Difference between revisions of "2007 AMC 12A Problems/Problem 12"
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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> | <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 1== |
| − | The only | + | 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)== | ||
| + | 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>\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>). | ||
| + | |||
| + | Our first subcase of case 2 has a chance of <math>\frac 1{16}</math> (same reasoning as above). | ||
| + | |||
| + | 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 subcase of case 2 has a <math>\frac 18</math> chance, because the 1st, 2nd, and either 3rd or 4th flip matter. | ||
| + | |||
| + | 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 1{16} + \frac 1{16} + \frac 14 + \frac 18 + \frac 18 = \boxed {\mathrm{(E )} \frac 58\ }.</math> | ||
| + | |||
| + | ~Dynosol | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/a2ZPFLkRrK4 | ||
| + | |||
| + | ~savannahsolver | ||
| + | |||
| + | ==See also== | ||
| + | {{AMC12 box|year=2007|ab=A|num-b=11|num-a=13}} | ||
| + | {{AMC10 box|year=2007|ab=A|num-b=15|num-a=17}} | ||
| + | |||
| + | [[Category:Introductory Number Theory Problems]] | ||
| + | {{MAA Notice}} | ||
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 
and 
, not necessarily distinct, are chosen independently and at random from 0 to 2007, inclusive. What is the probability that 
is even?
Solution 1
The only time when is even is when 
and 
are of the same parity. The chance of being odd is 
, since the only way to have be odd is to have both 
and be odd. As a result, has a 
probability of being even. also has a 
chance of being odd and a 
chance of being even. Therefore, the probability that will be even is 
.
Solution 2 (casework)
If we don't know our parity rules, we can check and see that is only even when and are of the same parity (as stated above). From here, we have two cases.
Case 1: 
(which must be 
). The probability for this to occur is 
, because each integer has a 
chance of being odd.
Case 2: 
(which occurs in 4 cases: 
), (
) (alternating of any kind), and (
) with its reverse, (
).
Our first subcase of case 2 has a chance of 
(same reasoning as above).
Our second subcase of case 2 has a chance, since only the 2nd and 4th flip matter (or 1st and 3rd).
Our third subcase of case 2 has a 
chance, because the 1st, 2nd, and either 3rd or 4th flip matter.
Our fourth subcase of case 2 has a chance, because it's the same, just reversed.
We sum these, and get our answer of 
~Dynosol
Video Solution
~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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
