empty_spaces wrote:

probability of choosing the first person : 12/12 = 1

probability of choosing the 2nd person = 10/11 (who is not married to the first one )

probability of choosing 3rnd person = 8/10 (who is not married to either of the first 2 )

probability of choosing 4th person = 6/9 (who is not married to either of the first 3 )

so combining ... 1 * 10/11 * 8/10 * 6/9 = 16/33

its kind of tricky but think about this a bit and you will understand.

Thanks. This is the right answer.

By the way, I was trying to solve it in the following manner:

number of ways of selecting 4 people out of 12 = 12C4 = 495.

now, number of ways of choosing 4 people in such a way that there are no couples

= 12*10*8*6.

This is greater than 495. What is my mistake?

I also tried, finding the number of ways we can pick couples among these 4 people.

First person can be chosen in 12 ways.

Second person can be chosen in only 1 ways.

third person can be chosen in 10 ways.

fourth person can be chosen in 1 way

so the total number of favourable ways = 495 -120=375

P=375/495.

could you please point out my errors?

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