BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Probability of white or even

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Junior | Next Rank: 30 Posts
Posts: 11
Joined: Tue Mar 17, 2009 11:27 am

Probability of white or even

by scorpion_phoenix » Tue Sep 22, 2009 7:10 pm
Each of the 25 balls in a certain box is either red blue or white and has a number from 1-10 painted on it. If one ball is 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.

OA = E
Last edited by scorpion_phoenix on Tue Sep 22, 2009 7:21 pm, edited 1 time in total.
Top
Source: — Data Sufficiency |
Junior | Next Rank: 30 Posts
Posts: 11
Joined: Tue Mar 17, 2009 11:27 am

by scorpion_phoenix » Tue Sep 22, 2009 7:12 pm
Not understanding where I went wrong on this one. Some help would be much appreciated.
Top

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

Re: Probability of white or even

by Brent@GMATPrepNow » Tue Sep 22, 2009 8:34 pm
scorpion_phoenix wrote:Each of the 25 balls in a certain box is either red blue or white and has a number from 1-10 painted on it. If one ball is 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.

OA = E
To solve this, we need P(A or B) = P(A)+P(B)-P(A&B)
So, P(white or even)=P(white)+P(even)-P(white&even)

(1) P(white&even)=0 --> P(A or B) = P(A)+P(B)-0
Need P(white) and P(even)
INSUFF

(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)
INSUFF

(1)&(2) 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
Brent Hanneson - Creator of GMATPrepNow.com
Image
Top
Junior | Next Rank: 30 Posts
Posts: 11
Joined: Tue Mar 17, 2009 11:27 am

by scorpion_phoenix » Wed Sep 23, 2009 8:15 am
When you put it that way it does make sense. However, my approach was different and I now realize the error in my logic. See below.

P(w)= w/25
P(e)= e/25

I misread the information and thought that 10 of the 25 balls were numbered 1-10 with each number being used only once. With this I came up with there being only 5 even numbers (2, 4, 6, 8, 10). And said therefore, P(e)= 5/25. With this I took statement 2 and constructed the following equation....

P(w) - P(e)= .20 ---> w/25 - 5/25 = .20 or 5/25

After careful review, I now realize that statement 1 is also sufficient with this error in thinking.

P(w) * P(e)= 0 ---> w/25 * 5/25 = 0

When I got here I would have realized that I went wrong somewhere as this suggest that there are no white balls which is clearly inconsistent with the information provided.

Thanks for your response.
Top
Post new topic Post Reply
• Page 1 of 1