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

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

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01 May 2008, 11:01
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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
[Reveal] Spoiler: OA

Last edited by Bunuel on 17 Jun 2014, 03:05, edited 1 time in total.
Renamed the topic, edited the question, added the OA and moved to DS forum.

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Re: Math: Probability - 25 balls [#permalink]

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01 May 2008, 13:37
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lexis 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 have an eve number painted on it is 0.2

I got E.

P(white) + P(even) - P(white&even) = ?

(1) is saying: P(white&even) = 0
INSUFFICIENT

(2) is saying: P(white) - P(even) = 0.2
We don't know P(white&even), INSUFFICIENT

Together, you have
P(white) - P(even) = 0.2
and want to find: P(white) + P(even)=?
cannot complete the calculation with information given.
INSUFFICIENT

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Re: Math: Probability - 25 balls [#permalink]

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01 May 2008, 13:43
I agree that the answer is E. Here is my logic:

The question is asking for what is P(W) or P(E).

Statement 1 tells you that P(W) and P(E) is mutually exclusively. Thus P(W+E) = 0 So not enough info on its own.

Statement 2 tells you that P(W)-P(E) is 0.2. That is not sufficient either. In order to find P(W) or P(E), we need P(W) + P(E). However there is no information given concerning P(W) and P(E).

Please let me know if my analysis is not correct. Probability is not my strong suit.

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Re: Math: Probability - 25 balls [#permalink]

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01 May 2008, 21:24
I am new to this but this is a probablity Math question. There are 25 balls. 8 of each color with the probability of
3 even numbers per color. 3 divided by 8 is 0.375 divided by 25 is 0.015 which is .02.

But where did that other ball go.

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

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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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13 Aug 2010, 03:42
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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.

Hope it's clear.
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13 Aug 2010, 09:27
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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 42518 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 42505 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 42500 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 42502 times ]

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08 Oct 2010, 16:31
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.

Hope it's clear.

how did you get this?
$$P(WorE)=2P(E)+0.2$$
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08 Oct 2010, 16:35
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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.

Hope it's clear.

how did you get this?
$$P(WorE)=2P(E)+0.2$$

From (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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Re: Math: Probability - 25 balls [#permalink]

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17 Jul 2011, 21:51
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linda577ford wrote:
I am new to this but this is a probablity Math question. There are 25 balls. 8 of each color with the probability of
3 even numbers per color. 3 divided by 8 is 0.375 divided by 25 is 0.015 which is .02.

But where did that other ball go.

It is not essential that there will be 8 balls of each color. Each ball is red, blue or white. Overall, we could have 20 red balls, 2 blue balls and 3 white balls or 10 red balls, 10 blue balls and 5 white balls or some other combination. The point is that it is not essential that there are an equal number of balls with the same color. Similarly, the numbers on the balls will also be random. Say 9 balls could have 1-9 written on them and the rest of the balls could have 10 written on them. So you cannot find the probability of selecting a white ball or an even numbered ball until and unless you have some other data.

