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

(Solution 2 (casework))
(Solution 2 (casework))
 
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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.
  
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).
+
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.
+
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>
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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 = \frac 58\ \mathrm{(E)}</math>
  
 
~Dynosol
 
~Dynosol

Latest revision as of 14:38, 3 June 2021

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

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 $\frac 1{16} + \frac 1{16} + \frac 14 + \frac 18 + \frac 18 = \frac 58\ \mathrm{(E)}$

~Dynosol

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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