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# ds probs

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VP
Joined: 22 Nov 2007
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01 Mar 2008, 23:25
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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.
Director
Joined: 18 Feb 2008
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02 Mar 2008, 03:09
i think E.
Need two condition to determine the question.
1. probability of both white and even
2. probability of white plus probability of ball.

you use 2 minus 1
VP
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18 Mar 2008, 09:59
billyjeans wrote:
i think E.
Need two condition to determine the question.
1. probability of both white and even
2. probability of white plus probability of ball.

you use 2 minus 1

OA is E. other explanations?
CEO
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18 Mar 2008, 11:43
construct examples:

Pw - the probability that the ball will be white
Pe - the probability that the ball have an even number painted on it
Pwe - the probability that the ball will both be white and have an even number painted on it
Pw|e - the probability that the ball selected will either be white or have an even number painted on it

both Pw=0.3, Pe=0.1 and Pw=0.5, Pe=0.3 satisfy two conditions but

Pw=0.3, Pe=0.1: Pw|e = Pw + Pe = 0.4
Pw=0.5, Pe=0.3: Pw|e = Pw + Pe = 0.8
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SVP
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18 Mar 2008, 18:31
i get e as well.

p(white or even) = p(white)+p(even)-p(white and even)

from stat 1, p(white and even) = 0, but we know nothing about p(white) or p(even). insuff.

from stat 2, p(white)-p(even)=0.2, but nothing about p(even and white). insuff.

together, all we know is that p(white or even) = p(even)+0.2+p(even). insuff.
Director
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20 Mar 2008, 19:28
As per question we need white OR even (we DON'T want white AND even - we have to be able to strip out those that fall into both categories). To calculate, can either do:
a) probability of "white and odd" + probability of "even and not white"
b) probability of white + probability of even - probability of white & even

Statetment 1:
Tells us there aren't any that are both white and even. This doesn't tell us how many are white or how many are even. Insufficient. Eliminate A and D.

Statement 2:
Tells us that Pwhite - Peven = 0.2. So, Pwhite could be 0.4 which would make Peven 0.2. Or Pwhite could be 0.3 which would make Peven 0.1. And (by itself) it doesn't tell me Prob of even & white, which I'd need to subtract, so... insufficient in many ways. Eliminate B.

Combining both statement
we know that Peven+white = 0. BUT, we still have multiple possibilities for Pwhite and Peven (see above). 0.4+0.2-0=0.6. 0.3+0.1-0=0.4. ?? Still insufficient.

Re: ds probs   [#permalink] 20 Mar 2008, 19:28
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