# Difference between revisions of "2007 AMC 12A Problems/Problem 12"

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The only times when <math>ad-bc</math> is even is when <math>ad</math> and <math>bc</math> are of the same [[parity]]. The chance of <math>ad</math> being odd is <math>\frac 12 \cdot \frac 12 = \frac 14</math>, so it has a <math>\frac 34</math> probability of being even. Therefore, the probability that <math>ad-bc</math> will be even is <math>\left(\frac 14\right)^2+\left(\frac 34\right)^2=\frac 58\ \mathrm{(E)}</math>. | The only times when <math>ad-bc</math> is even is when <math>ad</math> and <math>bc</math> are of the same [[parity]]. The chance of <math>ad</math> being odd is <math>\frac 12 \cdot \frac 12 = \frac 14</math>, so it has a <math>\frac 34</math> probability of being even. Therefore, the probability that <math>ad-bc</math> will be even is <math>\left(\frac 14\right)^2+\left(\frac 34\right)^2=\frac 58\ \mathrm{(E)}</math>. | ||

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## Revision as of 18:14, 28 May 2017

*The following problem is from both the 2007 AMC 12A #12 and 2007 AMC 10A #16, so both problems redirect to this page.*

## Problem

Integers and , not necessarily distinct, are chosen independently and at random from 0 to 2007, inclusive. What is the probability that is even?

## Solution

The only times when is even is when and are of the same parity. The chance of being odd is , so it has a probability of being even. Therefore, the probability that will be even is .