Question

A parking lot consists of a single row containing n parking spaces (n ? 2). Mary arrives when all spaces are free. Tom is the next person to arrive. Each person makes an equally likely choice among all available spaces at the time of arrival. Describe the sample space. Obtain P(A), the probability the parking spaces selected by Mary and Tom are at most 2 spaces apart.

Answer 1

Let P denote the probability that Tom and Mary are atmost 2 places apart.

This probability can be further broken into cases where Mary and Tom are placed 0 places apart, 1 place apart and 2 places apart.

Case 1. Tom and Mary are placed zero place apart(side by side).

If Mary parks on any end Tom have only one place where he can park. If Mary does not park in the end Tom can park in two places.

$P_1=\frac{2}{n}*\frac{1}{n-1}+\frac{n-2}{n}*\frac{2}{n-1}=\frac{2n-2}{n(n-1)}$

Case 2.Mary and Tom are 1 place apart.

Similarly if mary parks on the end or second last from the end Tom has only one parking spot in sample space. But if Mary parks elsewhere, Tom has 2 parking space in sample space.

$P_2=\frac{4}{n}*\frac{1}{n-1}+\frac{n-4}{n}*\frac{2}{n-1}=\frac{2n-4}{n(n-1)}$

Case 3. Mary and Tom are placed 2 places apart.

Similarly if Mary parks on the end or second last from the end or third last from the end Tom has only one parking spot in sample space. But if Mary parks elsewhere, Tom has 2 parking space in sample space.

$P_3=\frac{6}{n}*\frac{1}{n-1}+\frac{n-6}{n}*\frac{2}{n-1}=\frac{2n-6}{n(n-1)}$

Total Probability that Mary and Tom are placed two places apart is given by adding these probabilities.

$P=P_1+P_2+P_3=\frac{2n-2+2n-4+2n-6}{n(n-1)}=\frac{6(n-2)}{n(n-1)}$