Competition · AMC preparation · step 4 of 4

AMC 8 · 2000 · #21

Grade 7 probabilitycounting

Archetype Casework by Position / Vertex Type └ Sub-archetype Count of Type

probability-basiccombinations-basicsystematic-enumerationfraction-arithmetic caseworksystematic-enumerationtree-enumeration ↑ Prerequisites: probability-basiccombinations-basic

📏 Medium solution 💡 3 insights

Problem

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?

Pick an answer.

(A)

$\frac{1}{4}$

(B)

$\frac{3}{8}$

(C)

$\frac{1}{2}$

(D)

$\frac{2}{3}$

(E)

$\frac{3}{4}$

AMC 8 2000 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Make a Systematic List

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.

1STEP 1

Count the sample space

Three independent fair tosses give 2 × 2 × 2 = 8 equally likely H/T sequences, each of probability $\frac{1}{8}$ .

|sample space| = 2³ = 8

2STEP 2

Split into two cases

Split by Keiko's outcome: matching means Ephraim gets 0 heads when she does, or 1 head when she does — two disjoint cases.

P(match) = P(K = 0, E = 0) + P(K = 1, E = 1)

3STEP 3

Count the zero-heads case

Case 1 — both get 0 heads: Keiko 1 way (T), Ephraim 1 way (TT), giving 1 · 1 = 1 sequence (TTT).

Case 1 count = 1 · 1 = 1

4STEP 4

Count the one-head case

Case 2 — both get 1 head: Keiko 1 way (H), Ephraim's head in either slot (HT, TH), so 1 · 2 = 2 sequences (HHT, HTH).

Case 2 count = 1 · 2 = 2

5STEP 5

Add and divide by 8

Add the case counts and divide by the sample space: 1 + 2 = 3 favorable out of 8 equally likely sequences.

P(match) = (1 + 2)/8 = 3/8 → (B)

Answer

3/8

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 =

\frac{3}{8}

. The magnitude also makes sense: a match is the most likely single outcome but far from certain, so a value between

\frac{1}{4}

and

\frac{1}{2}

is expected, and

\frac{3}{8}

= 0.375 fits.

💡Key takeaway

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 $\frac{3}{8}$ .

Count the sample space
Split into two cases
Count the zero-heads case
Count the one-head case
Add and divide by 8

A parent dashboard for the family lives at sensimlab.com.

Count the sample space

Split into two cases

Count the zero-heads case

Count the one-head case

Add and divide by 8

More like this