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Pure luck

This topic has 6 expert replies and 6 member replies
lukaswelker Senior | Next Rank: 100 Posts Default Avatar
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Pure luck

Post 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

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ash4gmat Senior | Next Rank: 100 Posts Default Avatar
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Post Sun May 22, 2016 11:28 pm
Matt@VeritasPrep wrote:
ash4gmat wrote:
Experts please correct me where I'm wrong,

From statement 1Razz(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 2Razz(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.

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Post Fri May 13, 2016 3:06 pm
ash4gmat wrote:
Experts please correct me where I'm wrong,

From statement 1Razz(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.

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Post 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.

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Brent

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Post 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:
Quote:
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

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Matt@VeritasPrep GMAT Instructor
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Post 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).

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ash4gmat Senior | Next Rank: 100 Posts Default Avatar
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Post Wed May 11, 2016 10:02 pm
Experts please correct me where I'm wrong,

From statement 1Razz(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.

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binit Master | Next Rank: 500 Posts Default Avatar
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Post Mon May 04, 2015 10:45 pm
Sure. I shall TC.

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Post 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.

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binit Master | Next Rank: 500 Posts Default Avatar
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Post 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.

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ajaysingh24 Junior | Next Rank: 30 Posts Default Avatar
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Post 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

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Post Tue Apr 08, 2014 10:38 am
The site is acting up. It posted my solution twice.

Cheers,
Brent

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Post Tue Apr 08, 2014 10:37 am
Quote:
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

_________________
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