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 500,000 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 [#permalink]

Show Tags

11 Dec 2008, 08:03

00:00

A

B

C

D

E

Difficulty:

25% (medium)

Question Stats:

64% (02:06) correct
36% (00:59) wrong based on 97 sessions

HideShow timer Statistics

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.

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. _________________

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.

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? _________________

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.

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

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.

Probability--someone please post an explanation [#permalink]

Show Tags

22 Jul 2010, 11:46

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

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

Show Tags

30 Dec 2014, 07:37

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

This is the kickoff for my 2016-2017 application season. After a summer of introspect and debate I have decided to relaunch my b-school application journey. Why would anyone want...

Sometimes Mom comes into town, you meet her at the airport to surprise her. Shenanigans ensue. You grab dinner and chat. You don’t write a long blog post that...