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2007 AMC 12A Problems/Problem 12

Revision as of 13:35, 30 June 2022 by Agentmath (talk | contribs) (→‎Solution 2 (casework))
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 $a, b, c,$ and $d$, not necessarily distinct, are chosen independently and at random from 0 to 2007, inclusive. What is the probability that $ad-bc$ is even?

$\mathrm{(A)}\ \frac 38\qquad \mathrm{(B)}\ \frac 7{16}\qquad \mathrm{(C)}\ \frac 12\qquad \mathrm{(D)}\ \frac 9{16}\qquad \mathrm{(E)}\ \frac 58$

Solution 1

The only time when $ad-bc$ is even is when $ad$ and $bc$ are of the same parity. The chance of $ad$ being odd is $\frac 12 \cdot \frac 12 = \frac 14$, since the only way to have $ad$ be odd is to have both $a$ and $d$ be odd. As a result, $ad$ has a $\frac 34$ probability of being even. $bc$ also has a $\frac 14$ chance of being odd and a $\frac34$ chance of being even. Therefore, the probability that $ad-bc$ will be even is $\left(\frac 14\right)^2+\left(\frac 34\right)^2=\boxed {\mathrm{(E )} \frac 58\ }$.

Solution 2 (casework)

If we don't know our parity rules, we can check and see that $ad-bc$ is only even when $ad$ and $bc$ are of the same parity (as stated above). From here, we have two cases.

Case 1: $odd-odd$ (which must be $o \cdot o-o \cdot o$). The probability for this to occur is $\left(\frac 12\right)^4 = \frac 1{16}$, because each integer has a $\frac 12$ chance of being odd.

Case 2: $even-even$ (which occurs in 4 cases: $(e \cdot e-e \cdot e$), ($o \cdot e-o \cdot e$) (alternating of any kind), and ($e \cdot e-o \cdot e$) with its reverse, ($o \cdot e-e \cdot e$).

Our first subcase of case 2 has a chance of $\frac 1{16}$ (same reasoning as above).

Our second subcase of case 2 has a $\frac 14$ chance, since only the 2nd and 4th flip matter (or 1st and 3rd).

Our third subcase of case 2 has a $\frac 18$ chance, because the 1st, 2nd, and either 3rd or 4th flip matter.

Our fourth subcase of case 2 has a $\frac 18$ chance, because it's the same, just reversed.

We sum these, and get our answer of $\frac 1{16} + \frac 1{16} + \frac 14 + \frac 18 + \frac 18 = \boxed {\mathrm{(E )} \frac 58\ }$

~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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