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Each of the 25 balls in a certain box is either red, blue or [#permalink]
 13 Aug 2010, 03:03
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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 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 the ball will have an even number painted on it is 0.2
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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 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? Probability ball: white - P(W); Probability ball: even - P(E); Probability ball: white and even - P(W&E). Probability ball picked being white or even: P(WorE)=P(W)+P(E)-P(W&E). (1) The probability that the ball will both be white and have an even number painted on it is 0 --> P(W&E)=0 (no white ball with even number) --> P(WorE)=P(W)+P(E)-0. Not sufficient (2) The probability that the ball will be white minus the probability that the ball will have an even number painted on it is 0.2 --> P(W)-P(E)=0.2, multiple values are possible for P(W) and P(E) (0.6 and 0.4 OR 0.4 and 0.2). Can not determine P(WorE). (1)+(2) P(W&E)=0 and P(W)-P(E)=0.2 --> P(WorE)=2P(E)+0.2 --> multiple answers are possible, for instance: if P(E)=0.4 (10 even balls) then P(WorE)=1 BUT if P(E)=0.2 (5 even balls) then P(WorE)=0.6. Not sufficient. Answer: E. Hope it's clear.
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But how can this be solved in less than 2 mins???
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jananijayakumar wrote: But how can this be solved in less than 2 mins??? You can solve this problem in another way. Transform probability into actual numbers and draw the table. Given: Attachment: 
1.JPG [ 8.64 KiB | Viewed 3110 times ]
So we are asked to calculate \frac{a+b-c}{25} (we are subtracting c not to count twice even balls which are white). (1) The probability that the ball will both be white and have an even number painted on it is 0 --> c=0 --> \frac{a+b}{25}=?. Not sufficient. Attachment: 
4.JPG [ 8.7 KiB | Viewed 3113 times ]
(2) The probability that the ball will be white minus the probability that the ball will have an even number painted on it is 0.2 --> \frac{white}{25}-\frac{even}{25}=0.2 --> white-even=25*0.2=5 --> a-b=5 --> b=a-5 --> \frac{a+a-5-c}{25}=?. Not sufficient. Attachment: 
2.JPG [ 8.68 KiB | Viewed 3109 times ]
(1)+(2) c=0 and b=a-5 --> \frac{a+a-5+0}{25}=\frac{2a-5}{25}. Not sufficient. Attachment: 
3.JPG [ 8.82 KiB | Viewed 3112 times ]
Answer: E.
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great explanation bunuel.
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Wonderful Bunuel.  Thanks so much!
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Bunuel 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?
Probability ball: white - P(W);
Probability ball: even - P(E);
Probability ball: white and even - P(W&E).
Probability ball picked being white or even: P(WorE)=P(W)+P(E)-P(W&E).
(1) The probability that the ball will both be white and have an even number painted on it is 0 --> P(W&E)=0 (no white ball with even number) --> P(WorE)=P(W)+P(E)-0. Not sufficient
(2) The probability that the ball will be white minus the probability that the ball will have an even number painted on it is 0.2 --> P(W)-P(E)=0.2, multiple values are possible for P(W) and P(E) (0.6 and 0.4 OR 0.4 and 0.2). Can not determine P(WorE).
(1)+(2) P(W&E)=0 and P(W)-P(E)=0.2 --> P(WorE)=2P(E)+0.2 --> multiple answers are possible, for instance: if P(E)=0.4 (10 even balls) then P(WorE)=1 BUT if P(E)=0.2 (5 even balls) then P(WorE)=0.6. Not sufficient.
Answer: E.
Hope it's clear. how did you get this? P(WorE)=2P(E)+0.2
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BlitzHN wrote: Bunuel 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?
Probability ball: white - P(W);
Probability ball: even - P(E);
Probability ball: white and even - P(W&E).
Probability ball picked being white or even: P(WorE)=P(W)+P(E)-P(W&E).
(1) The probability that the ball will both be white and have an even number painted on it is 0 --> P(W&E)=0 (no white ball with even number) --> P(WorE)=P(W)+P(E)-0. Not sufficient
(2) The probability that the ball will be white minus the probability that the ball will have an even number painted on it is 0.2 --> P(W)-P(E)=0.2, multiple values are possible for P(W) and P(E) (0.6 and 0.4 OR 0.4 and 0.2). Can not determine P(WorE).
