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

 
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==Problem==
 
==Problem==
Integers a, b, c, and d, not necessarily distinct, are chosen independantly and at random from 0 to 2007, inclusive. What is the probability that ad-bc is even?
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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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<math>\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</math>
  
 
==Solution==
 
==Solution==
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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>\displaystyle \left(\frac 14\right)^2+\left(\frac 34\right)^2=\frac 58\ \mathrm{(E)}</math>.
  
The only times when ad-bc is even is when ad and bc are of the same [[parity]]. The chance of ad being odd (1/2)*(1/2)=1/4, and has a 3/4 probability that it will be even. Therefore, the probability that ad-bc will be even is (1/4)^2+(3/4)^2=5/8
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==See also==
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{{AMC12 box|year=2007|ab=A|num-b=11|num-a=13}}
  
==See also==
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[[Category:Introductory Combinatorics Problems]]
* [[2007 AMC 12A Problems/Problem 11 | Previous problem]]
 
* [[2007 AMC 12A Problems/Problem 13 | Next problem]]
 
* [[2007 AMC 12A Problems]]
 

Revision as of 15:33, 10 September 2007

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 $\displaystyle \left(\frac 14\right)^2+\left(\frac 34\right)^2=\frac 58\ \mathrm{(E)}$.

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
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