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
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
could you please point out my errors?
for every person who doesn't try because he is
afraid of loosing , there is another person who
keeps making mistakes and succeeds..