Pure luck

This topic has expert replies
Senior | Next Rank: 100 Posts
Posts: 65
Joined: Wed Mar 26, 2014 6:56 am
Followed by:1 members

Pure luck

by lukaswelker » Tue Apr 08, 2014 10:28 am
Hey guys, I nailed this question but it was pure luck. No I want to understand how I got lucky.

Here's the question.

Each of the 25 balls in a certain box is either red or 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 the ball will have an even number painted on it is 0.2

Can anybody explain me what is the reasoning behind it?

Many thanks
Lukas

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 16207
Joined: Mon Dec 08, 2008 6:26 pm
Location: Vancouver, BC
Thanked: 5254 times
Followed by:1268 members
GMAT Score:770

by Brent@GMATPrepNow » Tue Apr 08, 2014 10:37 am
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.

Answer: E

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
Image

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 16207
Joined: Mon Dec 08, 2008 6:26 pm
Location: Vancouver, BC
Thanked: 5254 times
Followed by:1268 members
GMAT Score:770

by Brent@GMATPrepNow » Tue Apr 08, 2014 10:38 am
The site is acting up. It posted my solution twice.

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
Image

Junior | Next Rank: 30 Posts
Posts: 17
Joined: Fri Feb 14, 2014 10:05 am
Thanked: 2 times

by ajaysingh24 » Tue Apr 08, 2014 10:52 am
lukaswelker wrote:Hey guys, I nailed this question but it was pure luck. No I want to understand how I got lucky.

Here's the question.

Each of the 25 balls in a certain box is either red or 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 the ball will have an even number painted on it is 0.2

Can anybody explain me what is the reasoning behind it?

Many thanks
Lukas
I think the answer should be (E)
I have used following approach the following combination of color and number can be
BE, BO, RE, RO, WE, WO (B-blue,R-Red , W-white , O-Odd , E-even )

question is to find --- no of (BE + RE + WO) balls / Total number of balls

1 statement does not give any info on the numerator part ... according to statement we get No of WE balls is 0
2 Statement tells about WO probability in turn WO number .. but still we do not get about BE and RE ,
So it can not be solved using both statement together

Master | Next Rank: 500 Posts
Posts: 111
Joined: Sat Mar 07, 2015 11:00 pm
Thanked: 8 times
Followed by:1 members

by binit » Sun May 03, 2015 4:01 am
Rather plain problem.. (E) was expected.. given so many data, some action was expected while combining St. 1 and 2.

~Binit.

GMAT Instructor
Posts: 2630
Joined: Wed Sep 12, 2012 3:32 pm
Location: East Bay all the way
Thanked: 625 times
Followed by:119 members
GMAT Score:780

by Matt@VeritasPrep » Sun May 03, 2015 11:48 pm
binit wrote:Rather plain problem.. (E) was expected.. given so many data, some action was expected while combining St. 1 and 2.

~Binit.
It's probably best not to revive these old threads unless there's a significant new development (e.g. the answers given are all actually wrong, or you have a question about some detail that hasn't been answered, etc.) Otherwise the new questions and the already solved questions get all mixed up.

Master | Next Rank: 500 Posts
Posts: 111
Joined: Sat Mar 07, 2015 11:00 pm
Thanked: 8 times
Followed by:1 members

by binit » Mon May 04, 2015 10:45 pm
Sure. I shall TC.

Senior | Next Rank: 100 Posts
Posts: 57
Joined: Thu Sep 17, 2015 2:57 am

by ash4gmat » Wed May 11, 2016 10:02 pm
Experts please correct me where I'm wrong,

From statement 1:P(W&E)=0 we get P(W).P(E)=0
When we use this information combined with statement 2 we can conclude P(E)=0
There for P(W)=0.2

Therefore answer should be P(W)+P(E)-P(W&E)=0.2+0+0 and henceforth C.

