Brent@GMATPrepNow wrote:A box contains 1 blue ball, 1 yellow ball, and 2 red balls.
Three balls are randomly selected (one after the other) without replacement.
What is the probability that the 2nd ball is yellow or the 3rd ball is NOT yellow?

A) 1/2
B) 3/4
C) 5/6
D) 11/12
E) 1

First of all, it's useful to recognize that P(2nd ball is yellow or the 3rd ball is NOT yellow) is the SAME as P(1st ball is yellow or the 2nd ball is NOT yellow)

For more on this, see my post at the bottom of https://www.beatthegmat.com/beat-this-pr ... 85719.html

Let's apply the OR probability rule: P(A or B) = P(A) + P(B) - P(A and B)
So, P(1st ball is yellow or the 2nd ball is NOT yellow) = P(1st ball is yellow) + P(2nd ball is NOT yellow) - P(1st ball is yellow AND the 2nd ball is NOT yellow)

P(1st ball is yellow) = 1/4

P(2nd ball is NOT yellow) = P(1st ball is NOT yellow) = 3/4

P(1st ball is yellow AND the 2nd ball is NOT yellow) = P(1st ball is yellow) x P(the 2nd ball is NOT yellow)
= 1/4 x 1
= 1/4

So, P(1st ball is yellow or the 2nd ball is NOT yellow) = P(1st ball is yellow) + P(2nd ball is NOT yellow) - P(1st ball is yellow AND the 2nd ball is NOT yellow)
= 1/4 + 3/4 - 1/4
= 3/4
= B

Cheers,
Brent

Brent@GMATPrepNow wrote: A box contains 1 blue ball, 1 yellow ball, and 2 red balls.
Three balls are randomly selected (one after the other) without replacement.
What is the probability that the 2nd ball is yellow or the 3rd ball is NOT yellow?

A) 1/2
B) 3/4
C) 5/6
D) 11/12
E) 1

Another solution is to list all of the possible outcomes for the 3 selections:
- RRY
- RRB
- RBR
- RYR
- RYB
- RBY
- BYR
- YBR
- BRY
- YRB
- BRR
- YRR

Of the 12 possible outcomes, 9 meet one or both of requirements.
So, P(2nd ball is yellow or the 3rd ball is NOT yellow) = 9/12 = 3/4 = B