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

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==Solution 1==
 
==Solution 1==
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The only time 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>, since the only way to have <math>ad</math> be odd is to have both <math>a</math> and <math>d</math> be odd. As a result, <math>ad</math> has a <math>\frac 34</math> probability of being even. <math>bc</math> also has a <math>\frac 14</math> chance of being odd and a <math>\frac34</math> chance 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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The only time 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>, since the only way to have <math>ad</math> be odd is to have both <math>a</math> and <math>d</math> be odd. As a result, <math>ad</math> has a <math>\frac 34</math> probability of being even. <math>bc</math> also has a <math>\frac 14</math> chance of being odd and a <math>\frac34</math> chance 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=\boxed {\mathrm{(E )} \frac 58\ }</math>.
  
 
==Solution 2 (casework)==
 
==Solution 2 (casework)==
 
If we don't know our parity rules, we can check and see that <math>ad-bc</math> is only even when <math>ad</math> and <math>bc</math> are of the same [[parity]] (as stated above). From here, we have two cases.
 
If we don't know our parity rules, we can check and see that <math>ad-bc</math> is only even when <math>ad</math> and <math>bc</math> are of the same [[parity]] (as stated above). From here, we have two cases.
  
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Case 1: <math>odd-odd</math> (which must be <math>o \cdot o-o \cdot o</math>). The probability for this to occur is <math>(\frac 12)^4 = \frac 1{16}</math>, because each flip has a <math>\frac 12</math> chance of being odd.
+
Case 1: <math>odd-odd</math> (which must be <math>o \cdot o-o \cdot o</math>). The probability for this to occur is <math>\left(\frac 12\right)^4 = \frac 1{16}</math>, because each integer has a <math>\frac 12</math> chance of being odd.
  
 
Case 2: <math>even-even</math> (which occurs in 4 cases: <math>(e \cdot e-e \cdot e</math>), (<math>o \cdot e-o \cdot e</math>) (alternating of any kind), and (<math>e \cdot e-o \cdot e</math>) with its reverse, (<math>o \cdot e-e \cdot e</math>).
 
Case 2: <math>even-even</math> (which occurs in 4 cases: <math>(e \cdot e-e \cdot e</math>), (<math>o \cdot e-o \cdot e</math>) (alternating of any kind), and (<math>e \cdot e-o \cdot e</math>) with its reverse, (<math>o \cdot e-e \cdot e</math>).
  
−
Our first case has a chance of <math>\frac 1{16}</math> (same reasoning as above).
+
Our first subcase of case 2 has a chance of <math>\frac 1{16}</math> (same reasoning as above).
  
−
Our second case has a <math>\frac 14</math> chance, since only the 2nd and 4th flip matter (or 1st and 3rd).
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Our second subcase of case 2 has a <math>\frac 14</math> chance, since only the 2nd and 4th flip matter (or 1st and 3rd).
  
−
Our third case has a <math>\frac 18</math> chance, because the 1st, 2nd, and either 3rd or 4th flip matter.
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Our third subcase of case 2 has a <math>\frac 18</math> chance, because the 1st, 2nd, and either 3rd or 4th flip matter.
  
−
Our fourth case has the same chance, because it's the same, just reversed. <math>\frac 18</math>
+
Our fourth subcase of case 2 has a <math>\frac 18</math> chance, because it's the same, just reversed.
  
−
We sum these, and get our answer of <math>\frac 58\ \mathrm{(E)}</math>
+
We sum these, and get our answer of <math>\frac 1{16} + \frac 1{16} + \frac 14 + \frac 18 + \frac 18 = \boxed {\mathrm{(E )} \frac 58\ }.</math>
  
 
~Dynosol
 
~Dynosol
 +
 +
==Video Solution==
 +
https://youtu.be/a2ZPFLkRrK4
 +
 +
~savannahsolver
  
 
==See also==
 
==See also==

Latest revision as of 07:59, 24 March 2023

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 1

The only time when $ad-bc$ is even is when $ad$ and $bc$ are of the same parity. The chance of $ad$ being odd is $\frac 12 \cdot \frac 12 = \frac 14$, since the only way to have $ad$ be odd is to have both $a$ and $d$ be odd. As a result, $ad$ has a $\frac 34$ probability of being even. $bc$ also has a $\frac 14$ chance of being odd and a $\frac34$ chance of being even. Therefore, the probability that $ad-bc$ will be even is $\left(\frac 14\right)^2+\left(\frac 34\right)^2=\boxed {\mathrm{(E )} \frac 58\ }$.

Solution 2 (casework)

If we don't know our parity rules, we can check and see that $ad-bc$ is only even when $ad$ and $bc$ are of the same parity (as stated above). From here, we have two cases.

Case 1: $odd-odd$ (which must be $o \cdot o-o \cdot o$). The probability for this to occur is $\left(\frac 12\right)^4 = \frac 1{16}$, because each integer has a $\frac 12$ chance of being odd.

Case 2: $even-even$ (which occurs in 4 cases: $(e \cdot e-e \cdot e$), ($o \cdot e-o \cdot e$) (alternating of any kind), and ($e \cdot e-o \cdot e$) with its reverse, ($o \cdot e-e \cdot e$).

Our first subcase of case 2 has a chance of $\frac 1{16}$ (same reasoning as above).

Our second subcase of case 2 has a $\frac 14$ chance, since only the 2nd and 4th flip matter (or 1st and 3rd).

Our third subcase of case 2 has a $\frac 18$ chance, because the 1st, 2nd, and either 3rd or 4th flip matter.

Our fourth subcase of case 2 has a $\frac 18$ chance, because it's the same, just reversed.

We sum these, and get our answer of $\frac 1{16} + \frac 1{16} + \frac 14 + \frac 18 + \frac 18 = \boxed {\mathrm{(E )} \frac 58\ }.$

~Dynosol

Video Solution

https://youtu.be/a2ZPFLkRrK4

~savannahsolver

See also

2007 AMC 12A (Problems • Answer Key • Resources)
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 (Problems • Answer Key • Resources)
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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