gmat prep probability

For the ball to be white OR Even the ball could be:
1) white AND even
2) not white and even
3) white and not even

the sum of all the above probabilities will give us the probabilty or White OR even.

so in (1) we are given the white and even - insuff
in (2) we are given white and not even - insuff

combined we still don't know how many not white and even so answer is E

When finding the Probability of Either/ Or:

Use the Formula
P(a) + p(b) = P(a or b)
(This is in Erics Flashcards on this site)
The first question tells you that it can not be both white and even, but no information on individual probabilities

The second tells you
p(a) - p(b) = .2

However; you need to know p(a) + p(b)
because the information in 2 can give you different answers.

If p(a) = .4 and p(b) = .2
then p(a or b) = .6

If p(a) = .3 and p(b)= .1
then p(a or b) = .4

Hence E

see Stuart's explanation here:

https://www.beatthegmat.com/probability- ... t8994.html

-BM-

I solved this one using a Venn diagram; maybe it will be clearer. Make a Venn diagram, the circles will be white and even and the middle, where they intersect, will be white and even combined. Now, to get either white or even (like the question asks), we will need to get the numbers for both the white and even circles minus the middle, which as you know, we can’t count because the middle is white and even.

So if answers 1 & 2 give us the below we can answer the question:

Prob. of being white + Prob. of being even – Prob. of being white and even.

(1)Give us Prob. of being white and even, so this gives us the middle of the Venn diagram. Insufficient by itself

(2)This doesn’t help us to complete our equation above. Insufficient.

Now if you look at the two together you don’t have enough info to complete the equation, hence the answer is E. Hope this helped, just wanted to show a different way of solving this question.