GMAT Instructor
Posts: 2630
Joined: Wed Sep 12, 2012 3:32 pm
Location: East Bay all the way
Thanked: 625 times
Followed by:119 members
GMAT Score:780

by Matt@VeritasPrep » Fri May 13, 2016 3:06 pm
ash4gmat wrote:Experts please correct me where I'm wrong,

From statement 1:P(W&E)=0 we get P(W).P(E)=0
When we use this information combined with statement 2 we can conclude P(E)=0
There for P(W)=0.2

Therefore answer should be P(W)+P(E)-P(W&E)=0.2+0+0 and henceforth C.
The issue here is the equation P(W&E) = P(W) * P(E). This is true if W and E are independent, but we don't know that those probabilities are independent here.

For instance, suppose that we have 10 red balls, all of which are even, and 15 white balls, all of which are odd. P(W&E) = 0, but it ISN'T true that P(W) = 0 or P(E) = 0.

Senior | Next Rank: 100 Posts
Posts: 57
Joined: Thu Sep 17, 2015 2:57 am

by ash4gmat » Sun May 22, 2016 11:28 pm
Matt@VeritasPrep wrote:
ash4gmat wrote:Experts please correct me where I'm wrong,

From statement 1:P(W&E)=0 we get P(W).P(E)=0
When we use this information combined with statement 2 we can conclude P(E)=0
There for P(W)=0.2

Therefore answer should be P(W)+P(E)-P(W&E)=0.2+0+0 and henceforth C.
The issue here is the equation P(W&E) = P(W) * P(E). This is true if W and E are independent, but we don't know that those probabilities are independent here.

For instance, suppose that we have 10 red balls, all of which are even, and 15 white balls, all of which are odd. P(W&E) = 0, but it ISN'T true that P(W) = 0 or P(E) = 0.
Thanks!!!

But statement 2 clearly tells that P(W)-P(E)=0.2
& statement 1 clearly tells P(W).P(E)= 0
So either P(W) is 0 or P(E) is 0.

From statement 2:P(W) cannot be 0 as this would result in -ve probability which is not possible.
Hence P(E)=0 and P(W)=0.2

Please correct my understanding in terms of given information.

GMAT Instructor
Posts: 2630
Joined: Wed Sep 12, 2012 3:32 pm
Location: East Bay all the way
Thanked: 625 times
Followed by:119 members
GMAT Score:780

by Matt@VeritasPrep » Thu May 26, 2016 3:19 pm
Ah, "clearly": the most dangerous word in a math proof! (As an old professor of mine used to say, "if I want to know where you've made a mistake in your work, I look for where you've written 'clearly' or 'obviously'".)

S1 tells us that P(W+E) = 0. It DOESN'T tell us that P(W) * P(E) = 0, since P(W+E) doesn't necessarily equal P(W) * P(E)! (This only happens if W and E are independent, but we can't assume independence here.)

In any event, we can use my earlier example. If we have 10 red, even balls and 15 white, odd balls, then

P(E) = 10/25
P(W) = 15/25
P(W&E) = 0

and as you can see, P(W&E) ≠ P(W) * P(E).

Junior | Next Rank: 30 Posts
Posts: 13
Joined: Fri Aug 11, 2017 1:40 am

by santhosh_katkurwar » Fri Nov 17, 2017 11:44 pm
Thanks for the wonderful explanation. I have one doubt though -what if the questions ask the probability of two white balls? Then it will be considered as independent event right?
Brent@GMATPrepNow 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.

Answer: E

Cheers,
Brent

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 16207
Joined: Mon Dec 08, 2008 6:26 pm
Location: Vancouver, BC
Thanked: 5254 times
Followed by:1268 members
GMAT Score:770

by Brent@GMATPrepNow » Sat Nov 18, 2017 7:15 am
santhosh_katkurwar wrote:Thanks for the wonderful explanation. I have one doubt though -what if the questions ask the probability of two white balls? Then it will be considered as independent event right?
If there are x balls in the box and we select two balls, then it comes down to whether we replace the first ball before we draw the second ball.
If we DO replace the first ball before we draw the second ball, then selecting a white ball on the 1st draw and selecting a white ball on the 2nd draw are INDEPENDENT events.
If we DON'T replace the first ball before we draw the second ball, then selecting a white ball on the 1st draw and selecting a white ball on the 2nd draw are NOT INDEPENDENT events.

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
Image