Most probability question that include the phrase 'at least' are easier to solve "backwards." that is, for such questions, it tends to be easier to find the probability of NOT getting what we want and then subtracting from one. (since, for example, if an event has a 3/4 prob of NOT occurring, then it has a 1/4 chance of occuring)
So for this question, we can start by finding the prob that NO couples are selected.
I like to imagine choosing each of the four people one at a time. in order to NOT get any couples here, the selection would proceed as follows:
1)Choose anybody
2)choose somebody not married to the person already selected)
3)choose somebody not married to either of the people selected
3)choose somebody not married to any of the three people already selected
Now we can find numbers for the 4 events:
1) there's a 100% chance that the first person selected will be selected. So prob for event 1 is 1.
2) Now there are 15 people left. But one of them is married to the person selected already. So there are 14 permissible people left. So prob of choosing somebody permissible is 14/15
3) On the third selection, there are 14 people left, 12 of whom are not married to any of those already selected. So the third event has a 12/14 prob.
4) On the fourth selection, there are 13 people left, 10 of whom are not married to any of those already selected. so the fourth event has a 10/13 prob.
Since we want the prob of all four of these events happening together, we multiply:
(1)(14/15)(12/14)(10/13) --> lots of canceling --> 8/13
So there is 8/13 chance of getting NO couples.
Therefore, there is a 5/13 chance of getting AT LEAST one couple.
