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:

1) P(W and E) = P(W)*P(E) = 0 --> P(W) or P(E) or both are zero. Insufficient.

Think about it like this. (Represent the given info in a tree structure as shown below). We are given that every ball is either Red or Blue or White. We are also given that ball of each color can have either an even number or an odd number written on them.

Now from statement 1, we can see that the number of white balls that have an even number written on them is 0. But this doesn't mean that the total number of white balls is 0. Similarly, this doesn't also mean that there are no even numbers written on other colored balls. (Referring to your inference that P(E) is 0).

The correct way to write the above formula would be: P(choosing a white ball with an even number) = P(choosing a white ball out of all balls) * P(choosing an even numbered ball out of all white balls).

Notice that the second term is not P(E). Since you chose the short notation, you incorrectly inferred that the P(E) is 0. In fact, the correct inference would be P(choosing Even numbered ball from white balls) = 0)

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

Show Tags

23 Aug 2015, 07:14

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.

Normally in probability problems we say "and" means multiply. Hence why are we not saying P(W) * P(EVEN) = 0 --> Either P(W) or P(EVEN) must be 0 ?

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

Show Tags

23 Aug 2015, 22:54

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

Answer: E.

I would like one clarification, have we assumed that there is a possibility that a each ball will have a unique number on it, so if we translate the statement the probability of a ball being whit and even is 0, that mean only 5 possible combination of white balls are present (1,3,5,7,9) there fore 10 blue and 10 red balls would be present. then A would be sufficient.

would it be explicitly written in the question that each ball has a unique number?

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

Show Tags

12 Dec 2015, 03:45

bkk145 wrote:

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

I have a query related this explanation. Doesn't statement 1 tell us that either P(white) or P(even) is ZERO? If it does, then from statement 2, we can conclude that P(white) + P(even) is also 0.2, and hence the overall answer becomes 0.2; both statements combined become sufficient (C).

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

Show Tags

23 May 2016, 13:45

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.

---

Isnt P(E) = 12 / 25 (each ball has numbers 1- 10, assuming 5 from first 10 balls, another 5 from second 10 balls, and 2 from the remaining 5 balls) which will let us calculate P(W), and then calculate P(W) + P(E) - (PWnE = 0)

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

I have a query related this explanation. Doesn't statement 1 tell us that either P(white) or P(even) is ZERO? If it does, then from statement 2, we can conclude that P(white) + P(even) is also 0.2, and hence the overall answer becomes 0.2; both statements combined become sufficient (C).

Explain please!

Statement 1 tells us that P(White AND Even) = 0 That is, there is no ball such that it is white and has an even number painted on it. Obviously, there could be balls that are white and there could be other balls with even numbers painted on them. So P(White) or P(Even) is not given 0.
_________________

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.

---

Isnt P(E) = 12 / 25 (each ball has numbers 1- 10, assuming 5 from first 10 balls, another 5 from second 10 balls, and 2 from the remaining 5 balls) which will let us calculate P(W), and then calculate P(W) + P(E) - (PWnE = 0)

What says that numbers are allotted in sequence? There could be 5 balls with number 1 on them, 10 balls with number 2 on them, 3 balls with number 3 on them and then rest of the 7 balls with numbers 4 to 10 on them. Or any other combination.
_________________

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

Show Tags

09 Oct 2016, 10:34

This is my approach. Please correct me where I went wrong. I chose A for this one.....

- Label from 1 to 10 >> maximum number of each balls = 10

My curious is, from statement 1, can't we conclude that white balls are only marked with odd numbers?

Then, the total number of balls is 25 thus there are 5 white balls, 10 blue balls, and 10 red balls. Blue balls and red balls have 5 even balls each. >>> Even balls = 10 White balls = 5

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

Show Tags

10 Oct 2016, 09:36

1

This post received KUDOS

ariiet wrote:

This is my approach. Please correct me where I went wrong. I chose A for this one.....

- Label from 1 to 10 >> maximum number of each balls = 10

My curious is, from statement 1, can't we conclude that white balls are only marked with odd numbers?

