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# A box contains 1 blue ball, 1 green ball, 1 yellow ball

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CEO
Joined: 12 Sep 2015
Posts: 3513
Location: Canada
A box contains 1 blue ball, 1 green ball, 1 yellow ball  [#permalink]

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10 Jan 2017, 12:41
2
Top Contributor
7
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Difficulty:

95% (hard)

Question Stats:

32% (03:19) correct 68% (02:27) wrong based on 66 sessions

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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

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##### Most Helpful Community Reply
Senior Manager
Joined: 13 Oct 2016
Posts: 367
GPA: 3.98
Re: A box contains 1 blue ball, 1 green ball, 1 yellow ball  [#permalink]

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10 Jan 2017, 13:27
3
2
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

* 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.
##### General Discussion
CEO
Joined: 12 Sep 2015
Posts: 3513
Location: Canada
Re: A box contains 1 blue ball, 1 green ball, 1 yellow ball  [#permalink]

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11 Jan 2017, 08:13
1
Top Contributor
1
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)

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
_________________

Test confidently with gmatprepnow.com

Intern
Joined: 29 Oct 2014
Posts: 37
Re: A box contains 1 blue ball, 1 green ball, 1 yellow ball  [#permalink]

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13 Jan 2017, 04:07
1
1
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

* Kudos for all correct solutions

As last ball must be yellow,
Ball 3 can be picked in only 1 way
2 ball cannot be red so only 2 options are left(total-1 yellow-2 red)
1st ball can be picked in3 ways(2red+1 b/g)
So total ways is 3*2*1=6
Now total number of ways without any conditions is =5!/2!=60
Required probability =6/60=1/10
Manager
Status: Gathering chakra
Joined: 05 Feb 2018
Posts: 129
A box contains 1 blue ball, 1 green ball, 1 yellow ball  [#permalink]

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17 Feb 2019, 09:41
1
probability: 3/3 * 2/4 * 1/5
Starting from the right slot because it's easier.
1 yellow out of 5 = 1/5
2 non red (green or blue) out of remaining 4 = 2/4 = 1/2
any of the 3 remaining balls (red and green or blue) = 3/3
1*1/2*1/5 = 1/10

combinations: 3C1*2C1*1C1 / 5P3
denominator is 5P3 = 60, we use permutations because order matters
Starting from the right of the 3 slots, ways to choose 1 yellow = 1C1
ways to choose a non-red from remaining 4 = 2C1
ways to choose a remaining ball out of 3 = 3C1
numerator = 3*2*1
3*2*1/60 = 6/60 = 1/10
Intern
Joined: 28 Jan 2019
Posts: 10
A box contains 1 blue ball, 1 green ball, 1 yellow ball  [#permalink]

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28 Feb 2019, 04:56
GMATPrepNow wrote:

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

On the highlighted text, is this always true? Applicable in all cases?
A box contains 1 blue ball, 1 green ball, 1 yellow ball   [#permalink] 28 Feb 2019, 04:56
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# A box contains 1 blue ball, 1 green ball, 1 yellow ball

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