AMC 10 · 2007 · #16
Grade 7 probabilityIntegers 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?
Pick an answer.
AMC 10 2007 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Four integers $a$, $b$, $c$, $d$ are each picked at random and independently from the whole numbers $0$ through $2007$. Find the probability that $ad - bc$ comes out even.
Givens: $a$, $b$, $c$, $d$ are chosen independently and uniformly from the integers $0$ to $2007$ inclusive; The $2008$ integers from $0$ to $2007$ split into exactly $1004$ even numbers ($0,2,\dots,2006$) and $1004$ odd numbers ($1,3,\dots,2007$); Answer choices: (A) $\tfrac38$, (B) $\tfrac7{16}$, (C) $\tfrac12$, (D) $\tfrac9{16}$, (E) $\tfrac58$
Unknowns: The probability that $ad - bc$ is even
Understand
Restated: Four integers $a$, $b$, $c$, $d$ are each picked at random and independently from the whole numbers $0$ through $2007$. Find the probability that $ad - bc$ comes out even.
Givens: $a$, $b$, $c$, $d$ are chosen independently and uniformly from the integers $0$ to $2007$ inclusive; The $2008$ integers from $0$ to $2007$ split into exactly $1004$ even numbers ($0,2,\dots,2006$) and $1004$ odd numbers ($1,3,\dots,2007$); Answer choices: (A) $\tfrac38$, (B) $\tfrac7{16}$, (C) $\tfrac12$, (D) $\tfrac9{16}$, (E) $\tfrac58$
Plan
Primary tool: #9 Solve an Easier Related Problem
Secondary: #16 Change Focus / Count the Complement, #7 Identify Subproblems
The actual values from $0$ to $2007$ do not matter for whether $ad-bc$ is even; only the parity (even or odd) of each number matters. So Tool #9 (Solve an Easier Related Problem) shrinks a question about $2008$ possible values down to a coin-flip question: each variable is just 'even' or 'odd' with probability $\tfrac12$. Tool #16 (Change Focus) then aims at the parity of the two products $ad$ and $bc$ instead of the numbers themselves, since a product is odd only when both of its factors are odd. Tool #7 (Identify Subproblems) splits the goal into two clean pieces — the chance each product is even and the chance the two products match in parity — which multiply and add to the final probability.
Execute — Answer: E
7.SP.C.5 Step 1 Shrink to a parity question
- Whether $ad - bc$ is even depends only on whether each of $a,b,c,d$ is even or odd, not on its exact value.
- Count the pool: from $0$ to $2007$ there are $2008$ integers, of which $1004$ are even and $1004$ are odd.
- Since $1004$ and $1004$ are equal, each variable is even with probability $\tfrac{1004}{2008} = \tfrac12$ and odd with probability $\tfrac12$, and the four picks are independent.
- So the whole problem is really four fair even-or-odd flips.
💡 Only the parity of each number changes whether $ad-bc$ is even, and the range splits perfectly in half, so every variable is a fair even-or-odd coin.
4.OA.B.4 Step 2 Parity of each product
- Look at one product, say $ad$.
- A product of two integers is odd only when both factors are odd; if either factor is even, the product is even.
- So $ad$ is odd only when $a$ is odd and $d$ is odd, which happens with probability $\tfrac12 \cdot \tfrac12 = \tfrac14$.
- That leaves $ad$ even with probability $1 - \tfrac14 = \tfrac34$.
- The product $bc$ works identically: odd with probability $\tfrac14$, even with probability $\tfrac34$.
💡 A single even factor drags a product to even, so a product is odd only in the one case where both factors are odd.
7.SP.C.7 Step 3 When is the difference even?
- A difference $ad - bc$ is even exactly when $ad$ and $bc$ share the same parity — both even, or both odd.
- (If one is even and the other odd, the difference is odd.) So the target event breaks into two non-overlapping subcases: 'both products even' and 'both products odd'.
