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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, 03:04
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Each of 25 balls in a certain box is either red, blue, or white and has a number from 110 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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Required prob= P(wUe)=Pw+pEP(w&e)
From A) last part of the equation is 0, so A is insuff
From B) PwPe=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 110 and are 25 so we have 12 evennumbered balls, so C is the ans



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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 110, 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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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 110, 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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Well, from 1 to 105E 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



Director
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No , this may not be true. It is an assumption.



Current Student
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yessuresh wrote: Its E.
Are you sure about that? Can you show your work, please?



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Re: DS: GMATPrep probability [#permalink]
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30 Aug 2006, 19:04
GMATT73 wrote: Each of 25 balls in a certain box is either red, blue, or white and has a number from 110 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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Re: DS: GMATPrep probability [#permalink]
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31 Aug 2006, 06:24
GMATT73 wrote: Each of 25 balls in a certain box is either red, blue, or white and has a number from 110 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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Re: DS: GMATPrep probability [#permalink]
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31 Aug 2006, 06:29
apollo168 wrote: GMATT73 wrote: Each of 25 balls in a certain box is either red, blue, or white and has a number from 110 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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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



Current Student
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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 bball is exciting to watch!



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got E as well. now what is chinese division???



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I feel the answer is C [#permalink]
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10 Jul 2007, 23:09
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.
Where am i going wrong ?? Please help..




I feel the answer is C
[#permalink]
10 Jul 2007, 23:09






