January 26, 2019 January 26, 2019 07:00 AM PST 09:00 AM PST Attend this webinar to learn how to leverage Meaning and Logic to solve the most challenging Sentence Correction Questions. January 27, 2019 January 27, 2019 07:00 AM PST 09:00 AM PST Attend this webinar to learn a structured approach to solve 700+ Number Properties question in less than 2 minutes.
Author 
Message 
TAGS:

Hide Tags

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8810
Location: Pune, India

Re: Each of the 25 balls in a certain box is either red, blue or
[#permalink]
Show Tags
02 Feb 2015, 19:04
pacifist85 wrote: 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. How did you get P(E) = 5/25? Did you say that there are 5 even numbers in 1 to 10 and that is why 5/25? What says that the numbers cannot be repeated. In fact, each ball has a number from 1 to 10 on it  i.e. there are 10 numbers but there are 25 balls. So numbers must be repeated. So we don't know exactly how many are even.
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >



Senior Manager
Status: Math is psychological
Joined: 07 Apr 2014
Posts: 416
Location: Netherlands
GMAT Date: 02112015
WE: Psychology and Counseling (Other)

Re: Each of the 25 balls in a certain box is either red, blue or
[#permalink]
Show Tags
02 Feb 2015, 23:12
VeritasPrepKarishma wrote: pacifist85 wrote: 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. How did you get P(E) = 5/25? Did you say that there are 5 even numbers in 1 to 10 and that is why 5/25? What says that the numbers cannot be repeated. In fact, each ball has a number from 1 to 10 on it  i.e. there are 10 numbers but there are 25 balls. So numbers must be repeated. So we don't know exactly how many are even. That's true. For some reason I thought that even if there are more even balls than 5 then the possibility would be the same. But I see it now.



Intern
Joined: 16 Mar 2014
Posts: 5

Re: Each of the 25 balls in a certain box is either red, blue or
[#permalink]
Show Tags
04 May 2015, 10:14
Why is the following logic wrong?
1) P(W and E) = P(W)*P(E) = 0 > P(W) or P(E) or both are zero. Insufficient. 2) P(W)  P(E) = .2. Insufficient. 1+2) P(W)=.2, P(E)=0. Therefore P(W) + P(E)  P(W and E) = .2



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8810
Location: Pune, India

Re: Each of the 25 balls in a certain box is either red, blue or
[#permalink]
Show Tags
04 May 2015, 20:18
swaggerer wrote: Why is the following logic wrong?
1) P(W and E) = P(W)*P(E) = 0 > P(W) or P(E) or both are zero. Insufficient. 2) P(W)  P(E) = .2. Insufficient. 1+2) P(W)=.2, P(E)=0. Therefore P(W) + P(E)  P(W and E) = .2 Think about it: Under what conditions is P(W and E) = P(W)*P(E)? Is it always true?
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >



eGMAT Representative
Joined: 04 Jan 2015
Posts: 2466

Each of the 25 balls in a certain box is either red, blue or
[#permalink]
Show Tags
05 May 2015, 03:38
swaggerer wrote: Why is the following logic wrong?
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) Hope this helps. Regards, Krishna
_________________
Register for free sessions Number Properties  Algebra Quant Workshop
Success Stories Guillermo's Success Story  Carrie's Success Story
Ace GMAT quant Articles and Question to reach Q51  Question of the week
Must Read Articles Number Properties – Even Odd  LCM GCD  Statistics1  Statistics2  Remainders1  Remainders2 Word Problems – Percentage 1  Percentage 2  Time and Work 1  Time and Work 2  Time, Speed and Distance 1  Time, Speed and Distance 2 Advanced Topics Permutation and Combination 1  Permutation and Combination 2  Permutation and Combination 3  Probability Geometry Triangles 1  Triangles 2  Triangles 3  Common Mistakes in Geometry Algebra Wavy line  Inequalities Practice Questions Number Properties 1  Number Properties 2  Algebra 1  Geometry  Prime Numbers  Absolute value equations  Sets
 '4 out of Top 5' Instructors on gmatclub  70 point improvement guarantee  www.egmat.com



Manager
Joined: 18 Aug 2014
Posts: 119
Location: Hong Kong

Re: Each of the 25 balls in a certain box is either red, blue or
[#permalink]
Show Tags
23 Aug 2015, 06:14
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 ?



Intern
Joined: 16 Jun 2015
Posts: 1

Re: Each of the 25 balls in a certain box is either red, blue or
[#permalink]
Show Tags
23 Aug 2015, 21:54
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 \(\frac{a+bc}{25}\) (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 > \(\frac{white}{25}\frac{even}{25}=0.2\) > \(whiteeven=25*0.2=5\) > \(ab=5\) > \(b=a5\) > \(\frac{a+a5c}{25}=?\). Not sufficient. Attachment: 2.JPG (1)+(2) \(c=0\) and \(b=a5\) > \(\frac{a+a5+0}{25}=\frac{2a5}{25}\). 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?



Intern
Joined: 06 Dec 2015
Posts: 1

Re: Each of the 25 balls in a certain box is either red, blue or
[#permalink]
Show Tags
12 Dec 2015, 02:45
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). Explain please!



