2007 AMC 12A Problems/Problem 12
- 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 |
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