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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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Updated on: 17 Jun 2014, 03:05
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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

Originally posted by lexis on 01 May 2008, 11:01.
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: GMAT prep DS  [#permalink]

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13 Aug 2010, 03:42
43
65
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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Re: Math: Probability - 25 balls  [#permalink]

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01 May 2008, 13:37
25
14
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.

The question is asking for:
P(white) + P(even) - P(white&even) = ?

(1) is saying: P(white&even) = 0
Still cannot find the answer
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
##### General Discussion
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Re: Math: Probability - 25 balls  [#permalink]

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01 May 2008, 13:43
1
1
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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Re: GMAT prep DS  [#permalink]

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13 Aug 2010, 09:27
14
3
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 53889 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 53871 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 53872 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 53869 times ]

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Re: GMAT prep DS  [#permalink]

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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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Re: GMAT prep DS  [#permalink]

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08 Oct 2010, 16:35
2
2
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: Probability - Data Sufficiency  [#permalink]

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15 May 2011, 10:50
2
probability of the ball to be white = A
probability of the ball to be an even numbered ball = B

A U B = A + B - (A intersection B)

a. A intersection B = 0. Not sufficient as A and B not given.

b. A-B = 0.2 not sufficient for solving the equation.

a+b

A U B = B+0.2 + B - 0

gives value in terms of B. Hence not sufficient.

E.
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Re: Probability - Data Sufficiency  [#permalink]

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16 May 2011, 13:03
2
We can also translate each statement into plain English:

1) There are no white balls with even numbers on them. Therefore, if we can find the chance of white and the chance of even, we can add them without having to worry about overlap. However, we don't have much to go on from this statement alone. If they asked for the chance that the ball was white AND even, clearly this would be sufficient.

2) The chance of white exceeds the chance of even by 20%. This isn't the same as saying 'The chance is 20% greater.' For instance, the chances could be white 50% and even 30%, but not white 12% and even 10%. In any case, we have no idea what the actual numbers are. Insufficient.

1 & 2) We know that the total chance = w + e. Since w = e + .2, we can say that the total chance = 2e + .2. However, since we have no idea what e is, this is insufficient. More simply, we can just say that we don't have an actual number for either probability, so there is no way to add them up.

I hope this helps!
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Re: Probability - Data Sufficiency  [#permalink]

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17 May 2011, 09:17
(1)

White balls don't have any even # on them

P(W) + P(E)

w/25 + e/25

Insufficient

(2)

w/25 - e/25 = 0.2

Insufficient

(1) + (2), Insufficient

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

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17 Jul 2011, 21:51
4
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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Re: Each of the 25 balls in a certain box is either red, blue or  [#permalink]

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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]

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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
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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 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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Re: GMAT prep DS  [#permalink]

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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.

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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Re: GMAT prep DS  [#permalink]

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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.

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?
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Re: GMAT prep DS  [#permalink]

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05 Aug 2013, 10:54
1
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.

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

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

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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.
Re: Each of the 25 balls in a certain box is either red, blue or &nbs [#permalink] 02 Feb 2015, 13:37

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