Question

A parking lot has $16$ spaces in a row. Twelve cars arrive, each of which requires one parking space, and their drivers choose their spaces at random from among the available spaces. Auntie Em then arrives in her SUV, which requires $2$ adjacent spaces. What is the probability that she is able to park?

$\dfrac{11}{20}$

$\dfrac{4}{7}$

$\dfrac{81}{140}$

$\dfrac{3}{5}$

$\dfrac{17}{28}$

Answer: E

Concepts:complementary counting combinations arrangements with restrictions

Difficulty rating: 2110

Solution:

After the $12$ cars park, $4$ spaces are empty, equally likely to be any $4$ of the $16,$ for $\binom{16}{4} = 1820$ equally likely sets.

Auntie Em fails exactly when no two empty spaces are adjacent. The number of ways to place $4$ non-adjacent empties among $16$ is $\binom{13}{4} = 715.$

So the probability she can park is $1 - \frac{715}{1820} = \frac{1105}{1820} = \frac{17}{28}.$

Thus, the correct answer is E.

2008 AMC 12B Problem 22

Answer: E

Difficulty rating: 2110

Solution:

Problem 22 in Other Years