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Each of the 25 balls in a certain box is either red, blue [#permalink]
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11 Dec 2008, 07:03
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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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Re: Probability  25 balls in a box [#permalink]
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11 Dec 2008, 13:31
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above720 wrote: Here is a DS probability question from GMATPrep, Practice Test 1:
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. p(w or e) = ? 1: p(w&e) = 0 ...nsf.. 2: p(w)  p(e) = 0.2 if p(w) = 0.2 and p(e) = 0, p(w)  p(e) = 0.2 if p(w) = 0.25 and p(e) = 0.05, p(w)  p(e) = 0.2 . . . . if p(w) = 1.00 and p(e) = 0.05, p(w)  p(e) = 0.2 ...nsff.. 1&2 also doesnot give a value that is p(w or e). So I would say E.
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Re: Probability  25 balls in a box [#permalink]
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12 Dec 2008, 13:04
I would go with B. here is the explanation.
First let get as much information from the question.
Total possibilities that a ball is picked at random = 25 (Since there are 25 balls) Let p(E)  probability of ball picked is painted in Even numbers. 2,4,5,6,10 are the possible out comes of a ball being painted with Even number.
So P(E) = possible outcomes / total outcomes = 5 / 25 = 1/5 = 0.2
Let P(W)  Probablity of ball picked is White.
The question asked is what is the probability of that a ball being picked is White. P(W) = ? To know this, one of the way is to get the White ball count from which we can get the Proability of picking a white ball.
Clue 1  Given that P(W and E) = 0 which states that the a ball painted in even number cannot be white. So we are left with 20 balls that some of them are possibly are white. But still we don't know how many are white. Clue is insufficient.
Clue 2  P(W)  P(E) = 0.2 ==> We already know that P(E) is 0.2. So p(W) = 0.2 + 0.2 = 0.4 Clue 2 alone is sufficient.



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Re: Probability  25 balls in a box [#permalink]
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12 Dec 2008, 14:38
mrsmarthi wrote: I would go with B. here is the explanation.
First let get as much information from the question.
Total possibilities that a ball is picked at random = 25 (Since there are 25 balls)
Let p(E)  probability of ball picked is painted in Even numbers. 2,4,5,6,10 are the possible out comes of a ball being painted with Even number.
So P(E) = possible outcomes / total outcomes = 5 / 25 = 1/5 = 0.2
Let P(W)  Probablity of ball picked is White.
The question asked is what is the probability of that a ball being picked is White. P(W) = ? To know this, one of the way is to get the White ball count from which we can get the Proability of picking a white ball.
Clue 1  Given that P(W and E) = 0 which states that the a ball painted in even number cannot be white. So we are left with 20 balls that some of them are possibly are white. But still we don't know how many are white. Clue is insufficient.
Clue 2  P(W)  P(E) = 0.2 ==> We already know that P(E) is 0.2. So p(W) = 0.2 + 0.2 = 0.4 Clue 2 alone is sufficient. how do you know that there 5 balls with even numbers? Why not 6 or 7 or 4 or 10 or even 15? If 5 balls have even numbers, what about the rest 20 balls? are they all with odd numbers? how?
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Re: Probability  25 balls in a box [#permalink]
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12 Dec 2008, 15:49
The info from qs stem and both statements is not sufficient, i would go for E on this one.



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Re: Probability  25 balls in a box [#permalink]
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12 Dec 2008, 23:48
I will also go for E.
The question asks for P(WUE)
From the first stmt, P(W intersection E) = 0. Hence, the question asks for the value of p(W) + p(E).
From the second question, p(W)  p(E) = 0.2 and the info is insufficient to get p(W) + p(E).



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Re: Probability  25 balls in a box [#permalink]
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13 Dec 2008, 08:19
I go with E. What is the OA?
We have to find: P(W or E) = P(W) + P(E)  P( W and E).
From 1: P(W and E) = 0 ( Not sufficient ) But we get, P(W or E) = P(W) + P(E) I
From 2: P(W)P(E) = 0.2 ( Not sufficient) II
Even from I and II we cannot get P(W or E) so answer is E.



