GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

 It is currently 01 Jun 2020, 05:54 ### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # Each of the 25 balls in a certain box is either red, blue or white and

Author Message
TAGS:

### Hide Tags

Manager  B
Joined: 19 Aug 2016
Posts: 67
Re: Each of the 25 balls in a certain box is either red, blue or white and  [#permalink]

### Show Tags

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.

Hi Bunuel..

How did u get Probability ball picked being white or even: P(W or E)=P(W)+P(E)-P(W&E)?
Math Expert V
Joined: 02 Sep 2009
Posts: 64153
Re: Each of the 25 balls in a certain box is either red, blue or white and  [#permalink]

### Show Tags

1
zanaik89 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.

Hi Bunuel..

How did u get Probability ball picked being white or even: P(W or E)=P(W)+P(E)-P(W&E)?

OR probability:
If Events A and B are independent, the probability that Event A OR Event B occurs is equal to the probability that Event A occurs plus the probability that Event B occurs minus the probability that both Events A and B occur: $$P(A \ or \ B) = P(A) + P(B) - P(A \ and \ B)$$.

This is basically the same as 2 overlapping sets formula:
{total # of items in groups A or B} = {# of items in group A} + {# of items in group B} - {# of items in A and B}.

Note that if event are mutually exclusive then $$P(A \ and \ B)=0$$ and the formula simplifies to: $$P(A \ or \ B) = P(A) + P(B)$$.

Also note that when we say "A or B occurs" we include three possibilities:
A occurs and B does not occur;
B occurs and A does not occur;
Both A and B occur.

AND probability:
When two events are independent, the probability of both occurring is the product of the probabilities of the individual events: $$P(A \ and \ B) = P(A)*P(B)$$.

This is basically the same as Principle of Multiplication: if one event can occur in $$m$$ ways and a second can occur independently of the first in $$n$$ ways, then the two events can occur in $$mn$$ ways.

22. Probability

For more:
ALL YOU NEED FOR QUANT ! ! !

Hope it helps.
_________________
Intern  B
Joined: 24 Oct 2017
Posts: 10
Location: United States (NY)
GMAT 1: 690 Q44 V40
GMAT 2: 710 Q44 V43
GPA: 3.25
Re: Each of the 25 balls in a certain box is either red, blue or white and  [#permalink]

### Show Tags

Can someone tell me where my logic went wrong? I selected A, with the thought that since P(W&E)=0, therefore there must be 5 white balls, all of which have an odd number on them (so 5 white balls total). Therefore, P(W)=5/25, P(E)=10/25, and P(W&E)=0. Is it wrong to assume that the numbers don't repeat on each ball color (i.e. the blue balls with all of 1 number cannot be true?).
Retired Moderator P
Joined: 21 Aug 2013
Posts: 1359
Location: India
Re: Each of the 25 balls in a certain box is either red, blue or white and  [#permalink]

### Show Tags

ulanky wrote:
Can someone tell me where my logic went wrong? I selected A, with the thought that since P(W&E)=0, therefore there must be 5 white balls, all of which have an odd number on them (so 5 white balls total). Therefore, P(W)=5/25, P(E)=10/25, and P(W&E)=0. Is it wrong to assume that the numbers don't repeat on each ball color (i.e. the blue balls with all of 1 number cannot be true?).

Hi

P(W&E) = 0, means that there is NO ball which is both white and has an even number on it. But there could be various white balls with odd numbers on them, and there could be various red/blue balls with even numbers on them. We need to take both these kinds of balls into account.

I think we cannot assume that there must be 5 white balls, its possible that there is only one white ball in the entire box. And yes, I think we also cannot assume that the numbers cannot repeat on same coloured balls (because its nowhere given). Its possible that all blue coloured balls have only number 1 painted on them.
GMAT Club Legend  V
Joined: 11 Sep 2015
Posts: 4879
GMAT 1: 770 Q49 V46
Re: Each of the 25 balls in a certain box is either red, blue or white and  [#permalink]

### Show Tags

Top Contributor
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 the ball will have an even number painted on it is 0.2

Target question: What is the value of P(white or even)?

To solve this, we'll use the fact that P(A or B) = P(A) + P(B) - P(A & B)
So, P(white or even) = P(white) + P(even) - P(white & even)

Statement 1: P(white & even) = 0
We can add this to our probability equation to get: P(white or even) = P(white) + P(even) - 0
Since we don't know the value of P(white) and P(even), we cannot determine the value of P(white or even)
NOT SUFFICIENT

Statement 2: P(white) - P(even)= 0.2
We have no idea about the sum of P(white) and P(even), and we don't know the value of P(white & even)
NOT SUFFICIENT

Statements 1 and 2 combined:
Given P(white) - P(even)= 0.2 does not tell us the individual values of P(white) and P(even), and it doesn't tell us the value of P(white) + P(even).

