There are n people in line, indexed by i=1,…,n , to enter a theater with n seats one by one. However, the first person ( i=1 ) in the line is drunk. This person has lost her ticket and decides to take a random seat instead of her assigned seat. That is, the drunk person decides to take any one of the seats 1 to n with equal probability. Every other person i=2,…,n that enters afterwards is sober and will take his assigned seat (seat i ) unless his seat i is already taken, in which case he will take a random seat chosen uniformly from the remaining seats.
Suppose that n=3 . What is the probability that person 2 takes seat 2?
(Enter a fraction or a decimal accurate to at least 3 decimal places.)
To solve this problem, we can analyze all the possible scenarios.
Scenario 1: The drunk person takes seat 1.
In this case, the probability that person 2 takes seat 2 is 0, as the seat is already taken by the drunk person.
Scenario 2: The drunk person takes seat 2.
In this case, person 2 will take seat 1, as his assigned seat (seat 2) is already taken. The probability of this happening is 1/2.
Scenario 3: The drunk person takes seat 3.
In this case, person 2 will take seat 2, as his assigned seat is available. The probability of this happening is also 1/2.
Now, let's calculate the probability of each scenario occurring:
Scenario 1: Probability = 1/n = 1/3
Scenario 2: Probability = 1/n * 1/(n-1) = 1/3 * 1/2 = 1/6
Scenario 3: Probability = 1/n * 1/(n-1) = 1/3 * 1/2 = 1/6
To find the total probability of person 2 taking seat 2, we sum the probabilities of Scenario 2 and Scenario 3:
Total probability = Scenario 2 probability + Scenario 3 probability
= 1/6 + 1/6
= 1/3
Therefore, the probability that person 2 takes seat 2 is 1/3.
1/2
1/2