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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 [#permalink] New post 28 Feb 2018, 12:43
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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?

Express your answer as a fraction.

Answer:
[Reveal] Spoiler:
1/10

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Re: A box contains 1 blue ball, 1 green ball, 1 yellow ball [#permalink] New post 01 Mar 2018, 15:11
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Since there is only one yellow ball, of course neither the 1st nor the 2nd can be yellow. But there are two scenarios which can result in the 2nd ball being "not red."

In scenario 1, the 1st ball chosen is red, and the 2nd ball is either blue or green. The odds of three events happening should multiplied. So, the odds of the 1st being red are 2/5, the 2nd being either blue or green are 2/4, and the odds of the 3rd being yellow are 1/3:

2/5 x 2/4 x 1/3 = 1/15

In scenario 2, the 1st ball chosen is either blue or green and the 2nd ball is whichever out of blue or green didn't get chosen the first time. So, the odds of the 1st being either blue or green are 2/5, the 2nd being the other one are 1/4, and the odds of the 3rd being yellow are 1/3:

2/5 x 1/4 x 1/3 = 1/30

Adding these two probabilities, 1/15 and 1/30, results in an answer of 1/10.
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Re: A box contains 1 blue ball, 1 green ball, 1 yellow ball [#permalink] New post 01 Mar 2018, 17:32
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GreenlightTestPrep 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?

Express your answer as a fraction.

Answer:
[Reveal] Spoiler:
1/10


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

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
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Re: A box contains 1 blue ball, 1 green ball, 1 yellow ball [#permalink] New post 02 Mar 2018, 14:52
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Great question, BTW.

Regards
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Re: A box contains 1 blue ball, 1 green ball, 1 yellow ball   [#permalink] 02 Mar 2018, 14:52
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