Question

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

Answer 1

Answer:

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.