AMC 8 · 2000 · #21
Grade 7 probabilitycountingProblem
Keiko tosses one penny and Ephraim tosses two pennies. The probability that Ephraim gets the same number of heads that Keiko gets is
Pick an answer.
Toolkit + CCSS Solution
Understand
Restated: Keiko tosses $1$ fair penny. Ephraim tosses $2$ fair pennies. What is the probability that Ephraim's number of heads equals Keiko's number of heads?
Givens: Keiko makes $1$ toss, so her head count is $0$ or $1$; Ephraim makes $2$ tosses, so his head count is $0$, $1$, or $2$; Every penny is fair, so each of the $2^3 = 8$ length-$3$ H/T sequences is equally likely; Answer choices: (A) $\tfrac{1}{4}$, (B) $\tfrac{3}{8}$, (C) $\tfrac{1}{2}$, (D) $\tfrac{2}{3}$, (E) $\tfrac{3}{4}$
Unknowns: $P(\text{Ephraim's heads} = \text{Keiko's heads})$
Understand
Restated: Keiko tosses $1$ fair penny. Ephraim tosses $2$ fair pennies. What is the probability that Ephraim's number of heads equals Keiko's number of heads?
Givens: Keiko makes $1$ toss, so her head count is $0$ or $1$; Ephraim makes $2$ tosses, so his head count is $0$, $1$, or $2$; Every penny is fair, so each of the $2^3 = 8$ length-$3$ H/T sequences is equally likely; Answer choices: (A) $\tfrac{1}{4}$, (B) $\tfrac{3}{8}$, (C) $\tfrac{1}{2}$, (D) $\tfrac{2}{3}$, (E) $\tfrac{3}{4}$
Plan
Primary tool: #2 Make a Systematic List
Secondary: #7 Identify Subproblems
There are only $8$ equally likely outcomes for the three coins combined, so Tool #2 (Make a Systematic List) lets us write them all down and count the matches directly. Tool #7 (Identify Subproblems) gives the cleaner path: split by Keiko's result (her $0$ heads vs. her $1$ head), and for each case count how many of Ephraim's $4$ outcomes match. Two small counts beat one big enumeration.
Execute — Answer: B
7.SP.C.8 Step 1 - Count the sample space.
- Three independent fair tosses (Keiko's $1$ plus Ephraim's $2$) produce $2 \times 2 \times 2 = 8$ equally likely length-$3$ H/T sequences, each with probability $\tfrac{1}{8}$.
💡 Listing outcomes for a compound event is the Grade 7 standard for counting independent trials.
7.SP.C.8 Step 2 - Split into cases by Keiko's outcome.
- Either Keiko gets $0$ heads or Keiko gets $1$ head.
- The matching condition becomes 'Ephraim gets $0$ heads' in the first case and 'Ephraim gets $1$ head' in the second case.
- These two cases are disjoint.
💡 Breaking the event into Keiko's two possible outcomes turns one hard count into two easy ones.
7.SP.C.8 Step 3 - Case 1: Keiko gets $0$ heads, Ephraim gets $0$ heads.
- Keiko has $1$ way (T) and Ephraim has $1$ way (TT).
- That is $1 \cdot 1 = 1$ favorable sequence: $TTT$.
💡 With $0$ heads forced everywhere, only the all-tails sequence works.
7.SP.C.8 Step 4 - Case 2: Keiko gets $1$ head, Ephraim gets $1$ head.
- Keiko has $1$ way (H).
- Ephraim's $1$ head can fall in either of his $2$ positions: HT or TH.
- That is $1 \cdot 2 = 2$ favorable sequences: $HHT$ and $HTH$.
💡 Choosing which of Ephraim's $2$ positions is the head is the same systematic-list move as $\binom{2}{1} = 2$.
7.SP.C.7 Step 5 - Add the case counts and divide by the sample-space size.
- Total favorable sequences: $1 + 2 = 3$ out of $8$ equally likely sequences.
💡 Probability $=$ favorable outcomes $\div$ total equally likely outcomes — the Grade 7 uniform-model formula.
7.SP.C.8 Count the sample space. Three independent fair tosses (Keiko's $1$ plus Ephraim' 7.SP.C.8 Split into cases by Keiko's outcome. Either Keiko gets $0$ heads or Keiko gets $ 7.SP.C.8 Case 1: Keiko gets $0$ heads, Ephraim gets $0$ heads. Keiko has $1$ way (T) and 7.SP.C.8 Case 2: Keiko gets $1$ head, Ephraim gets $1$ head. Keiko has $1$ way (H). Ephra 7.SP.C.7 Add the case counts and divide by the sample-space size. Total favorable sequenc Review
Reasonableness: Sanity check by listing all $8$ sequences (Keiko first, then Ephraim's two): TTT, TTH, THT, THH, HTT, HTH, HHT, HHH. Counting heads in each gives matches at TTT ($0\!=\!0$), HTH ($1\!=\!1$), HHT ($1\!=\!1$) — exactly $3$ of $8$, so $P = \tfrac{3}{8}$. The magnitude also makes sense: a match is the most likely single outcome but far from certain, so a value between $\tfrac{1}{4}$ and $\tfrac{1}{2}$ is expected, and $\tfrac{3}{8} = 0.375$ fits.
Alternative: Tool #13 (Convert to Algebra) using the law of total probability. Let $K(n)$ and $E(n)$ be the probabilities that Keiko and Ephraim get exactly $n$ heads. Then $K(0) = K(1) = \tfrac{1}{2}$ and $E(0) = \tfrac{1}{4}$, $E(1) = \tfrac{1}{2}$, $E(2) = \tfrac{1}{4}$. Since Keiko and Ephraim are independent, $P(\text{match}) = K(0)E(0) + K(1)E(1) = \tfrac{1}{2} \cdot \tfrac{1}{4} + \tfrac{1}{2} \cdot \tfrac{1}{2} = \tfrac{1}{8} + \tfrac{2}{8} = \tfrac{3}{8}$. Same answer (B).
CCSS standards used (min grade 7)
7.SP.C.7Develop a probability model and use it to find probabilities of events (Treating each of the $8$ length-$3$ H/T sequences as equally likely, so probability $=$ favorable count $\div\, 8$.)7.SP.C.8Find probabilities of compound events using organized lists, tables, tree diagrams, and simulation (Splitting by Keiko's outcome and counting Ephraim's matching sequences ($1$ for $0$ heads, $2$ for $1$ head) to get $3$ favorable outcomes.)6.RP.A.3Use ratio and rate reasoning to solve real-world and mathematical problems (Expressing the probability as the ratio $\tfrac{3}{8}$ of favorable outcomes to total outcomes.)
⭐ With only $8$ outcomes for $3$ coins, split by Keiko's result and count Ephraim's matches in each case — $1 + 2 = 3$ matches out of $8$ gives $\tfrac{3}{8}$.
⭐ With only $8$ outcomes for $3$ coins, split by Keiko's result and count Ephraim's matches in each case — $1 + 2 = 3$ matches out of $8$ gives $\tfrac{3}{8}$.