Brent@GMATPrepNow wrote:
A box contains 1 blue ball, 1 green 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 NOT red, and the 3rd ball is yellow?
A) 1/30
B) 1/20
C) 1/10
D) 3/20
E) 1/5
First of all, it's useful to recognize that P(
2nd ball is NOT red and the
3rd ball is yellow) is the SAME as P(
1st ball is NOT red and the
2nd ball is yellow)
For more on this, see my post at the bottom of
https://www.beatthegmat.com/beat-this-pr ... 85719.html
P(1st ball is NOT red
AND the 2nd ball is yellow) = P(1st ball is NOT red)
x P(the 2nd ball is yellow)
= 2/5
x 1/4
= 1/10
= C
Aside: The first probability, P(1st ball is NOT red), equals 2/5, because there are 5 balls to choose from, and we cannot choose a red ball (because that's not allowed) AND we cannot choose a yellow ball (because, that ball must be available for the next selection). So, of the 5 possible balls to choose from on the first selection, only 2 balls (the blue and green balls) are permissible.
Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
