Difference between revisions of "2007 AMC 12A Problems/Problem 12"
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==Solution 1== | ==Solution 1== | ||
− | 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. 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>. | + | 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>. |
==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. | ||
− | Case 1: odd-odd (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 | + | 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: even-even (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 | + | 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 |
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 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
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.