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

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

by Gmat_mission » Sun Apr 01, 2018 10:03 am
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

[spoiler]OA=C[/spoiler].

How can I solve this problem? I don't know what formulas should I use. Combinations or permutations, probability...? <i class="em em-tired_face "></i>
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by Jay@ManhattanReview » Sun Apr 01, 2018 11:14 pm
Gmat_mission 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

[spoiler]OA=C[/spoiler].

How can I solve this problem? I don't know what formulas should I use. Combinations or permutations, probability...? <i class="em em-tired_face "></i>
Since there are 2 Red balls, the first ball may or may not be Red.

Case 1: 1st ball: Red, 2nd ball: Not Read; 3rd ball: Yellow

Ways of drawing the 1st ball = 2/5; (There are 2 Red balls);
Ways of drawing the 2nd ball = 2/4; (It can only be blue or green since Yellow is reserved for the 3rd draw);
Ways of drawing the 3rd ball = 1/3; (There is only 1 Yellow ball)

Total number of ways for Case 1 = 2/5*2/4*1/3 = 1/15

Case 2: 1st ball: Not Red, 2nd ball: Not Read; 3rd ball: Yellow

Ways of drawing the 1st ball = 2/5; (It can only be blue or green since Yellow is reserved for the 3rd draw);
Ways of drawing the 2nd ball = 1/4; (It can only be ONE between blue and green since Yellow is reserved for the 3rd draw);
Ways of drawing the 3rd ball = 1/3; (There is only 1 Yellow ball)

Total number of ways for Case 2 = 2/5*1/4*1/3 = 1/30

Total number of ways = 1/15 + 1/30 = 3/30 = 1/10

The correct answer: C

Hope this helps!

-Jay
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by chetan.sharma » Mon Apr 02, 2018 7:07 pm
Gmat_mission 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

[spoiler]OA=C[/spoiler].

How can I solve this problem? I don't know what formulas should I use. Combinations or permutations, probability...? <i class="em em-tired_face "></i>
Hi..
Another approach, short and may be a bit easier...
_ _ _
In the third place it can be yellow
So 1/5
In the second place any one of the red, so 2/4
In the first place any of the remaining three, so 3/3=1

So prob = 1*2/4*1/5=1/10..

It's always better to fill the restrictions first..
As for Question on whether probability or combination or permutation, probability too is finally ( some combination/all combinations)..
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by Jeff@TargetTestPrep » Thu Apr 05, 2018 4:23 pm
Gmat_mission 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
Since there is only one yellow ball and the selection is without replacement, neither the first nor the second ball can be yellow if the third one must be yellow. So the first ball can be the blue ball, the green ball, or one of the two red balls. Let's say the first ball is blue or green; then we have:

P(1st ball is blue or green) x P(2nd ball is neither yellow nor red) x P(3rd ball is yellow)

2/5 x 1/4 x 1/3

2/60

Now, let's say the first ball is red; then we have:

P(1st ball is red) x P(2nd ball is not yellow or red) x P(3rd ball is yellow)

2/5 x 2/4 x 1/3

4/60

Thus the overall probability is 2/60 + 4/60 = 6/60 = 1/10.

Alternate Solution:

Let's interpret the selections as orderings of the letters B, G, Y and R. Without any restrictions, 3 of these 4 letters can be ordered in 5P3 = 5!/(5-3)! = 5 x 4 x 3 = 60 ways.

Let's list all the orderings where the second letter is not R and the third letter is Y. We have two cases of RGY (one for each red ball), two cases of RBY, one case of BGY and one case of GBY. In total, there are 6 cases where the second letter is not red and the third letter is yellow.

Thus, the probability of a selection with the desired properties is 6/60 = 1/10.

Answer: C

Jeffrey Miller
Head of GMAT Instruction
[email protected]

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