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

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

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30 Aug 2006, 02:04
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Each of 25 balls in a certain box is either red, blue, or white and has a number from 1-10 painted on it. If one ball is to be selected at random from the box what is the probability that the selected ball will either be white or have an even number painted on it?

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

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

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Director
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30 Aug 2006, 02:24
Required prob= P(wUe)=Pw+pE-P(w&e)
From A) last part of the equation is 0, so A is insuff
From B) Pw-Pe=0,2 or Pw=0,2+Pe, but we have no info about P(w&e) So NOT suff
From both we get P(wUe)=0,2+Pe+Pe. The balls are numbeeren from 1-10 and are 25 so we have 12 even-numbered balls, so C is the ans

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30 Aug 2006, 02:49
Assumed they are 1,2,3,4 to 10. This may not be true, however. Eventually, the ans could be E) but it depends , afterall , on the questionmaker.
In this line of reasoning,the author could have used only the word"NUMBER" without explicitly stating that the numbers are from 1-10, going even furhter, we may say that the prob Pe=0 because NUMBERS may not be integers.
Is this reasoning , however, necessary?

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30 Aug 2006, 03:00
BG wrote:
Assumed they are 1,2,3,4 to 10. This may not be true, however. Eventually, the ans could be E) but it depends , afterall , on the questionmaker.
In this line of reasoning,the author could have used only the word"NUMBER" without explicitly stating that the numbers are from 1-10, going even furhter, we may say that the prob Pe=0 because NUMBERS may not be integers.
Is this reasoning , however, necessary?

I agree that they are 1,2,3,...,10. But how do we know how many are even?

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30 Aug 2006, 03:12
Well, from 1 to 10-5E and 5O, SInce balls are 25 then 2 times 1 to 10 and one time 1 to 5 which gives 12 E and 13 O in total. Then Pe=12/25

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30 Aug 2006, 04:20
No , this may not be true. It is an assumption.

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Current Student
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30 Aug 2006, 05:37
yessuresh wrote:
Its E.

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30 Aug 2006, 18:04
GMATT73 wrote:
Each of 25 balls in a certain box is either red, blue, or white and has a number from 1-10 painted on it. If one ball is to be selected at random from the box what is the probability that the selected ball will either be white or have an even number painted on it?

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

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

what is P(w) or p(e)
since these are compatible events: p(w) + P(e) - P(w and e)

1. p(w and e) = 0 independent events
but we don;t have any info on p(w) or p(e) so INSUFF

2. p(w) - p(e) = 0.2 or p(w) = 0.2 + p(e) still insuff since we don;t know either

combined: INSUFF

so E?

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31 Aug 2006, 05:24
GMATT73 wrote:
Each of 25 balls in a certain box is either red, blue, or white and has a number from 1-10 painted on it. If one ball is to be selected at random from the box what is the probability that the selected ball will either be white or have an even number painted on it?

Hi Matt got E as well whats the OA?

P(White) U P(Even)= P(white)+P(even)-P(white)(even)

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

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

Hi Matt got E as well whats the OA?

P(White) U P(Even)= P(white)+P(even)-P(white)P(even)

From 1 P(W) or P(E)= P(white) +P(even) insufficient

From 2 P(W)-P(E)=.2--> P(W)=P(E)+.2 insufficient

Combined we can't ascertain anything more E

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Current Student
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31 Aug 2006, 05:29
apollo168 wrote:
GMATT73 wrote:
Each of 25 balls in a certain box is either red, blue, or white and has a number from 1-10 painted on it. If one ball is to be selected at random from the box what is the probability that the selected ball will either be white or have an even number painted on it?

Hi Matt got E as well whats the OA?

P(White) U P(Even)= P(white)+P(even)-P(white)(even)

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

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

Hi Matt got E as well whats the OA?

P(White) U P(Even)= P(white)+P(even)-P(white)P(even)

From 1 P(W) or P(E)= P(white) +P(even) insufficient

From 2 P(W)-P(E)=.2--> P(W)=P(E)+.2 insufficient

Combined we can't ascertain anything more E

OA is (E) Apollo. Good work.

In addition to the binomial theourum and chinese division, I will add this formula to my notes: P(X) U P(Y)= P(X)+P(Y)-P(X)P(Y)

BTW: Did you figure out the trick for that Exponential remainder problem. The answer should be 0, just don't know the shortcut.

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31 Aug 2006, 05:42
With regard to the exponential problem, I emailed my math major friend the problem, and he gave me a curt reply saying that it is permissible to add the elements in an expression raised to an odd power. He didnt elaborate so I don't really know the underlying reason So D turns out to be the correct answer. I'll just take comfort knowing that that particularly concept won't be tested on the gmat.

Hey since your in Japan, are you a basketball fan by chance? The world championships is taking place there

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Current Student
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31 Aug 2006, 05:44
apollo168 wrote:
With regard to the exponential problem, I emailed my math major friend the problem, and he gave me a curt reply saying that it is permissible to add the elements in an expression raised to an odd power. He didnt elaborate so I don't really know the underlying reason So D turns out to be the correct answer. I'll just take comfort knowing that that particularly concept won't be tested on the gmat.

Hey since your in Japan, are you a basketball fan by chance? The world championships is taking place there

Not really bro, I'm a snowboarder... but b-ball is exciting to watch!

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Director
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06 Sep 2006, 17:25
got E as well. now what is chinese division???

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10 Jul 2007, 22:09
Here's how i worked it out.

P(a)xP(b)=0

P(a) - P(b) =0.2

Solving these two equations we get P(a)=0.2 and P(b) =0.

Hence P(a) + P(b) =0 + 0.2 = 0.2

Hence, OA is C.

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