2007 AMC 12A Problem 12

Question

Integers

a,

b,

c,

and

d,

not necessarily distinct, are chosen independently and at random from

0

to

2007,

inclusive. What is the probability that

ad-bc

is even?

\dfrac{3}{8}

\dfrac{7}{16}

\dfrac{1}{2}

\dfrac{9}{16}

\dfrac{5}{8}

ad-bc

is even exactly when

ad

and

bc

have the same parity

A product like

ad

is odd with probability

\tfrac14

and even with probability

\tfrac34

Exactly half of the integers from

0

to

2007

are odd.

A product

ad

is odd only when both factors are odd, with probability

\tfrac12\cdot\tfrac12=\tfrac14,

and even with probability

\tfrac34.

The same holds for

bc.

Then

ad-bc

is even when both products are odd or both are even:

\tfrac14\cdot\tfrac14+\tfrac34\cdot\tfrac34=\tfrac{10}{16}=\tfrac58.

Thus, the correct answer is E.

Problem 12 in other years