Answer:
a) The total sample space of possible parking = n (n-1)
b)pr(at most 2 spaces) = 6/n
Step-by-step explanation:
a) Since Mary was the first person to arrive at the parking lot, she has n available parking spaces.
On arrival, Tom would meet a parking space that is already occupied by Mary, so he has (n-1) available parking spaces
The total sample space of possible parking = n (n-1)
b) Probability that the parking spaces selected by Mary and Tom are at most 2 spaces apart
Pr(at most 2) = pr(0 space) + pr(1 space) + pr(2 spaces)
The number of possible placings for 0 space = 2(n-0) = 2n
The factor 2 caters for the reversal in the order in which the vehicles are placed.
pr(0 space) = 2n/n(n-1) = 2/(n-1)
Number of possible placings for 1 space = 2(n-1)
pr(1 space) = 2(n-1)/n(n-1) = 2/n
The number of possible placings for 2 spaces = 2(n-2)
pr(2 spaces) = 2(n-2)/n(n-1)
pr(at most 2 spaces) = 2/n-1 + 2/n + 2(n-2)/n(n-1)
pr(at most 2 spaces) = [ 2n + 2(n-1) + 2(n-2)]/[n(n-1)]
pr(at most 2 spaces) = 6/n