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

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

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

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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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 =\(1-\frac{1}{3}=\frac{2}{3}\)

3 people out of 10: \(^{10}{C_3}\) = 120

There are 5 couples out of which we will select 3 couples:

=> \(^5{C_3}\) = 10 ways.

Out of every 2 people in a couple, we will select 1: \(^2{C_1}\) = 2 ways

For 3 people: 2 * 2 * 2 = 8 ways.

Total favourable: 10 * 8 = 80

Probability: \(\frac{80 }{ 120}\) =\(\frac{ 2 }{ 3}\)

Answer B

We don't need to calculate any nCr.

We start with ten people. What's the probability that we select SOMEONE with the first selection? 1
Once that person is selected, we have nine people remaining and can choose eight of them. What's the probability of success on the second selection? 8/9
Once that person is selected, we have eight people remaining and can choose six of them. What's the probability of success on the third selection? 6/8

\(1*\frac{8}{9}*\frac{6}{8} = \frac{6}{9} = \frac{2}{3}\)

Answer choice E.

Can someone explain to me why it is not (10C1 x 8C1 x 6C1) / (10C3) ? I get that this is wrong but I am not totally understanding why my numerator is wrong. First we choose out of 10 people and then 8 and then 6.

­This is explained HERE.

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Hope it helps.­

Thanks! That helped me.

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