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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]  01 May 2008, 10:01
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 random 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 0.

2). The probability that the ball will be white minus the probability that have an eve number painted on it is 0.2
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Re: Math: Probability - 25 balls [#permalink]  01 May 2008, 12:37
3
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lexis 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 random 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 0.

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

I got E.

P(white) + P(even) - P(white&even) = ?

(1) is saying: P(white&even) = 0
INSUFFICIENT

(2) is saying: P(white) - P(even) = 0.2
We don't know P(white&even), INSUFFICIENT

Together, you have
P(white) - P(even) = 0.2
and want to find: P(white) + P(even)=?
cannot complete the calculation with information given.
INSUFFICIENT
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Re: Math: Probability - 25 balls [#permalink]  01 May 2008, 12:43
I agree that the answer is E. Here is my logic:

The question is asking for what is P(W) or P(E).

Statement 1 tells you that P(W) and P(E) is mutually exclusively. Thus P(W+E) = 0 So not enough info on its own.

Statement 2 tells you that P(W)-P(E) is 0.2. That is not sufficient either. In order to find P(W) or P(E), we need P(W) + P(E). However there is no information given concerning P(W) and P(E).

Please let me know if my analysis is not correct. Probability is not my strong suit.
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Re: Math: Probability - 25 balls [#permalink]  01 May 2008, 16:23
bkk145 wrote:
lexis 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 random 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 0.

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

I got E.

P(white) + P(even) - P(white&even) = ?

(1) is saying: P(white&even) = 0
INSUFFICIENT

(2) is saying: P(white) - P(even) = 0.2
We don't know P(white&even), INSUFFICIENT

Together, you have
P(white) - P(even) = 0.2
and want to find: P(white) + P(even)=?
cannot complete the calculation with information given.
INSUFFICIENT

same approach and answer for me
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Re: Math: Probability - 25 balls [#permalink]  01 May 2008, 18:22
bkk145 wrote:
lexis 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 random 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 0.

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

I got E.

P(white) + P(even) - P(white&even) = ?

(1) is saying: P(white&even) = 0
INSUFFICIENT

(2) is saying: P(white) - P(even) = 0.2
We don't know P(white&even), INSUFFICIENT

Together, you have
P(white) - P(even) = 0.2
and want to find: P(white) + P(even)=?
cannot complete the calculation with information given.
INSUFFICIENT

Intern
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Re: Math: Probability - 25 balls [#permalink]  01 May 2008, 20:24
I am new to this but this is a probablity Math question. There are 25 balls. 8 of each color with the probability of
3 even numbers per color. 3 divided by 8 is 0.375 divided by 25 is 0.015 which is .02.

But where did that other ball go.
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Re: Math: Probability - 25 balls [#permalink]  02 May 2008, 00:54
bkk145 wrote:
lexis 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 random 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 0.

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

I got E.

P(white) + P(even) - P(white&even) = ?

(1) is saying: P(white&even) = 0
INSUFFICIENT

(2) is saying: P(white) - P(even) = 0.2
We don't know P(white&even), INSUFFICIENT

Together, you have
P(white) - P(even) = 0.2
and want to find: P(white) + P(even)=?
cannot complete the calculation with information given.
INSUFFICIENT

I love this way too.

OA is E
Manager
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Re: Math: Probability - 25 balls [#permalink]  11 Jun 2011, 21:46
Try it , good question Answer is E
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Re: Math: Probability - 25 balls [#permalink]  09 Jul 2011, 01:08
Great explanation, thanks!
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Location: Pune, India
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Re: Math: Probability - 25 balls [#permalink]  17 Jul 2011, 20:51
Expert's post
linda577ford wrote:
I am new to this but this is a probablity Math question. There are 25 balls. 8 of each color with the probability of
3 even numbers per color. 3 divided by 8 is 0.375 divided by 25 is 0.015 which is .02.

But where did that other ball go.

It is not essential that there will be 8 balls of each color. Each ball is red, blue or white. Overall, we could have 20 red balls, 2 blue balls and 3 white balls or 10 red balls, 10 blue balls and 5 white balls or some other combination. The point is that it is not essential that there are an equal number of balls with the same color. Similarly, the numbers on the balls will also be random. Say 9 balls could have 1-9 written on them and the rest of the balls could have 10 written on them. So you cannot find the probability of selecting a white ball or an even numbered ball until and unless you have some other data.

We know that P(Even OR White) = P(Even) + P(White) - P(Even AND White)
Both statements together don't give us the value of P(Even OR White).
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Re: Math: Probability - 25 balls   [#permalink] 17 Jul 2011, 20:51
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