Data Sufficiency

Hello, All--

Can someone please explain the rationale behind this question (#8 on one of the GMATPrep practice exams):

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.

.....

I selected (A) only the first statement is sufficient, because statement (1) is telling me that the two events are mutually exclusive. Therefore, the probability of the ball being white is 1/3 and the probability of the number on the ball being even is 1/2. Add those to probabilities together (since we're trying to find the union), and we get 5/6.

I had no idea what statement (2) was telling me, let alone if it's useful or not.

HOWEVER, the correct answer is (E) - both statements together are not sufficient.

My question is, Is (E) the answer because no info about the number of red, blue, or white balls given and no info about which exact numbers from 1 and 10 were on all the balls?

Any help/enlightenment is appreciated.

Thanks!

Kansonne wrote: My question is, Is (E) the answer because no info about the number of red, blue, or white balls given and no info about which exact numbers from 1 and 10 were on all the balls?

Any help/enlightenment is appreciated.

Thanks!

Yes, you cannot assume that probability of selecting a white ball is 1/3
since there's no mention of the number of balls of each type or that
there's equal probability of a ball being white, red or blue.

We are asked to find P(w) + P(even) - P(w n even)

1 tells us that P(w n even) = 0. So, it doesn't help

2 tells us that P(w) - P(even) = 0.2. Doesn't help

Combining 1 and 2, we can't arrive at P(w) + P(even)

So, E

Thanks, jayhawk2001!