AMC 10 · 2005 · #9
Grade 7 probabilityOne fair die has faces 1,1,2,2,3,3 and another has faces 4,4,5,5,6,6. The dice are rolled and the numbers on the top faces are added. What is the probability that the sum will be odd?
Pick an answer.
AMC 10 2005 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: Die A has faces $1,1,2,2,3,3$ and die B has faces $4,4,5,5,6,6$. Both are rolled and the two top numbers are added. Find the probability that this sum is odd.
Givens: Die A shows $1,1,2,2,3,3$ (each of $1,2,3$ appears twice); Die B shows $4,4,5,5,6,6$ (each of $4,5,6$ appears twice); Both dice are fair, so each of the six faces is equally likely; The score is the sum of the two top faces; Answer choices: (A) $\frac{1}{3}$, (B) $\frac{4}{9}$, (C) $\frac{1}{2}$, (D) $\frac{5}{9}$, (E) $\frac{2}{3}$
Unknowns: The probability that the sum of the two top faces is odd
Understand
Restated: Die A has faces $1,1,2,2,3,3$ and die B has faces $4,4,5,5,6,6$. Both are rolled and the two top numbers are added. Find the probability that this sum is odd.
Givens: Die A shows $1,1,2,2,3,3$ (each of $1,2,3$ appears twice); Die B shows $4,4,5,5,6,6$ (each of $4,5,6$ appears twice); Both dice are fair, so each of the six faces is equally likely; The score is the sum of the two top faces; Answer choices: (A) $\frac{1}{3}$, (B) $\frac{4}{9}$, (C) $\frac{1}{2}$, (D) $\frac{5}{9}$, (E) $\frac{2}{3}$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #2 Make a Systematic List, #3 Eliminate Possibilities
A sum is odd exactly when one addend is odd and the other is even, so Tool #7 (Identify Subproblems) splits the event into two clean, non-overlapping cases: (A odd, B even) and (A even, B odd). Tool #2 (Make a Systematic List) does the counting each case needs — tally how many of each die's six faces are odd versus even to get the single-die probabilities. Tool #3 (Eliminate Possibilities) is a quick backstop: since die A is odd more often than not while die B is even more often than not, an odd sum is a bit more likely than $\frac{1}{2}$, which already points past choices (A), (B), and (C).
Execute — Answer: D
7.SP.C.7 Step 1 Count odd and even faces
- Sort each die's six faces by parity.
- Die A's faces $1,1,3,3$ are odd and $2,2$ are even, so $4$ of $6$ faces are odd.
- Die B's faces $5,5$ are odd and $4,4,6,6$ are even, so only $2$ of $6$ faces are odd.
- Turn each count into a probability out of the $6$ equally likely faces.
💡 With equally likely faces, a probability is just favorable faces over all six faces.
7.SP.C.8 Step 2 Split the odd sum into cases
- A sum is odd only when the two numbers have different parity: odd $+$ even $=$ odd, and even $+$ odd $=$ odd, while odd $+$ odd and even $+$ even are both even.
- So the sum is odd in exactly two situations — die A odd with die B even, or die A even with die B odd.
- These two cases cannot happen at the same time, so their probabilities can be added later.
💡 Odd plus even is odd, so exactly one of the two rolls must be odd.
5.NF.B.4 Step 3 Find each case's probability
- The two rolls are independent, so within a case the single-die probabilities multiply.
- Case 1 (A odd, B even) is $\frac{2}{3}\times\frac{2}{3}=\frac{4}{9}$.
- Case 2 (A even, B odd) is $\frac{1}{3}\times\frac{1}{3}=\frac{1}{9}$.
💡 Independent events happening together means multiplying their probabilities.
4.NF.B.3 Step 4 Add the two cases
- Because the two cases are mutually exclusive, add their probabilities to get the total chance of an odd sum.
- The denominators already match, so add the numerators: $4+1=5$ over $9$.
- That gives $\frac{5}{9}$, which is choice (D).
💡 For separate, non-overlapping cases, the chances simply add up.
7.SP.C.7 Sort each die's six faces by parity. Die A's faces $1,1,3,3$ are odd and $2,2$ a 7.SP.C.8 A sum is odd only when the two numbers have different parity: odd $+$ even $=$ o 5.NF.B.4 The two rolls are independent, so within a case the single-die probabilities mul 4.NF.B.3 Because the two cases are mutually exclusive, add their probabilities to get the Review
Reasonableness: The result $\frac{5}{9}$ is a valid probability between $0$ and $1$, and it is just over $\frac{1}{2}$ — sensible, since die A leans odd while die B leans even, making the mismatched (odd sum) pairing slightly favored. A full check confirms it: an even sum needs both odd ($\frac{2}{3}\cdot\frac{1}{3}=\frac{2}{9}$) or both even ($\frac{1}{3}\cdot\frac{2}{3}=\frac{2}{9}$), totaling $\frac{4}{9}$, and $\frac{5}{9}+\frac{4}{9}=1$ as it must. Choice (C) $\frac{1}{2}$ is the trap for anyone who assumes odd and even sums are equally likely; here the loaded parities break that symmetry.
Alternative: Count outcomes directly with a grid (Tool #2). Each die has $3$ distinct values each appearing twice, so treat it as a $3\times3$ table of value-pairs, each equally likely. Odd sums come from A$\in\{1,3\}$ with B$=5$, or A$=2$ with B$\in\{4,6\}$: that is $2\cdot1+1\cdot2=4$ odd-sum pairs... but weight by the doubled faces — better yet, use the $6\times6=36$ equally likely face pairs. Odd sums: $4$ odd A-faces $\times$ $4$ even B-faces $=16$, plus $2$ even A-faces $\times$ $2$ odd B-faces $=4$, giving $20$ of $36=\frac{5}{9}$, the same (D).
CCSS standards used (min grade 7)
7.SP.C.7Develop probability models and use them to find probabilities of events (Turning the count of odd/even faces on each die into single-die probabilities over six equally likely outcomes.)7.SP.C.8Find probabilities of compound events using organized lists, tables, and simulation (Recognizing the odd sum as a compound event and splitting it into the two disjoint parity cases.)5.NF.B.4Apply and extend understanding of multiplication to multiply a fraction by a fraction (Multiplying independent single-die probabilities within each case, e.g. $\frac{2}{3}\cdot\frac{2}{3}=\frac{4}{9}$.)4.NF.B.3Understand a fraction with numerator greater than one as sum of unit fractions (Adding the two mutually exclusive case probabilities $\frac{4}{9}+\frac{1}{9}=\frac{5}{9}$.)
⭐ A sum is odd only when exactly one die is odd, so multiply the matching odd/even chances for each case and add: $\frac{4}{9}+\frac{1}{9}=\frac{5}{9}$.
⭐ A sum is odd only when exactly one die is odd, so multiply the matching odd/even chances for each case and add: $\frac{4}{9}+\frac{1}{9}=\frac{5}{9}$.
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