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

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

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04 Apr 2007, 08:12
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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 from the box, what is the probability 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 zero.

2. The probability that the ball will be white minus the probability that the ball have an even number painted on it is 0.2
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04 Apr 2007, 17:04
1. P(A & B) =0

2. P(A) - P(B)=0.2

We cannot find P (A or B) which is equal to P(A) + P(B) {since P(A & B) =0}

Is it E?
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04 Apr 2007, 23:04
Let A = ball selected is white
Let B = ball selected is even

1. The probability that the ball will both be white and have an even number painted on it is zero.

Insuff. We don't know how many odd numbered white balls there are.

2. The probability that the ball will be white minus the probability that the ball have an even number painted on it is 0.2

Insuff. 0.2 can be the results of 0.5 - 0.3, 0.4 - 0.2, etc.

1+2
From 1 we know that P(AnB) = 0
From 2 we know that P(A)-P(B) = 0.2

We need to know P(AuB) which is equal to P(A)+P(B)-P(AnB).
We only know P(AnB) and the difference of P(A) and P(B) where P(A) > P(B). We don't know P(A) and P(B) for sure. So Insuff.

Is the OA E?
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08 Apr 2007, 09:25
Guys in this question

from (I) we have the probablity of white or a even number are indepedent events right ? which means

P(White)*P(Even) = 0

Am I thinking right?
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08 Apr 2007, 11:46
trivikram wrote:
Guys in this question

from (I) we have the probablity of white or a even number are indepedent events right ? which means

P(White)*P(Even) = 0

Am I thinking right?

No. you are wrong in my opinion. What 1) says is that P(White and Even) is 0. That doesnt meen that events are independent.

P(White)*P(Even) = 0 <=> events are independent.

What do others thinks?
VP
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08 Apr 2007, 12:01
OK

So from (I) we have P(W and E) = 0

So p(PUE) = P(E)+P(W)-P(EandW)

Since we do not know P(E) and P(W)

its E

Am I correct?
08 Apr 2007, 12:01
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