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 Your Progress

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.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Each of the 25 balls in a certain box is either red, blue or [#permalink]
01 May 2008, 10:01

10

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

65% (hard)

Question Stats:

59% (02:13) correct
41% (02:39) wrong based on 157 sessions

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

Re: Math: Probability - 25 balls [#permalink]
01 May 2008, 12:37

4

This post received KUDOS

1

This post was BOOKMARKED

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

Re: Math: Probability - 25 balls [#permalink]
01 May 2008, 12:43

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.

Re: Math: Probability - 25 balls [#permalink]
01 May 2008, 20:24

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.

Each of the 25 balls in a certain box is either red, blue or [#permalink]
13 Aug 2010, 02:03

1

This post received KUDOS

2

This post was BOOKMARKED

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

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)=1BUT if P(E)=0.2 (5 even balls) then P(WorE)=0.6. Not sufficient.

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 8356 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 8361 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 8358 times ]

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

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)=1BUT 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 _________________

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)=1BUT 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

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;

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). _________________

Re: Each of the 25 balls in a certain box is either red, blue or [#permalink]
03 Mar 2012, 04:48

Expert's post

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 _________________

Re: Each of the 25 balls in a certain box is either red, blue or [#permalink]
03 Mar 2012, 06:32

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,

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)=1BUT 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. _________________

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)=1BUT 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

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)=1BUT 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? _________________

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)=1BUT 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) _________________

Great to know you are joining Kellogg. A lot was being talked about your last minute interview on Pagalguy (all good though). It was kinda surprise that you got the...

This is a long overdue post! A lot of Indian applicants, having scheduled interviews in March, reached out to me asking about my interview experience with Kellogg. I had a...

A critical phase of the MBA application, concurrent to researching your target schools, is “researching yourself” and building your profile. What are your unique traits? Where do you want to be...