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Each of 25 balls in a box are either blue, red, white and

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Each of 25 balls in a box are either blue, red, white and [#permalink] New post 10 Oct 2005, 08:10
Each of 25 balls in a box are either blue, red, white and has a # from 1-10 on it. If one ball is selected at random, what is the probability that it will either be white OR have an even # on it?

1. The probability that it will be both white and even is 0
2. the probability that it will be white minus the probability that it will be even is .2

Please explain your answer. thx!
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 [#permalink] New post 10 Oct 2005, 08:24
Yikes! Is the answer to this question possible?

Is this a Data Sufficiency question?

(I think I've forgotten all my probability background. Will have to go searching for some site that covers it. Anybody have a nice site?)
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 [#permalink] New post 10 Oct 2005, 08:26
I will chose E.

n = 25

p(W or Even) = P(W) + P(Even) - P (W and Even)

(1) P(W and Even) = 0
as we dont know indivisual probabilities. Insufficient.

(2) P(W) - p(Even) = 0.2
Insufficient.

together, insufficient as we still cannot find P(w) + p(even) to get P(W or Even)
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Correct...but [#permalink] New post 10 Oct 2005, 08:40
E is the correct answer but I had C

if the prob. of getting white and even is 0, this means that either the prob. of getting white OR the prob. of getting even is 0 since it's one * the other.
right?

Going with this, move on to B. If white minues even is .2 and we know one of them has a probability of 0, then why can't we say that the probabiliity of either is .2?
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 [#permalink] New post 10 Oct 2005, 09:00
I got (E)

I. Here you are only dealing with prob white * prob even white. This says there are either no white balls or no even white balls. We're looking for either white OR even of any color. All the other balls might be even, or all but 1 etc.

II. The prob white - Prob even could work for many instances. With out knowing anything about the amounts of the other balls.

Combining them does not help at all.

B.
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 [#permalink] New post 10 Oct 2005, 09:12
Still not convinced why it can't be C!
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 [#permalink] New post 10 Oct 2005, 09:34
Because you are falling for the trap in I. it is simply stating there are no white + even balls. It has nothing to do with the overall probability of choosing an even numbered ball.

There is a global probability of choosing even or odd just as there is a global probability of choosing each color. But within each color group there is a further probability of choosing even or odd.

We are concerned with the global probability of choosing even and I. just gives us the % of evens in the white group, what if all the blues were even and 2/3 the reds even too?

make sense?

B.
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Re: Correct...but [#permalink] New post 10 Oct 2005, 10:20
Jennif102 wrote:
E is the correct answer but I had C

if the prob. of getting white and even is 0, this means that either the prob. of getting white OR the prob. of getting even is 0 since it's one * the other.
right?



Your are assuming:
P(A and B) = P(A) * P(B)

This is ONLY true iff both A & B are Independent. Otherwise, it would be:
P(A and B) = P (A) * P(B/A) = P(B) * P(A/B)

from the question stem we do not know both A and B are independent.
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got it! [#permalink] New post 10 Oct 2005, 10:54
great explanation guys! I understand now. Thanks!
got it!   [#permalink] 10 Oct 2005, 10:54
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