- Handle each with the probabilities from the previous step, then add them because the two subcases cannot happen at the same time.
💡 Subtracting two numbers cancels evenness only when both carry the same parity, so matching parities is the whole game.
7.SP.C.8 Step 4 Multiply and add the cases
- The parities of $ad$ and $bc$ are independent because they use different variables ($a,d$ versus $b,c$), so their probabilities multiply within each subcase.
- Both even: $\tfrac34 \cdot \tfrac34 = \tfrac{9}{16}$.
- Both odd: $\tfrac14 \cdot \tfrac14 = \tfrac{1}{16}$.
- Add the two disjoint subcases: $\tfrac{9}{16} + \tfrac{1}{16} = \tfrac{10}{16} = \tfrac58$.
- So the probability that $ad - bc$ is even is $\tfrac58$, which is choice (E).
💡 Independent parities multiply inside each case, and disjoint cases add, so $\tfrac{9}{16}+\tfrac1{16}$ collects every way the difference lands even.
7.SP.C.5 Whether $ad - bc$ is even depends only on whether each of $a,b,c,d$ is even or o 4.OA.B.4 Look at one product, say $ad$. A product of two integers is odd only when both f 7.SP.C.7 A difference $ad - bc$ is even exactly when $ad$ and $bc$ share the same parity 7.SP.C.8 The parities of $ad$ and $bc$ are independent because they use different variabl Review
Reasonableness: The answer $\tfrac58 = 0.625$ is a valid probability (between $0$ and $1$) and sits just above $\tfrac12$, which fits: 'even' products are far more common than 'odd' ones (each product is even $\tfrac34$ of the time), so the two products agreeing in parity should be a bit more likely than a coin flip. A cross-check by complement confirms it: $ad-bc$ is odd when the products differ in parity, with probability $\tfrac34\cdot\tfrac14 + \tfrac14\cdot\tfrac34 = \tfrac{3}{16}+\tfrac{3}{16} = \tfrac{6}{16} = \tfrac38$, and $\tfrac58 + \tfrac38 = 1$ as it must. Choice (A) $\tfrac38$ is exactly this complement (the 'odd' probability), a natural trap; (C) $\tfrac12$ would be right only if products were equally likely even or odd, which they are not.
Alternative: Count the complement directly instead. $ad - bc$ is odd only when one product is even and the other odd. That is two mirror cases: $ad$ even and $bc$ odd, or $ad$ odd and $bc$ even, each with probability $\tfrac34\cdot\tfrac14 = \tfrac{3}{16}$, totaling $\tfrac{6}{16} = \tfrac38$. Subtract from $1$: the probability of even is $1 - \tfrac38 = \tfrac58$, matching (E).
CCSS standards used (min grade 7)
7.SP.C.5Understand that the probability of a chance event is between 0 and 1 (Showing each variable is even or odd with probability exactly $\tfrac12$ because $0$ to $2007$ splits into $1004$ even and $1004$ odd.)4.OA.B.4Find all factor pairs and recognize multiples; determine prime or composite (Using the parity rule that a product is odd only when both factors are odd to get $P(ad\text{ odd}) = \tfrac14$.)7.SP.C.7Develop probability models and use them to find probabilities of events (Modeling the target event 'difference is even' as the two disjoint subcases where $ad$ and $bc$ share parity.)7.SP.C.8Find probabilities of compound events using organized lists, tables, and simulation (Multiplying independent parity probabilities within each case and adding the disjoint cases: $\tfrac{9}{16}+\tfrac1{16}=\tfrac58$.)
⭐ Forget the real numbers and track only even-or-odd: a difference is even when both products match in parity, and since a product is even $\tfrac34$ of the time, the chances stack to $\tfrac58$.
⭐ Forget the real numbers and track only even-or-odd: a difference is even when both products match in parity, and since a product is even $\tfrac34$ of the time, the chances stack to $\tfrac58$.
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