Intern
Joined: 23 May 2016
Posts: 10

Re: Each of the 25 balls in a certain box is either red, blue or
[#permalink]
Show Tags
23 May 2016, 12:45
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)



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8810
Location: Pune, India

Re: Each of the 25 balls in a certain box is either red, blue or
[#permalink]
Show Tags
23 May 2016, 18:31
alizainulabedeen wrote: 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). 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.
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8810
Location: Pune, India

Re: Each of the 25 balls in a certain box is either red, blue or
[#permalink]
Show Tags
23 May 2016, 18:34
rjivani 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.
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.
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >



Intern
Joined: 02 Oct 2013
Posts: 5

Re: Each of the 25 balls in a certain box is either red, blue or
[#permalink]
Show Tags
09 Oct 2016, 09:34
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
P(white)+P(Even) = 10/25+5/25 = 3/5



Board of Directors
Status: Stepping into my 10 years long dream
Joined: 18 Jul 2015
Posts: 3626

Each of the 25 balls in a certain box is either red, blue or
[#permalink]
Show Tags
10 Oct 2016, 08:36
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.
_________________
My GMAT Story: From V21 to V40 My MBA Journey: My 10 years long MBA Dream My Secret Hacks: Best way to use GMATClub  Importance of an Error Log! Verbal Resources: All SC Resources at one place  All CR Resources at one place Blog: Subscribe to Question of the Day Blog GMAT Club Inbuilt Error Log Functionality  View More. New Visa Forum  Ask all your Visa Related Questions  here. New! Best Reply Functionality on GMAT Club! Find a bug in the new email templates and get rewarded with 2 weeks of GMATClub Tests for free Check our new About Us Page here.



Target Test Prep Representative
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 4612
Location: United States (CA)

Re: Each of the 25 balls in a certain box is either red, blue or
[#permalink]
Show Tags
11 Oct 2016, 15:55
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 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 evennumbered 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 WoodburyStewart
Founder and CEO
GMAT Quant SelfStudy Course
500+ lessons 3000+ practice problems 800+ HD solutions



Current Student
Joined: 08 Feb 2016
Posts: 69
Location: India
Concentration: Technology
GPA: 4

Re: Each of the 25 balls in a certain box is either red, blue or
[#permalink]
Show Tags
04 Feb 2017, 04:49
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.
Answer: E.
Hope it's clear.



Board of Directors
Status: Stepping into my 10 years long dream
Joined: 18 Jul 2015
Posts: 3626

Re: Each of the 25 balls in a certain box is either red, blue or
[#permalink]
Show Tags
04 Feb 2017, 04:55
ajay2121988 wrote: How do you get to this formula.:
\(P(WorE)=P(W)+P(E)P(W&E)\)
Any underlying concept ? Yes, it is a general formula to determine the probability of Events A or B when both A and B are NOT mutually exclusive. For example, say I have the numbers from 1 to 10. Event A : All even numbers Event B : All multiples of 3. Probability of A or B will be All even + All multiple of 3  Those that have been repeated(only 6 in this case). I hope it is clear now.
_________________
My GMAT Story: From V21 to V40 My MBA Journey: My 10 years long MBA Dream My Secret Hacks: Best way to use GMATClub  Importance of an Error Log! Verbal Resources: All SC Resources at one place  All CR Resources at one place Blog: Subscribe to Question of the Day Blog GMAT Club Inbuilt Error Log Functionality  View More. New Visa Forum  Ask all your Visa Related Questions  here. New! Best Reply Functionality on GMAT Club! Find a bug in the new email templates and get rewarded with 2 weeks of GMATClub Tests for free Check our new About Us Page here.



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8810
Location: Pune, India

Re: Each of the 25 balls in a certain box is either red, blue or
[#permalink]
Show Tags
07 Feb 2017, 08:49
ajay2121988 wrote: How do you get to this formula.:
\(P(WorE)=P(W)+P(E)P(W&E)\)
Any underlying concept ?
Exactly same as that of two overlapping sets. n(A or B) = n(A) + n(B)  n(A and B) Why do you subtract n(A and B) out of it? Because it is double counted  once in n(A) and another time in n(B). Similarly, P(A and B) is double counted  once in probability of A and another time in probability of B.
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >



Manager
Joined: 23 Dec 2013
Posts: 147
Location: United States (CA)
GMAT 1: 710 Q45 V41 GMAT 2: 760 Q49 V44
GPA: 3.76

Each of the 25 balls in a certain box is either red, blue or
[#permalink]
Show Tags
25 Jun 2017, 17:34
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 evennumbered. 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.



Intern
Joined: 23 Sep 2017
Posts: 1

Re: Each of the 25 balls in a certain box is either red, blue or
[#permalink]
Show Tags
09 Oct 2017, 00:16
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 we=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?



Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 6834
GPA: 3.82

Re: Each of the 25 balls in a certain box is either red, blue or
[#permalink]
Show Tags
10 Oct 2017, 08:55
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 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.
_________________
MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The oneandonly World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only $149 for 3 month Online Course" "Free Resources30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons  try it yourself"




Re: Each of the 25 balls in a certain box is either red, blue or &nbs
[#permalink]
10 Oct 2017, 08:55



Go to page
Previous
1 2 3
Next
[ 48 posts ]