So, since we can't determine the value of P(white) + P(even) - P(white & even), the statements combined are NOT SUFFICIENT.

Cheers,
Brent
_________________
Manager  P
Joined: 31 Jul 2017
Posts: 191
Location: India
GMAT 1: 500 Q47 V15
GPA: 3.4
WE: Information Technology (Computer Software)
Re: Each of the 25 balls in a certain box is either red, blue or white and  [#permalink]

### Show Tags

Bunuel wrote:
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

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

(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

(1)+(2) $$c=0$$ and $$b=a-5$$ --> $$\frac{a+a-5+0}{25}=\frac{2a-5}{25}$$. Not sufficient.
Attachment:
3.JPG

Hi Bunuel,

This is an interesting approach to this problem.

Please let me know when can we use this Transformation of probability into actual numbers.

And also if you have any similar questions in which I can apply this technique for practice.

Math Expert V
Joined: 02 Sep 2009
Posts: 64153
Re: Each of the 25 balls in a certain box is either red, blue or white and  [#permalink]

### Show Tags

SudhanshuSingh wrote:
Bunuel wrote:
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

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

(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

(1)+(2) $$c=0$$ and $$b=a-5$$ --> $$\frac{a+a-5+0}{25}=\frac{2a-5}{25}$$. Not sufficient.
Attachment:
3.JPG

Hi Bunuel,

This is an interesting approach to this problem.

Please let me know when can we use this Transformation of probability into actual numbers.

And also if you have any similar questions in which I can apply this technique for practice.

PS Overlapping + Probability questions

DS Overlapping + Probability questions

Hope it helps.
_________________
Intern  B
Joined: 24 May 2018
Posts: 8
Re: Each of the 25 balls in a certain box is either red, blue or white and  [#permalink]

### Show Tags

P(White)=W/25 , P(Even)=E/25, P(White&Even)=WE/25. P(W) or P(E) =W/25 + E/25 + WE/25
Question=what is W+ E + WE/25?
(1) P(WE)=0, So, W+E+0/25? Insuff.
(2) P(W)-P(E)=0.2 ---> P(W)=0.2+P(E). So, 0.2+ E + E + WE /25? Insuff
(1)(2) together, 0.2 + 2E/25? Still cant get the value.
"E"
Intern  B
Joined: 10 Feb 2019
Posts: 1
Re: Each of the 25 balls in a certain box is either red, blue or white and  [#permalink]

### Show Tags

Hate to re-open this thread 9 years later, but my instinct is to do this problem a different way and I do not understand why it is incorrect. I understand Bunuel's approach, and why Statements (1) and (2) are insufficient on their own, but with my approach, I am getting C.

(1) P(W&E)=0 --> P(W)*P(E)=0 --> Either P(W) or P(E) must =0
(2) P(W) - P(E) =0.2

Combining (1) and (2). Since either P(W) or P(E) must =0, and P(W) - P(E) =0.2, then P(W) = 0.2 and P(E) = 0. Therefore, P(W) + P(E) = 0.2 + 0 = 0.2

Can someone tell me what I'm missing here?

Many thanks!
Intern  B
Joined: 22 Aug 2019
Posts: 2
Location: Belgium
GMAT 1: 710 Q48 V39
GPA: 3.61
Re: Each of the 25 balls in a certain box is either red, blue or white and  [#permalink]

### Show Tags

1
I have a question. From statement (1), couldn't we deduct that there are no even numbers? Assuming that there is at least one ball from each colour, the fact that it is impossible (P(0)) to get W + E means that there are no even numbers? Thus, combining statement (1) and (2), we could determine that p(W) is 0.2. I really don't understand the logic here, so can anyone explain where I went wrong?
Math Expert V
Joined: 02 Sep 2009
Posts: 64153
Re: Each of the 25 balls in a certain box is either red, blue or white and  [#permalink]

### Show Tags

Klacet wrote:
I have a question. From statement (1), couldn't we deduct that there are no even numbers? Assuming that there is at least one ball from each colour, the fact that it is impossible (P(0)) to get W + E means that there are no even numbers? Thus, combining statement (1) and (2), we could determine that p(W) is 0.2. I really don't understand the logic here, so can anyone explain where I went wrong?

(1) says that there are no white balls with even number on them. But cannot there be red or blue balls with even numbers? Why not?

I suggest you to re-read the discussion to understand the question better.

Hope it helps.
_________________ Re: Each of the 25 balls in a certain box is either red, blue or white and   [#permalink] 12 Nov 2019, 04:45

Go to page   Previous    1   2   3   [ 51 posts ]

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