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Re: Probability  25 balls in a box [#permalink]
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30 Jan 2009, 19:48
Why can't it be C? Reasoning is: statement 1: W * even = 0 statement 2: W  even = 0.2 then by substitution we can find W+E.
What do you think?
Thanks.



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Probability Question [#permalink]
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08 Oct 2009, 05:53
Anybody? thanks!!
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Re: Probability Question [#permalink]
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08 Oct 2009, 06:46
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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) \(P(W&E)=0\) (no white ball with even number) > \(P(WorE)=P(W)+P(E)0\). Not sufficient (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: \(P(E)=0.4\) (10 even balls) > \(P(WorE)=1\) BUT \(P(E)=0.2\) (5 even balls) > \(P(WorE)=0.6\). Not sufficient. Answer: E.
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Each of the 25 balls in a certain box is either red, blue, [#permalink]
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26 Mar 2010, 14:14
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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 OPEN DISCUSSION OF THIS QUESTION IS HERE: eachofthe25ballsinacertainboxiseitherredblueor63290.html



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Re: Each of the 25 balls in a certain box is either red, blue, [#permalink]
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changhiskhan 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?
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.
Thanks in advance. P(White or Even) = P(White) + P(Even)  P(White & Even) st 1) P(White & Even) = 0 .. still dont know individual probabilities . Not sufficient st 2) P(White)  P(Even) = 0.2 .. still dont know abouit the combined probability and also individual probabilities. Not sufficient combinging also we cannot figure out P(White) + P(Even) as we cannot find out the number of white balls or the number of even balls E



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Re: Each of the 25 balls in a certain box is either red, blue, [#permalink]
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27 Mar 2010, 08:29
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changhiskhan 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?
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.
Thanks in advance. It has to be E. 1: p(W & E) = 0 to get p(W or E) = p(w) + p(E) cant find individual probability hence insufficient. 2: p(W)  p(E) = 0.2 cant find individual probability hence insufficient. Combining, also we cant find anything more hence E



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Re: Probability  25 balls in a box [#permalink]
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27 Apr 2010, 03:22
xALIx wrote: Why can't it be C? Reasoning is: statement 1: W * even = 0 statement 2: W  even = 0.2 then by substitution we can find W+E.
What do you think?
Thanks. It doesn't work because: P(even & white) =/= P(even) * P(white). Hence, the translation of statement 1 couldn't stand.



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DS : Probability [#permalink]
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09 Jul 2010, 23:49
Each of the 25 balls in a box is either Red, Blue or white and has a number 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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Re: DS : Probability [#permalink]
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10 Jul 2010, 01:20
25 balls. R, B or W with a number from 1 to 10 on it. i.e. 3 colors with one of the 5 even and 5 odd numbers. What is the probability that a ball selected is white OR even?
1. Probability of white AND even = 0. This means that there are no white balls that have even numbers on them. But we do not know how many balls are just white, or how many balls are just even.
INSUFFICIENT
2. P(white)  P(even) = 0.2
This won't help us calculate P(white) + P(even)
INSUFFICIENT
Both 1 and 2.
Again, we only know that no white balls are even and that P(white)  P(even) = 0.2
We cannot calculate the prob of either white or even.
INSUFFICIENT
Pick E



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Probabilitysomeone please post an explanation [#permalink]
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22 Jul 2010, 10:46
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 be both 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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Re: Probabilitysomeone please post an explanation [#permalink]
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22 Jul 2010, 11:19
Bunuel wrote: Merging similar topics. Thanks again, got it!



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Re: Each of the 25 balls in a certain box is either red, blue, [#permalink]
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24 Jul 2014, 22:57
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. OPEN DISCUSSION OF THIS QUESTION IS HERE: eachofthe25ballsinacertainboxiseitherredblueor63290.html
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