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10 Jan 2017, 12:41
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38% (02:56) correct
62%
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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
* Kudos for all correct solutions
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
* Kudos for all correct solutions
10 Jan 2017, 13:27
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GMATPrepNow
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
* Kudos for all correct solutions
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
* Kudos for all correct solutions
Hi
The second picked ball can't be red and the first and second also can't be yellow, otherwise our probability of the third ball to be yellow will be 0. Hence there are only three possible cases:
Blue Green Yellow + Green Blue Yellow + Red (Blue/Green) Yellow (or 4 if we write blue and green for the second ball separately)
\(\frac{1}{5}*\frac{1}{4}*\frac{1}{3} + \frac{1}{5}*\frac{1}{4}*\frac{1}{3} + \frac{2}{5}*\frac{2}{4}*\frac{1}{3} = \frac{1+1+4}{60} = \frac{6}{60} = \frac{1}{10}\)
Answer C.
11 Jan 2017, 08:13
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GMATPrepNow
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
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)
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
