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

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{{duplicate|[[2007 AMC 12A Problems|2007 AMC 12A #12]] and [[2007 AMC 10A Problems/Problem 16|2007 AMC 10A #16]]}}
 
==Problem==
 
==Problem==
 
Integers <math>a, b, c,</math> and <math>d</math>, not necessarily distinct, are chosen independently and at random from 0 to 2007, inclusive. What is the [[probability]] that <math>ad-bc</math> is [[even]]?
 
Integers <math>a, b, c,</math> and <math>d</math>, not necessarily distinct, are chosen independently and at random from 0 to 2007, inclusive. What is the [[probability]] that <math>ad-bc</math> is [[even]]?
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==Solution==
 
==Solution==
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>\displaystyle \left(\frac 14\right)^2+\left(\frac 34\right)^2=\frac 58\ \mathrm{(E)}</math>.
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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>.
  
 
==See also==
 
==See also==
 
{{AMC12 box|year=2007|ab=A|num-b=11|num-a=13}}
 
{{AMC12 box|year=2007|ab=A|num-b=11|num-a=13}}
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{{AMC10 box|year=2007|ab=A|num-b=15|num-a=17}}
  
 
[[Category:Introductory Combinatorics Problems]]
 
[[Category:Introductory Combinatorics Problems]]

Revision as of 18:50, 5 January 2008

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 $a, b, c,$ 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?

$\mathrm{(A)}\ \frac 38\qquad \mathrm{(B)}\ \frac 7{16}\qquad \mathrm{(C)}\ \frac 12\qquad \mathrm{(D)}\ \frac 9{16}\qquad \mathrm{(E)}\ \frac 58$

Solution

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

See also

2007 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions
2007 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions
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