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## 25 balls in a certain box is either red

This topic has 2 expert replies and 9 member replies
abhi332 Master | Next Rank: 500 Posts
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#### 25 balls in a certain box is either red

Wed Feb 24, 2010 5:05 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.

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kstv Legendary Member
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Wed Feb 24, 2010 6:40 am
IMO [spoiler]B[/spoiler

1) P(White) X P(Even) = 0 We know that no white ball has even nos. or all balls even even nos is either red or blue.
But a ball may still be Red or blue and have odd no on it. Insuff.
2) P(W) - P(E) = 0.2 . Probability of getting a even no is P(E) is 1/2. so P(W) is 0.7. Suff. to calculate P(W)+P(E)

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Testluv GMAT Instructor
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Wed Feb 24, 2010 10:30 am
We need to figure out the probability of getting a white ball OR a ball with an even number.

(1) tells us that there are no white balls with even numbers but gives us no info on the number of (or likelihood of picking)a white or even-numbered ball.

Insufficient.

(2) tells us that there are more white balls than there are even-numbered balls but doesn't yield the necesary info.

Insufficient.

Together, you still don't know probability of getting white or even.

Choose E.

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firdaus117 Master | Next Rank: 500 Posts
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Wed Feb 24, 2010 11:41 am
kstv wrote:
IMO [spoiler]B[/spoiler

1) P(White) X P(Even) = 0 We know that no white ball has even nos. or all balls even even nos is either red or blue.
But a ball may still be Red or blue and have odd no on it. Insuff.
2) P(W) - P(E) = 0.2 . Probability of getting a even no is P(E) is 1/2. so P(W) is 0.7. Suff. to calculate P(W)+P(E)
Where did you get this information shown above in Red color from?Probably,you assumed equal number of odd and even balls.But as number of balls is 25,the probability of getting an even numbered ball can never be 0.5.

shweta.aec Junior | Next Rank: 30 Posts
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Mon Jun 21, 2010 3:37 am
I am still not clear with the explanations . Can someone explain the problem in more detail

brijesh Senior | Next Rank: 100 Posts
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Mon Jun 21, 2010 5:25 am
shweta.aec wrote:
I am still not clear with the explanations . Can someone explain the problem in more detail
Lets understand the problem first:

There are 25 balls.

Balls are of Red, Blue and White colour but we dont know the actual no of balls of individual colours.

Again the coloured balls are numbered from 1 to 10 but again we dont know how they are marked (How many 1s, 2s etc)

The Q`s ask the following (i.e. abt the following formula)

[P(W) U P(E)]= P(w) + P(E)- P(W) intersection P(E)

Option 1 : Talks about P(W) Î  P(E) which is Zero
Option 2 : Talks about P(W)-P(E)= 0.2
So neither individual nor together the options are sufficient to answer
Option - E

shweta.aec Junior | Next Rank: 30 Posts
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Tue Jun 22, 2010 2:06 am
thanks a lot .. that helps

YellowSapphire Master | Next Rank: 500 Posts
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Tue Aug 03, 2010 10:39 am
P(while or even) = ?
We can find it in two ways.
P(while or even) = P(white) + P(even) - P(white and even)
P(while or even) = P(white and odd) + P(even and not white)

Statement 1:
P(white and even) = 0; we still do not know how many white and how many even are.
Not Sufficient

Statement 2:
P(white) - P(even) = 0.2; P(white) and (even) can take multiple values.
Not Sufficient

Statement 1 and 2:
P(white and even) = 0; P(white) - P(even) = 0.2; P(white) and (even) can take multiple values.
Not Sufficient

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dario.brignone Junior | Next Rank: 30 Posts
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Sat Apr 07, 2012 6:40 am
This is how I solved it:

there are 6 types of balls which summed are 25:
white even
white odd
red even
red odd
blue even
blue odd

So:

25 = WE + WO + RE + RO + BE + BO

We are asked probability of picking WE or WO or RE or BE

Statement 1)
WE = 0
insufficient

Statement 2)
WE + WO - WE - RE - BE = 0.2
Again not sufficient as RO and BO are not mentioned

Togher again not suffiecient as we still don't know RO and BO

lisa.scherer86@web.de Junior | Next Rank: 30 Posts
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Sun Oct 14, 2012 3:05 pm
Hi all,

statement 1 says:
P(White) X P(Even) = 0 Therefore we know that

- either P(White) or P(Even) has to be zero
- both P(White) P(Even) can be zero

=> Insufficient

From statement 2 I know that P(White)is def NOT zero

Both together, I know that P(even) HAS TO BE zero

What is wrong in my calculation?
Thanks

vaibhav108 Newbie | Next Rank: 10 Posts
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Tue Oct 16, 2012 12:20 am
lisa.scherer86@web.de wrote:
Hi all,

statement 1 says:
P(White) X P(Even) = 0 Therefore we know that

- either P(White) or P(Even) has to be zero
- both P(White) P(Even) can be zero

=> Insufficient

From statement 2 I know that P(White)is def NOT zero

Both together, I know that P(even) HAS TO BE zero

What is wrong in my calculation?
Thanks
The first statement only mentions that P(W) intersection P(E) is zero. That is no ball painted white will have even number on it or vice versa. It does not mean that the product of the probabilities is zero.
It can be that P(W)=0.4 and P(E)=0.2 (just an example)

Have a look at solution by brijesh...he has explained very well.

Hope this helps.

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Scott@TargetTestPrep GMAT Instructor
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Tue Dec 05, 2017 5:02 pm
abhi332 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 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.

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