seven people are seated in a row. They all got up and sit down again in random order. What is the probability that the two originally seated at the two end are no longer at the ends

P (A&B) = 0.4

Explanation:

Because it is a random process and there are no special constraints the probability for everybody is the same, the probability of choosing a particular site is 1/7, the person originally seated in chair number seven has 5/7 chance of not seating in chair number six and seven, the same goes for the person originally seated in chair number six; Because we want the probability of the two events happening, we want the probability of the intersection of the two events, and because the selection of a chair change the probability for the others (Dependents events) the probability P (A&B) = P (A) * P (B/A) where P (A) is 5/7 and the probability of choosing the right chair after the event A is 4/7, therefore, P (A&B) = 4/7*5/7 = 0.4.

If the events were independent the probability would be 0.51.