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A box contains 1 blue ball, 1 green ball

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A box contains 1 blue ball, 1 green ball

by Brent@GMATPrepNow » Tue Jan 10, 2017 11:40 am
Here's a probability question I created.
Difficulty level: 700+
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
Answer: C
Brent Hanneson - Creator of GMATPrepNow.com
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GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 16207
Joined: Mon Dec 08, 2008 6:26 pm
Location: Vancouver, BC
Thanked: 5254 times
Followed by:1268 members
GMAT Score:770

by Brent@GMATPrepNow » Wed Jan 11, 2017 7:12 am
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
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