(1)+(2) P(W&E)=0 and P(W)-P(E)=0.2 --> P(WorE)=2P(E)+0.2 --> multiple answers are possible, for instance: if P(E)=0.4 (10 even balls) then P(WorE)=1 BUT if P(E)=0.2 (5 even balls) then P(WorE)=0.6. Not sufficient.
Answer: E.
Hope it's clear. how did you get this? P(WorE)=2P(E)+0.2From (1) P(WorE)=P(W)+P(E)-0 --> P(WorE)=P(W)+P(E); From (2) P(W)-P(E)=0.2 --> P(W)=P(E)+0.2; Substitute P(W) --> P(WorE)=P(W)+P(E)=P(E)+0.2+P(E)=2P(E)+0.2.
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very cool.. thanks!
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Re: Each of the 25 balls in a certain box is either red, blue or [#permalink]
03 Mar 2012, 05:12
Hi
But can we not say that 12 out of the 25 balls were even ?
If we can then we already get the answer with only B !.
Thanks & Regards
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Re: Each of the 25 balls in a certain box is either red, blue or [#permalink]
03 Mar 2012, 05:48
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Re: Each of the 25 balls in a certain box is either red, blue or [#permalink]
03 Mar 2012, 07:32
Dont know why I have already put 1000 Dollars behind GMAT - lolz I calculated 12 even numbers as follows - first 10 balls can be painted with 5 even numbers viz - 2,4,6,8,10 likewise for next 10 - 2,4,6,8,10 and next 5 - 2,4, then from Statement B P(W) - P(E) = 0.2 So P (W) = 0.2 + P(E) P(E) = 12/25 P(W) = 0.2 + 12/25 
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Bunuel 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?
Probability ball: white - P(W);
Probability ball: even - P(E);
Probability ball: white and even - P(W&E).
Probability ball picked being white or even: P(WorE)=P(W)+P(E)-P(W&E).
(1) The probability that the ball will both be white and have an even number painted on it is 0 --> P(W&E)=0 (no white ball with even number) --> P(WorE)=P(W)+P(E)-0. Not sufficient
(2) The probability that the ball will be white minus the probability that the ball will have an even number painted on it is 0.2 --> P(W)-P(E)=0.2, multiple values are possible for P(W) and P(E) (0.6 and 0.4 OR 0.4 and 0.2). Can not determine P(WorE).
(1)+(2) P(W&E)=0 and P(W)-P(E)=0.2 --> P(WorE)=2P(E)+0.2 --> multiple answers are possible, for instance: if P(E)=0.4 (10 even balls) then P(WorE)=1 BUT if P(E)=0.2 (5 even balls) then P(WorE)=0.6. Not sufficient.
Answer: E.
Hope it's clear. I found this method simple although it might look complicated. Took me exactly 2 minutes.
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Bunuel 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?
Probability ball: white - P(W);
Probability ball: even - P(E);
Probability ball: white and even - P(W&E).
Probability ball picked being white or even: P(WorE)=P(W)+P(E)-P(W&E).
(1) The probability that the ball will both be white and have an even number painted on it is 0 --> P(W&E)=0 (no white ball with even number) --> P(WorE)=P(W)+P(E)-0. Not sufficient
(2) The probability that the ball will be white minus the probability that the ball will have an even number painted on it is 0.2 --> P(W)-P(E)=0.2, multiple values are possible for P(W) and P(E) (0.6 and 0.4 OR 0.4 and 0.2). Can not determine P(WorE).
(1)+(2) P(W&E)=0 and P(W)-P(E)=0.2 --> P(WorE)=2P(E)+0.2 --> multiple answers are possible, for instance: if P(E)=0.4 (10 even balls) then P(WorE)=1 BUT if P(E)=0.2 (5 even balls) then P(WorE)=0.6. Not sufficient.
Answer: E.
Hope it's clear. The interesting part about this explanation is particularly helpful expression " multiple values are possible for P(W) and P(E) (0.6 and 0.4 OR 0.4 and 0.2). Can not determine P(WorE)." I did not think of that at all (I just thought, well this is a minus probabilities and we need plus, so not sufficient- I am not even sure I understood any relevance of the 2nd option), when I finally thought of the problem in the same way. Now, may be it is quite late here and my brain refuses to come up with something  , but are there instances in which multiple values are not possible and hence the answer would be B? Sort of a "what if" on this problem...
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Each of the 25 balls in acertain box is red, blue or white and has a number from 1 to 10 painted on it. If one ball is 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 ? a. the probability that the ball selected will be white and an even number painted on it is 0 b.The probability that the ball will be white minus an even number painted on it is 0.2
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