We know that P(Even OR White) = P(Even) + P(White) - P(Even AND White)
Both statements together don't give us the value of P(Even OR White).
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Kudos [?]: 17858 [3], given: 235 Manager Joined: 25 Dec 2011 Posts: 54 Kudos [?]: 56 [0], given: 29 GMAT Date: 05-31-2012 Re: Each of the 25 balls in a certain box is either red, blue or [#permalink] ### Show Tags 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 Kudos [?]: 56 [0], given: 29 Math Expert Joined: 02 Sep 2009 Posts: 42326 Kudos [?]: 133109 [0], given: 12412 Re: Each of the 25 balls in a certain box is either red, blue or [#permalink] ### Show Tags 03 Mar 2012, 05:48 morya003 wrote: 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 Well since the OA is E then apparently we cannot. Also how did you get 12? Anyway even if we knew from (2) that # of even balls is 12 the answer still wouldn't be B since we would need # of balls which are both white and have an even number painted on them, so in this case answer would C. Please refer to: each-of-the-25-balls-in-a-certain-box-is-either-red-blue-or-99044.html#p763481 _________________ Kudos [?]: 133109 [0], given: 12412 Manager Joined: 25 Dec 2011 Posts: 54 Kudos [?]: 56 [0], given: 29 GMAT Date: 05-31-2012 Re: Each of the 25 balls in a certain box is either red, blue or [#permalink] ### Show Tags 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 Kudos [?]: 56 [0], given: 29 Manager Joined: 06 Feb 2013 Posts: 59 Kudos [?]: 54 [0], given: 35 Re: GMAT prep DS [#permalink] ### Show Tags 14 Apr 2013, 18:37 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... _________________ There are times when I do not mind kudos...I do enjoy giving some for help Kudos [?]: 54 [0], given: 35 Manager Joined: 16 Feb 2012 Posts: 217 Kudos [?]: 415 [0], given: 121 Concentration: Finance, Economics Re: GMAT prep DS [#permalink] ### Show Tags 04 Aug 2013, 06:05 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. You say in the second statement that multiple values are possible (0.6 and 0.4 or 0.4 and 0.2) Are these values only possible? If they are please explain why. Why they cannot be 0.3 and 0.1 or 0.5 and 0.3? _________________ Kudos if you like the post! Failing to plan is planning to fail. Kudos [?]: 415 [0], given: 121 Senior Manager Joined: 10 Jul 2013 Posts: 326 Kudos [?]: 426 [1], given: 102 Re: GMAT prep DS [#permalink] ### Show Tags 05 Aug 2013, 10:54 1 This post received KUDOS Stiv 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. You say in the second statement that multiple values are possible (0.6 and 0.4 or 0.4 and 0.2) Are these values only possible? If they are please explain why. Why they cannot be 0.3 and 0.1 or 0.5 and 0.3? solution: White balls = w Red = R Blue = b Total ball = 25 Sum total of even numbered ball = E We have to evaluate = w/25 + (E)/25 From st(1) , we only know there is no white ball which contains even number. Even we still don’t know about red and blue balls have how many even numbered balls in them. So all are in mystery and doubt. From st(2), w/25 – E/25 = 0.2 , again all unknown. From both statement, we can assume several different things, I mean double case. So both are insufficient. Answer is (E) _________________ Asif vai..... Kudos [?]: 426 [1], given: 102 Intern Joined: 28 May 2014 Posts: 16 Kudos [?]: 2 [0], given: 0 Re: Each of the 25 balls in a certain box is either red, blue or [#permalink] ### Show Tags 17 Jun 2014, 03:41 I probably did this the wrong way. But... The first statement is telling me that there are no white balls with even numbers. Hence, insufficient. Crossing off A and D. The second statement is telling me that there are more than, or equal to, 20% of balls that's white. But I do not know how many of the red and blue balls that are given an even number on them. It could be 20% for a white ball and 0% for any colored even ball. It could be 40% for a white ball and 20% for any colored even ball. Hence, E. Kudos [?]: 2 [0], given: 0 Senior Manager Status: Math is psycho-logical Joined: 07 Apr 2014 Posts: 437 Kudos [?]: 141 [0], given: 169 Location: Netherlands GMAT Date: 02-11-2015 WE: Psychology and Counseling (Other) Re: Each of the 25 balls in a certain box is either red, blue or [#permalink] ### Show Tags 02 Feb 2015, 13:37 I also think that [1] in insufficient for the reasons outlined above. But I don't know why [2] is insufficient. That because we have that: P(W) - P(E) = 20/100 P(E) = 5/25 or 1/5. Is there a reason why we cannot find this probability? Is it that we need to know how many the even balls are? If not we could replace P(E) for 1/5 and solve for P(W). This is why I chose B. Kudos [?]: 141 [0], given: 169 Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7747 Kudos [?]: 17858 [0], given: 235 Location: Pune, India Re: Each of the 25 balls in a certain box is either red, blue or [#permalink] ### Show Tags 02 Feb 2015, 20:04 pacifist85 wrote: I also think that [1] in insufficient for the reasons outlined above. But I don't know why [2] is insufficient. That because we have that: P(W) - P(E) = 20/100 P(E) = 5/25 or 1/5. Is there a reason why we cannot find this probability? Is it that we need to know how many the even balls are? If not we could replace P(E) for 1/5 and solve for P(W). This is why I chose B. How did you get P(E) = 5/25? Did you say that there are 5 even numbers in 1 to 10 and that is why 5/25? What says that the numbers cannot be repeated. In fact, each ball has a number from 1 to 10 on it - i.e. there are 10 numbers but there are 25 balls. So numbers must be repeated. So we don't know exactly how many are even. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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

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03 Feb 2015, 00:12
VeritasPrepKarishma wrote:
pacifist85 wrote:
I also think that [1] in insufficient for the reasons outlined above. But I don't know why [2] is insufficient.

That because we have that:
P(W) - P(E) = 20/100
P(E) = 5/25 or 1/5. Is there a reason why we cannot find this probability? Is it that we need to know how many the even balls are?

If not we could replace P(E) for 1/5 and solve for P(W).

This is why I chose B.

How did you get P(E) = 5/25?
Did you say that there are 5 even numbers in 1 to 10 and that is why 5/25? What says that the numbers cannot be repeated. In fact, each ball has a number from 1 to 10 on it - i.e. there are 10 numbers but there are 25 balls. So numbers must be repeated. So we don't know exactly how many are even.

That's true. For some reason I thought that even if there are more even balls than 5 then the possibility would be the same. But I see it now.

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Re: Each of the 25 balls in a certain box is either red, blue or   [#permalink] 03 Feb 2015, 00:12

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