AMC 10 · 2008 · #16
Grade 7 probabilityTwo fair coins are to be tossed once. For each head that results, one fair die is to be rolled. What is the probability that the sum of the die rolls is odd? (Note that if no die is rolled, the sum is 0.)
Pick an answer.
AMC 10 2008 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: Two fair coins are tossed. For every coin that lands heads, one fair die is rolled. Add up the numbers on the dice that were rolled (if no die is rolled, the total is $0$). Find the probability that this total is odd.
Givens: Two fair coins are tossed once, so each coin is heads or tails with probability $\frac{1}{2}$.; Each head triggers one roll of a fair six-sided die.; The 'sum' is the total of all the die rolls; if no die is rolled the sum is $0$.
Unknowns: The probability that the sum of the die rolls is odd.
Understand
Restated: Two fair coins are tossed. For every coin that lands heads, one fair die is rolled. Add up the numbers on the dice that were rolled (if no die is rolled, the total is $0$). Find the probability that this total is odd.
Givens: Two fair coins are tossed once, so each coin is heads or tails with probability $\frac{1}{2}$.; Each head triggers one roll of a fair six-sided die.; The 'sum' is the total of all the die rolls; if no die is rolled the sum is $0$.
Plan
Primary tool: #7 Identify Subproblems
Secondary: #2 Make a Systematic List, #5 Look for a Pattern
How many dice get rolled is itself random, so tool #7 (Identify Subproblems) leads: split the problem by the number of heads — $0$, $1$, or $2$ — because each case has a different, much simpler sub-question about parity. Tool #2 (Make a Systematic List) pins the chance of each case from the coin toss: $0$ heads and $2$ heads each happen $\frac{1}{4}$ of the time, $1$ head happens $\frac{1}{2}$ of the time. Tool #5 (Look for a Pattern) handles the two-dice case, where a clean parity fact — the sum of two dice is odd exactly when the two dice disagree in parity — gives $\frac{1}{2}$ without listing all $36$ outcomes. Weighting each case's odd-probability by how often the case occurs and adding gives the answer.
Execute — Answer: A
7.SP.C.8 Step 1 Split by number of heads
- The number of dice rolled equals the number of heads.
- Two fair coins give $0$ heads with probability $\frac{1}{4}$ (TT), $1$ head with probability $\frac{1}{2}$ (HT or TH), and $2$ heads with probability $\frac{1}{4}$ (HH).
- Treat these three cases separately, then combine.
💡 The randomness has two layers — first the coins decide how many dice, then the dice decide the sum — so sort by the coins first.
7.SP.C.7 Step 2 Zero heads: sum is 0
- If both coins are tails, no die is rolled and the sum is $0$.
- Zero is even, so this case can never make the sum odd.
- Its chance of an odd sum is $0$.
💡 No dice means the total is $0$, and $0$ is even, so an odd total is impossible here.
7.SP.C.7 Step 3 One head: one die
- With exactly one head, exactly one die is rolled.
- A fair die shows an odd number in three of six faces ($1,3,5$), so the sum (just that one die) is odd with probability $\frac{3}{6}=\frac{1}{2}$.
💡 Half of a die's faces are odd, so a single die is odd half the time.
7.SP.C.8 Step 4 Two heads: two dice
- With two heads, two dice are rolled and we need their sum to be odd.
- A sum of two whole numbers is odd exactly when one is odd and the other is even.
- Whatever the first die shows, the second die has the opposite parity with probability $\frac{1}{2}$.
- So the sum is odd with probability $\frac{1}{2}$, no need to list all $36$ outcomes.
💡 Two numbers add to an odd total only when their parities disagree, and the second die disagrees with the first exactly half the time.
5.NF.B.4 Step 5 Weight each case
- Multiply each case's chance of occurring by its chance of an odd sum.
- Zero heads: $\frac{1}{4}\times 0=0$.
- One head: $\frac{1}{2}\times\frac{1}{2}=\frac{1}{4}$.
- Two heads: $\frac{1}{4}\times\frac{1}{2}=\frac{1}{8}$.
💡 A case only helps as often as it actually happens, so scale each odd-chance by how likely the case is.
5.NF.A.1 Step 6 Add the cases
- Add the three weighted pieces to get the overall probability of an odd sum.
- Using a common denominator of $8$: $0+\frac{2}{8}+\frac{1}{8}=\frac{3}{8}$.
- So the probability is $\frac{3}{8}$, which is choice $(A)$.
💡 The three cases are separate ways to succeed, so their probabilities simply add up.
7.SP.C.8 The number of dice rolled equals the number of heads. Two fair coins give $0$ he 7.SP.C.7 If both coins are tails, no die is rolled and the sum is $0$. Zero is even, so t 7.SP.C.7 With exactly one head, exactly one die is rolled. A fair die shows an odd number 7.SP.C.8 With two heads, two dice are rolled and we need their sum to be odd. A sum of tw 5.NF.B.4 Multiply each case's chance of occurring by its chance of an odd sum. Zero heads 5.NF.A.1 Add the three weighted pieces to get the overall probability of an odd sum. Usin Review
Reasonableness: The answer $\frac{3}{8}$ is below $\frac{1}{2}$, which makes sense: the $0$-heads case (probability $\frac{1}{4}$) can never give an odd sum, dragging the total down from the $\frac{1}{2}$ you would get if a die were always rolled. It is a legal probability between $0$ and $1$, and it matches choice $(A)$. As a sanity check, the chance of an even sum is $1-\frac{3}{8}=\frac{5}{8}$, exactly choice $(D)$; the extra weight on 'even' comes entirely from the no-die case, confirming the arithmetic.
Alternative: Count the complement (tool #16). An even sum happens in the $0$-heads case (always, probability $\frac{1}{4}\times 1=\frac{1}{4}$), plus half of the $1$-head case ($\frac{1}{2}\times\frac{1}{2}=\frac{1}{4}$), plus half of the $2$-heads case ($\frac{1}{4}\times\frac{1}{2}=\frac{1}{8}$), totalling $\frac{1}{4}+\frac{1}{4}+\frac{1}{8}=\frac{5}{8}$. Then odd $=1-\frac{5}{8}=\frac{3}{8}$, the same result.
CCSS standards used (min grade 7)
7.SP.C.8Find probabilities of compound events using organized lists, tables, and simulation (Splitting the two-stage experiment into cases by number of heads and finding the two-dice odd-sum probability.)7.SP.C.7Develop probability models and use them to find probabilities of events (Assigning the odd-sum probability within each case ($0$ for no die, $\frac{1}{2}$ for one die).)5.NF.B.4Apply and extend understanding of multiplication to multiply a fraction by a fraction (Weighting each case's odd-sum chance by the probability that the case occurs.)5.NF.A.1Add and subtract fractions with unlike denominators (Adding the three weighted case contributions $0+\frac{1}{4}+\frac{1}{8}$ to get $\frac{3}{8}$.)
⭐ When the number of dice is itself random, split by how many dice get rolled, find the odd-sum chance in each case, and weight each by how often that case happens.
⭐ When the number of dice is itself random, split by how many dice get rolled, find the odd-sum chance in each case, and weight each by how often that case happens.
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