Parking problems

Probability Level 4

A parking lot has 10 empty spaces in a row.

6 cars arrive, each of which fills exactly 1 parking spot, chosen at random from among the available spaces. Robbie then arrives in his pick-up truck, which requires 2 empty adjacent spaces to park.

If the probability that Robbie will be able to park is $\frac{a}{b},$ where $a$ and $b$ are coprime positive integers, then what is $a+b?$

The answer is 11.

1 solution

Andy Hayes
Dec 12, 2016

One can establish a bijection between the available parking spaces and the combinations of 4 objects out of 10. Each parking space corresponds to one of 10 distinct objects. Each empty parking space corresponds to one of the 4 objects selected as the combination.

First consider the arrangements in which Robbie will not be able to park. The 4 empty spaces will always be adjacent to filled spaces. This can be represented with stars and bars :

$\_\star\_\star\_\star\_\star\_\star\_\star\_$

In the above diagram, the stars represent the placement of the cars, and the spaces represent where empty spaces can go between them. There are 7 spots that an empty space can go, so the number of ways to arrange the cars such that there are no adjacent empty spaces is:

$\binom{7}{4}=35.$

Then, the total number of ways to select 4 empty spaces out of 10 is:

$\binom{10}{4}=210.$

Thus, the probability that Robbie will be able to park is:

$1-\frac{35}{210}=\frac{5}{6}.$

So, $a+b=\boxed{11}.$

Very nice! For some reason, it feels to me like thinking about it as parking spaces made me want to move pieces of paper around, instead of writing out the stars and bars argument. I think that's why this problem has such a low correct rate, despite having an obvious (after-the-fact) solution.

Calvin Lin Staff - 4 years, 6 months ago

@Andy Hayes @Calvin Lin Yes indeed, very neatly works out into the stars and bars mould. Neat explanation!

Siva Bathula - 3 years, 11 months ago

Parking problems

The answer is 11.

1 solution

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