Difference between revisions of "2007 AMC 12A Problems/Problem 12"
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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 integer 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>). | ||
Revision as of 13:25, 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 $(\left\frac 12\right)^4 = \frac 1{16}$ (Error compiling LaTeX. Unknown error_msg), 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 case has a chance of 
(same reasoning as above).
Our second case has a chance, since only the 2nd and 4th flip matter (or 1st and 3rd).
Our third case has a 
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.
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. 
