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 times 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>, so it 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 times 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>, so it 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>. | ||
Revision as of 01:10, 16 February 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 times when is even is when 
and 
are of the same parity. The chance of being odd is 
, so it has a 
probability 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: odd-odd (which must be 
). The probability for this to occur is 
, because each flip has a 
chance of being odd.
Case 2: even-even (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. 
