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
GMAT PREP - Q 8 - Probability
This topic has expert replies
- gabriel
- Legendary Member
- Posts: 986
- Joined: Wed Dec 20, 2006 11:07 am
- Location: India
- Thanked: 51 times
- Followed by:1 members
Use the probability formula ..kajcha 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
Let the probability.. of the ball being white be P(w)
of the ball being even be P(e)
of the ball being both even and white be P(w&e)
So, the probability of the ball picked being white or even is given by the probability formula ..
P(w or e) = P(w)+P(e) - P(w & e) ..
Now, the first statement says P(w&e) = 0, which simply means that there is no ball that is white and has an even number painted on it .. but the statement says nothing about P(w) and P(e).. so statement 1 is insufficient ..
The second statement says that P(w)-P(e)= 0.2 , which again cannot be used to find the P(w or e) .. so insufficient ..
Combine the 2 statements and we have P(w&e)=0 and P(w)-P(e)=0.2 .. which does not help in finding out the probability P(w or e) .. so the answer is E ..