2008 AMC 12B Problem 22

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

Small hint:

The

4

empty spaces are equally likely to be any

4

of the

16;

she fails exactly when no two empties are adjacent

Big hint:

The number of ways to choose

4

empty spaces among

16

with no two adjacent is

\binom{13}{4};

subtract that probability from

1

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.