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There are 3 blue disks, 5 green disks, and nothing else in a container
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17 May 2018, 05:45
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Difficulty:
65% (hard)
Question Stats:
58% (03:00) correct 42% (02:36) wrong based on 68 sessions
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There are 3 blue disks, 5 green disks, and nothing else in a container. Disks will be taken one at a time, and at random without replacement, from the container, until all 3 blue disks have been chosen. What is the probability that the third blue disks will be the seventh disk chosen?
Re: There are 3 blue disks, 5 green disks, and nothing else in a container
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17 May 2018, 16:07
There are 3 blue disks, 5 green disks, and nothing else in a container. Disks will be taken one at a time, and at random without replacement, from the container, until all 3 blue disks have been chosen. What is the probability that the third blue disks will be the seventh disk chosen?
A) 3/28
B) 1/8
C) 5/28
D) 15/56
E) 3/8
Probability 3rd Blue is seventh chosen = (# combos with 3rd blue seventh)/(total possible combos of where 3rd blue can be)
1) The third blue can be in spots 3, 4, 5, 6, 7, or 8:
_ _ _B3
_ _ _ _B3
_ _ _ _ _B3
_ _ _ _ _ _B3
_ _ _ _ _ _ _B3
_ _ _ _ _ _ _ _B3
2) Number of arrangements for each scenario depends on what can be placed before:
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There are 3 blue disks, 5 green disks, and nothing else in a container
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Updated on: 18 May 2018, 06:28
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2
gmatbusters wrote:
There are 3 blue disks, 5 green disks, and nothing else in a container. Disks will be taken one at a time, and at random without replacement, from the container, until all 3 blue disks have been chosen. What is the probability that the third blue disks will be the seventh disk chosen?
A) 3/28 B) 1/8 C) 5/28 D) 15/56 E) 3/8
Let's pick disks one by one, let's start picking from green 4 and then 2 blue (actually order doesn't matter, you can start with blue as well), and the 3rd blue in the end. Probability 5/8*4/7*3/6*2/5*3/4*2/3*1/2
Now we have to find in how many ways first 6 are oganized. Note since 7th (blue) has to be last we can't organize it. So first 6 can be organized 6!/4!*2! = 15 (we have 2 blue and 4 green, and they are the same).
Re: There are 3 blue disks, 5 green disks, and nothing else in a container
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17 May 2018, 20:02
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Since there are a total of 8 discs, 3 of them blue, there is 8C3 different ways of picking discs. We are asked what is the probability that the third blue disks will be the seventh disk chosen, as a result we need to count the number of ways we can have G,G,G,G,B,B,B. Here the last blue disc is fixed but the order of selecting the other 6 are not. As a result, there is 6C2 different ways. As a result, answer is 6C2/8C3 = 15 / 56
Re: There are 3 blue disks, 5 green disks, and nothing else in a container
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06 Aug 2018, 09:45
3 blue, 5 green
Number of ways to arrange them in a row: 8!/3!5! = 56 Number of ways to arrange them in a row so that the 7th spot is occupied by the 3rd blue ---> means we have 2 blue and 4 green in the first 6 spots, 1 blue in the 7th spot, and 1 green in the 8th spot --> (# of ways to arrange 2 blue and 4 green) * (1 way to fill the 7th spot) * (1 way to fill the 8th spot) = 6!/2!4! = 15
Probability to have 3rd blue in the 7th spot = 15/56
Re: There are 3 blue disks, 5 green disks, and nothing else in a container
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08 Aug 2018, 22:55
We do not have to count the 8th disk as per question and 7th will always be Blue. We stop when all blue disks are chosen. That means 8th disk is Green. So we are left with 6 places from starting to work with. (2 Blue and 4 Green ) No of ways to place : 6!/(2!*4!) =15 Total No of ways :- 8!/(3!*5!)=56
Answer:- 15/56. Option D.
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58% (03:00) correct 
