The Parking Lot Problem
In the new Vrindavan Temple, a parking lot has 2020 parking spaces in a row.
There are 674 cars, each two parking spaces wide, arriving at the parking lot one by one. Each car parks in a pair of consecutive vacant spaces selected uniformly at random over all such pairs; for example, the first car can park in 2019 ways, all with equal probability. If no pair of consecutive vacant spaces remain when a car arrives, it leaves disappointingly. If the probability that all 674 cars successfully park can be expressed as where and are prime numbers, submit your answer as .
The answer is 676.
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1 solution
The numbers the parking spaces can take is 1, 2, . . . , 2020.
The first 673 cars will always be able to park;
The final car is ONLY UNABLE to park when the first 673 cars are in the maximally inefficient configuration, with cars parked in spaces 2 − 3, 5 − 6, . . . , 2018 − 2019.
Let us find the probability of the first 673 cars parked in this configuration.
The first car can park in 2019 ways, of which 673 are in the configuration. If the first car parks in one of those ways, the second car can park in 2016 ways, of which 672 are in the configuration.
Continuing this reasoning, the first 673 cars park in the maximally inefficient configuration with probability
(673/2019)(672/2016)(671/2013)....(1/3) = 1/(3^673)
The probability the last car can park is the complement of this probability
So a + b = 676.