbW - ProbEven = .2
we dont know the probability of even balls OR white ball drawn so 2 unknowns INSUFF
Combined:
No even white balls doesnt help use derive ProbW or ProbEven so INSUFF

E.
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Kudos [?]: 270 [0], given: 52

Manager
Joined: 10 Aug 2009
Posts: 129

Kudos [?]: 74 [0], given: 10

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03 Sep 2009, 13:49
P(white or Even)=P(white)+P(even)-P(white and even)

1) statement 1 says that events "white" and even" are mutually exclusive or P(white and even)=0
From here: P(white or even)= P(white)+P(even)
Not sufficient since we do not now P(white) and P(even)

2) statement 2
P(white)-P(even)=0.22
since we do not know P(white), P(even) and P(white and even)

1)+2)

we know:
P(white or even)= P(white)+P(even)
P(white)-P(even)=0.22
We can't solve since we need to know either P(even) or P(white)...

Kudos [?]: 74 [0], given: 10

Manager
Joined: 14 Dec 2008
Posts: 161

Kudos [?]: 30 [0], given: 39

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03 Sep 2009, 21:22
shaselai/LenaA tried well, i am still figuring out, will update with OA ,

By the way how u folks add 'Kudos' Button?

Kudos [?]: 30 [0], given: 39

Current Student
Joined: 12 Jun 2009
Posts: 1837

Kudos [?]: 270 [0], given: 52

Location: United States (NC)
Concentration: Strategy, Finance
Schools: UNC (Kenan-Flagler) - Class of 2013
GMAT 1: 720 Q49 V39
WE: Programming (Computer Software)

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04 Sep 2009, 07:00
manojgmat wrote:
shaselai/LenaA tried well, i am still figuring out, will update with OA ,

By the way how u folks add 'Kudos' Button?

just click on the +1kudos button on that person's reply
_________________

Kudos [?]: 270 [0], given: 52

Manager
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06 Sep 2009, 11:33
From statement 1 we can conclude that white ball and even number painted are two mutually exclusive events. Combing statement 1 and 2 we can come with probability for white ball but probability of ball with even number on it is still a question. So answer is E

Kudos [?]: 114 [0], given: 8

Re: DS - GMATPREP   [#permalink] 06 Sep 2009, 11:33
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