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Give that there are 5 married couples. If we select only 3 people out

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Give that there are 5 married couples. If we select only 3 people out  [#permalink]

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New post 02 Sep 2018, 10:30
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Q. Given that there are 5 married couples. If we select only 3 people out of the 10, what is the probability that none of them are married to each other?

I want to solve this question using reverse probability approach, can someone post the solution pls.
OA - \(\frac{2}{3}\)

Source: https://gmatclub.com/forum/math-probability-87244.html
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Give that there are 5 married couples. If we select only 3 people out  [#permalink]

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New post 02 Sep 2018, 12:17
Total ways to choose 3 out of 10 people is 10C3 = 120


Ways to choose 3 out of 10 where no 3 are married to each other = #ways to choose 3 out of 5 * 2^3 (as each chosen person can be replaced by his partner) = 5C3 × 8 = 80

Probability = 80/120 = 2/3

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Re: Give that there are 5 married couples. If we select only 3 people out  [#permalink]

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New post 02 Sep 2018, 17:08
aggvipul wrote:
Q. Given that there are 5 married couples. If we select only 3 people out of the 10, what is the probability that none of them are married to each other?

I want to solve this question using reverse probability approach, can someone post the solution pls.
OA - \(\frac{2}{3}\)

Source: https://gmatclub.com/forum/math-probability-87244.html


I don't think using the reverse probability(probability none are married = 1 - probabilty at least one pair is married) is the best approach. The reason for this is because we have to take into account the fact that there are multiple ways of getting a couple ---> pick 1 is married to pick 2, pick 2 is married to pick 3, pick 1 is married to pick 3.

Probability no couples = (Number of 3 person selections with no couples)/Total number of selections of 3 people from 10


Number of 3 person selections with no couples = (10 * 8 * 6)/3! ---->80
-Note: We have to divide by 3! to get rid of repeats(ie ABC is the same team as BAC)

Total number of selections of 3 people from 10 ----> 10C3 = 10!/(7! * 3!) = 120

80/120 = 2/3
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adam@alldaytestprep.com

Unlimited private GMAT Tutoring in Chicago for less than the cost a generic prep course. No tracking hours. No watching the clock.
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Give that there are 5 married couples. If we select only 3 people out  [#permalink]

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New post 03 Sep 2018, 00:31
arosman wrote:
aggvipul wrote:
Q. Given that there are 5 married couples. If we select only 3 people out of the 10, what is the probability that none of them are married to each other?

I want to solve this question using reverse probability approach, can someone post the solution pls.
OA - \(\frac{2}{3}\)

Source: https://gmatclub.com/forum/math-probability-87244.html


I don't think using the reverse probability(probability none are married = 1 - probabilty at least one pair is married) is the best approach. The reason for this is because we have to take into account the fact that there are multiple ways of getting a couple ---> pick 1 is married to pick 2, pick 2 is married to pick 3, pick 1 is married to pick 3.

Probability no couples = (Number of 3 person selections with no couples)/Total number of selections of 3 people from 10


Number of 3 person selections with no couples = (10 * 8 * 6)/3! ---->80
-Note: We have to divide by 3! to get rid of repeats(ie ABC is the same team as BAC)

Total number of selections of 3 people from 10 ----> 10C3 = 10!/(7! * 3!) = 120

80/120 = 2/3


Hi arosman while I agree that reverse probability approach may not be the best here but as mentioned in my post the purpose to get the answer through reverse probability is to learn the concept more thoroughly.
bb can you pls help out here
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Affiliations: All Day Test Prep
Joined: 08 May 2018
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Location: United States (IL)
Schools: Booth '20 (A)
GMAT 1: 770 Q51 V49
GRE 1: Q167 V167
GPA: 3.58
Give that there are 5 married couples. If we select only 3 people out  [#permalink]

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New post 03 Sep 2018, 17:35
aggvipul wrote:
arosman wrote:
aggvipul wrote:
Q. Given that there are 5 married couples. If we select only 3 people out of the 10, what is the probability that none of them are married to each other?

I want to solve this question using reverse probability approach, can someone post the solution pls.
OA - \(\frac{2}{3}\)

Source: https://gmatclub.com/forum/math-probability-87244.html


I don't think using the reverse probability(probability none are married = 1 - probabilty at least one pair is married) is the best approach. The reason for this is because we have to take into account the fact that there are multiple ways of getting a couple ---> pick 1 is married to pick 2, pick 2 is married to pick 3, pick 1 is married to pick 3.

Probability no couples = (Number of 3 person selections with no couples)/Total number of selections of 3 people from 10


Number of 3 person selections with no couples = (10 * 8 * 6)/3! ---->80
-Note: We have to divide by 3! to get rid of repeats(ie ABC is the same team as BAC)

Total number of selections of 3 people from 10 ----> 10C3 = 10!/(7! * 3!) = 120

80/120 = 2/3


Hi arosman while I agree that reverse probability approach may not be the best here but as mentioned in my post the purpose to get the answer through reverse probability is to learn the concept more thoroughly.
bb can you pls help out here



Start at highest level possible.

Probability no couples picked + probability at least one couple picked = 1

Therefore -----> probability no couples picked = 1 - probability at least one couple picked

Since only three people are picked total, there is no way to select more than one couple. Therefore ---> probability at least one couple picked = probability exactly one couple picked

Probability exactly one couple picked = probability picks 1 & 2 are a couple + probability picks 1 & 3 are a couple + probability picks 2 & 3 are a couple

Since all 3 of these options are going to have the same probability we can just think of this as ---> probability exactly one couple picked = 3 * probability picks 1 & 2 are a couple

Imagine we have 3 slots.

Slot 1: Slot 1 can be any person as long as slot two is that person's partner. Probability of selecting any person is just 10/10 = 1

Slot 2: There are nine people left but only 1 of them is Slot 1's partner. Therefore probability here is 1/9

Slot 3: There are 8 people left and since we already have out couple any 8 of them can be selected. 8/8 = 1

Therefore, the exact probability of this exact sequence(Partner 1 ---> Partner 2 ---> Anybody else) is 1 * 1/9 * 1 = 1/9.

Accounting for the 3 different scenarios ------> Probability at least one couple = 3 * 1/9 = 1/3

Probability no couples = 1 - probability at least one couple = 1 - 1/3 = 2/3

Lemme know if you have any questions!

Best,
Adam
_________________

Adam Rosman, MD
University of Chicago Booth School of Business, Class of 2020
adam@alldaytestprep.com

Unlimited private GMAT Tutoring in Chicago for less than the cost a generic prep course. No tracking hours. No watching the clock.
https://www.alldaytestprep.com


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Joined: 15 Nov 2017
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Location: India
Concentration: Operations, Marketing
GPA: 4
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Re: Give that there are 5 married couples. If we select only 3 people out  [#permalink]

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New post 04 Sep 2018, 01:30
Thanks a lot arosman that does solve the purpose :thumbup:
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Re: Give that there are 5 married couples. If we select only 3 people out &nbs [#permalink] 04 Sep 2018, 01:30
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