Problem Solving

3 persons (1 couple and 1 single) are seated at random in a raw of 5 chairs. What is the probability that the couple DOES NOT sit together?

5/7
4/5
2/5
3/5
11/18

any help is welcome!

IMO

Total Number of ways the 3 people can be seated on 5 chairs is 5C3 = 10

The number of ways they can be seated with the couple sitting together is 4C2

The probability that the couple will SIT TOGETHER is 4C2/5C3 = 6/10

so the probability that the couple DOES NOT SIT TOGETHER should be
1-6/10 = 4/10 => 2/5

Please correct me if am wrong.

I think you have to take order into account, since, for instance, if the three people sit on chairs 1, 4 and 5, you'll have different cases depending on where the loner sits.
Number of ways in which the 3 people can be seated:

5*4*3 = 60

Number of ways in which the couple will be seated together:

4*3*2 = 24

If A and B are the members of the couple and S the other chairs:
A B S S S
S A B S S
S S A B S
S S S A B
B A S S S
etc...
We have 8 of those, and for every one of them the loner can be seated on 1 of the three different chairs left, and therefore we need to multiply by 3...
The probability that the couple sits together is then 24/60 = 2/5

and the probability that the couple DOES NOT sit together: 3/5

Thanks avenus that makes more sense.