Then, the total number of balls is 25 thus there are 5 white balls, 10 blue balls, and 10 red balls. Blue balls and red balls have 5 even balls each. >>> Even balls = 10 White balls = 5

P(white)+P(Even) = 10/25+5/25 = 3/5

Yes, we can conclude that white balls are only marked with odd numbers. But we have numbers on the balls from 1 to 10 only and we have total 25 balls. SO, it is obvious that numbers are repeated. Now, We don't have any information whether we do have only 5 ODD numbered white balls or more than 5 ODD numbered white balls. So, we cannot conclude the number of white balls. Hence, Statement 1 is also insufficient.

I hope it is clear now. Let me know if you have more doubts.
_________________

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

We are given that there are 25 balls in a box and each one is either red, blue, or white and has a number from 1 to 10 painted on it. We need to determine the probability of selecting a white ball or an even-numbered ball.

Since each ball is colored and has a number painted on it, selecting a ball with a number or color is not mutually exclusive. Thus, to determine the probability of selecting a white or even ball we use the following formula:

P(white or even ball) = P(white ball) + P(even ball) – P(white and even ball)

Statement One Alone:

The probability that the ball will both be white and have an even number painted on it is 0.

Using the information in statement one, we know that P(white and even ball) = 0. However, we still cannot determine the probability of selecting a white or even ball. We can eliminate answer choices A and D.

Statement Two Alone:

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.

Statement two does not provide enough information to determine the probability of selecting a white or even ball. We can eliminate answer choice B.

Statements One and Two Together:

Using the information from statements one and two we still only know the following:

P(white or even ball) = P(white ball) + P(even ball) – P(white and even ball)

P(white or even ball) = P(white ball) + P(even ball) – 0

Without knowing the sum of P(white ball) and P(even ball), we cannot answer the question.

Answer: E
_________________

Scott Woodbury-Stewart Founder and CEO

GMAT Quant Self-Study Course 500+ lessons 3000+ practice problems 800+ HD solutions

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

Show Tags

04 Feb 2017, 05:49

How do you get to this formula.:

\(P(WorE)=P(W)+P(E)-P(W&E)\)

Any underlying concept ?

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.

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

Show Tags

25 Jun 2017, 18:34

1

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

The goal is to find the probability of picking a ball that is either white or even-numbered. In other words, P(W) + P(E) - P(W&E).

Statement 1) P(W&E) = 0.

We don't know P(W) or P(E), so this statement is insufficient.

Statement 2) P(W) - P(E) = 0.2

We don't know the value of P(W) + P(E) or P(W&E). Insufficient.

Statements 1+2) P(W&E) = 0 and P(W) - P(E) = 0.2.

We still don't know the value of P(W) + P(E). Insufficient.

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

Show Tags

09 Oct 2017, 01:16

guys correct me where am i going wrong and lets not get into the grammar details of the question,lets solve it the pure mathematical way: we need to find =P(w) + P(e) - [P(w)*P(e)]

1) tells us that P(w)* P(e)=0 ,, so either P(w)=0 or P(e)=0 or both =0 {basically w=0 or e=0 or both 0} 2)tells us that P(w)-P(e)=0 or w-e=5

clearly 1) and 2) alone are insufficient

but when we combine both we get w=5. and e=0,hence we can get the answer....

i dont get the flaw in reasoning here ..anyone please?

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

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

...........| White | Not White ------------------------------ .....Even.| a | b --------------------------------- Not Even.| c | d

=> Q: ( a + b + c ) / 25 ?

a + b + c + d = 25

Since we have 4 variables and 1 equation, E could be the answer most likely.

1) & 2) From the condition 1), we have a = 0. From the condition 2), ( a + c ) / 25 - ( a + b ) / 25 = ( c - b ) / 25 = 0.2 c - b = 5

Since we have only 2 equations a = 0 and c - b = 5, we can't figure out b + c and solve this question.

Therefore, the answer is E as expected.

For cases where we need 3 more equations, such as original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 80 % chance that E is the answer, while C has 15% chance and A, B or D has 5% chance. Since E is most likely to be the answer using 1) and 2) together according to DS definition. Obviously there may be cases where the answer is A, B, C or D. Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.
_________________