25 balls in a certain box is either red

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abhi332: Master | Next Rank: 500 Posts; Posts: 117; Joined: Wed Dec 30, 2009 1:57 pmFollowed by: 1 member; Location: India; Thanked: 4 times

by abhi332 » 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.

What you think, you become.

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Source: Beat The GMAT — Data Sufficiency | Wed Feb 24, 2010 5:05 am

kstv: Legendary Member; Posts: 610; Joined: Fri Jan 15, 2010 12:33 amFollowed by: 2 members; Thanked: 47 times

by kstv » 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; Posts: 1302; Joined: Mon Oct 19, 2009 2:13 pmFollowed by: 164 members; Location: Toronto; Thanked: 539 times; GMAT Score:800

by Testluv » 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.

Kaplan Teacher in Toronto

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firdaus117: Master | Next Rank: 500 Posts; Posts: 102; Joined: Sat Feb 20, 2010 5:38 am; Location: IIM Ahmedabad; Thanked: 10 times

by firdaus117 » 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.

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shweta.aec: Junior | Next Rank: 30 Posts; Posts: 24; Joined: Thu Jun 03, 2010 1:45 am; GMAT Score:700

by shweta.aec » Mon Jun 21, 2010 3:37 am

I am still not clear with the explanations . Can someone explain the problem in more detail

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brijesh: Senior | Next Rank: 100 Posts; Posts: 84; Joined: Mon Apr 26, 2010 2:51 amFollowed by: 1 member; Thanked: 6 times

by brijesh » 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 don`t know the actual no of balls of individual colours.

Again the coloured balls are numbered from 1 to 10 but again we don`t know how they are marked (How many 1`s, 2`s 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

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shweta.aec: Junior | Next Rank: 30 Posts; Posts: 24; Joined: Thu Jun 03, 2010 1:45 am; GMAT Score:700

by shweta.aec » Tue Jun 22, 2010 2:06 am

thanks a lot .. that helps

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YellowSapphire: Master | Next Rank: 500 Posts; Posts: 117; Joined: Fri Jul 23, 2010 7:57 pm; Location: India; Thanked: 1 times

by YellowSapphire » 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

Yellow Sapphire

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dario.brignone: Junior | Next Rank: 30 Posts; Posts: 19; Joined: Wed Mar 30, 2011 6:12 am; Thanked: 1 times

by dario.brignone » 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

Answer (E)

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[email protected]: Junior | Next Rank: 30 Posts; Posts: 20; Joined: Sun Oct 14, 2012 1:34 pm

by [email protected] » 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

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vaibhav108: Newbie | Next Rank: 10 Posts; Posts: 6; Joined: Fri Jun 15, 2012 11:48 pm; Thanked: 1 times

by vaibhav108 » Tue Oct 16, 2012 12:20 am

[email protected] 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; Posts: 8142; Joined: Sat Apr 25, 2015 10:56 amFollowed by: 29 members; Location: Los Angeles, CA; Thanked: 43 times

by Scott@TargetTestPrep » 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.

Answer: E

Scott Woodbury-Stewart
Founder and CEO
[email protected]

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