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?

Solution:
Since we do not want couples, we would have to ensure that the 3 people belong to 3 different couples.

The selection of the 3 couples form the 5 couples can be done in 5c3 ways = 10 ways

From these 3 couples, we need to select 1 person from each couple, which can be done in 2c1 ways for each couple. So total ways = 2c1 x 2c1 x 2c1 = 8 ways

Thus, favourable cases = 10 x 8 = 80

Total cases = 10c3 = 120

=> Probability = 80/120 = 2/3

Answer B

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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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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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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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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
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Thanks a lot arosman that does solve the purpose
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Total number of ways of choosing 3 people is 10C3.
Total number of ways of picking 3 couples out of 5 couples is 5C3.
Now number of ways of picking one couple out of three couple is 2C1 X 2C1 X2C1.
So, probability of picking no married couples is 5C3(2C1)^3 / 10C3 = 2/3

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?



A) 1/3
B) 2/3
C) 1/9
D) 7/9
E) 1/12




PS:

Can someone show how to solve this using reverse probability method because i'm getting 1-(1*1/9*1)=8/9 as the answer.

Answer: B

Probability of selecting 1 person out of 10= 10C1/10C1 = 1

Since married people are not to be selected, we have to choose 1 person from remaining 9 people out of which only 8 are valid options (since partner of 1st chosen person cannot be selected). Probability of selecting another person from these 9 = 8C1/9C1

Probability of choosing last person from remaining 8 people (partner of the above selected 2 people cannot be chosen so we are left with only 6 options)=6C1/8C1

Total probability= 1*(8C1/9C1)*(6C1/8C1)
=2/3

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Probability of no couples
= 1 - Probability of 1 couple
(There cannot be more than one couple since 2 couples would mean 4 people, whereas we need only 3 people)

To have one couple, we need to select one of the 5 couples (and both people in that couple) and one person from the remaining 8 people
= 5c1 x 8c1 = 40 ways

Total number of ways to select = 10c3 = 120

Probability of having 1 couple = 40/120 = 1/3

Thus, required probability

= 1 - 1/3 = 2/3

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Can anybody please help me with what is wrong in this approach?

we have in total 10 people-
Total sample space = 10C3 (getting 3 people from 10 people)

Favourable:
1) So, for first person we have all 10 options available making my choices (10C1)
2) For second person, I can not select the 1st person's partner, thus I have only 8 choices available (8C1)
3) For third person again, I can't select either 1st person's partner or 2nd person's partner, thus only 6 choices available (6C1)

Probability = favourable/sample space

P = (10C1 * 8C1 * 6C1) / 10C3 = 4

which is obviously wrong.

Please help.

Hi arya251294

When you say 10C1*8C1*6C1, you are picking each of the three people in a certain order but here the order of selection does not matter. So you just have to divide by 3!

So the number of favorable cases is 10C1*8C1*6C1/3! = 80

Total sample space is 10C3=120

Probability = 80/120 = 2/3

Hope this helps!

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Ok let's say that the 10 people are -

AB CD EF GH IJ

when I say 10C1 it means that I am choosing any one of the 10 - say I chose 'A'
then I do 8C1, say I chose - C
then I do 6C1, say I chose - E

Now this selection could have been anything right - ACE or AEC or CAE or CEA or EAC or ECA (3! possibilities)
So, while doing 10C1 * 8C1 * 6C1 I am not selecting anything in a certain order right?

I am still not able to wrap my head around it.

Please help.

As you said, the selection could have been anything - ACE or AEC or CAE or CEA or EAC or ECA and so that includes 6 arrangements but we are not concerned about the order (or arrangement) in which they are selected. Our concern is just that A, C and E are among our selection. That is why we need to divide by 3!.
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Hi chetan2u Bunuel alldaytestprep
Can this approach be taken if I want to solve it taking couples (for the learning purpose) using reverse probability. I am using the below methodology but the answer is not correct

Prob of choosing 1 couple = 5/5
Prob of choosing 2 people from one selected couple = 2/2
Prob of choosing 3rd person from a different couple = \(\frac{4}{4}∗\frac{1}{2}\)
Possible combination = 2 (CCO, OCC)

Hence, Prob. =\(\frac{5}{5}∗\frac{2}{2}∗\frac{4}{4}∗\frac{1}{2}∗2\) = 1
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That may not be the correct method to do such a question.
Ways 1 couple is chosen - \(5C1*8=5*8\)
Total ways =\(10C3=\frac{10*9*8}{3*2}\)
\(P=\frac{5*8}{\frac{10*9*8}{3*2}}=\frac{3*2}{2*9}=\frac{1}{3}\)..

So the opposite that we are looking for =\(1-\frac{1}{3}=\frac{2}{3}\)
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out of 5 married couples i.e 10 people we need to select 3 people such that none are married to each other
10*8*6/ 10*9*8 ; 6/9 ; 2/3
OPTION E

Thanks chetan2u for the reply, I do understand the combinatoric method. Can you pls help determine error in my above posted method (using reverse prob. approach)
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I picture the five couples as this:

* * * * * * * * * *

Task: we will select three people.

1. First person selected

Can be any of them: 10 possibilities
* * * * * * * * * *

2. Second person selected:

* * * * * * * * * *

Can be any except the first person's couple:

8 possibilities.

3. Third person selected:

Can be any except the two first people's partners.

* * * * * * * * * *

Hence six possibilities.

TOTAL: 10·8·6 possibilities.

4. Calculate total number of possible selections:

10·9·8

5. Calculate probability:

( 10·8·6 ) / (10·9·8 )
At this step, note that 10 and 8 cancel out; this is why I didn't multiply 10·8·6 or 10·9·8 in the earlier steps.

Result: 6 / 9 = 2 / 3