AMC 10 · 2004 · #10
Grade 7 probabilityCoin A is flipped three times and coin B is flipped four times. What is the probability that the number of heads obtained from flipping the two fair coins is the same?
Pick an answer.
AMC 10 2004 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: A fair coin $A$ is flipped $3$ times and a fair coin $B$ is flipped $4$ times. Find the probability that coin $A$ shows the same number of heads as coin $B$.
Givens: Coin $A$ is fair and is flipped exactly $3$ times.; Coin $B$ is fair and is flipped exactly $4$ times.; The two coins are flipped independently of each other.; Answer choices: (A) $\frac{19}{128}$, (B) $\frac{23}{128}$, (C) $\frac14$, (D) $\frac{35}{128}$, (E) $\frac12$
Unknowns: The probability that the number of heads from coin $A$ equals the number of heads from coin $B$.
Understand
Restated: A fair coin $A$ is flipped $3$ times and a fair coin $B$ is flipped $4$ times. Find the probability that coin $A$ shows the same number of heads as coin $B$.
Givens: Coin $A$ is fair and is flipped exactly $3$ times.; Coin $B$ is fair and is flipped exactly $4$ times.; The two coins are flipped independently of each other.; Answer choices: (A) $\frac{19}{128}$, (B) $\frac{23}{128}$, (C) $\frac14$, (D) $\frac{35}{128}$, (E) $\frac12$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #2 Make a Systematic List
"Same number of heads" is not one event but several rolled together: both show $0$, both show $1$, both show $2$, or both show $3$. That is a signal to break the problem into cases (Tool #7) — one subproblem per shared head-count — and add the results, since the cases cannot happen at the same time. Inside each case the coins are independent, so the count of matching flip-sequences is (ways $A$ gets that many heads) $\times$ (ways $B$ gets that many heads). A systematic list (Tool #2) sweeping $k = 0, 1, 2, 3$ makes sure no case is skipped or double-counted. Working in whole counts of equally likely sequences, then dividing by the total at the end, keeps the arithmetic clean.
Execute — Answer: D
7.SP.C.7 Step 1 Count all equally likely outcomes
- Coin $A$'s $3$ flips give $2\times2\times2 = 8$ equally likely head/tail sequences.
- Coin $B$'s $4$ flips give $2\times2\times2\times2 = 16$.
- The two coins are flipped independently, so each of $A$'s $8$ sequences can pair with each of $B$'s $16$ sequences: $8\times16 = 128$ equally likely combined outcomes.
- That $128$ is the bottom of every probability here.
💡 When each flip is a fair coin, every full sequence is equally likely, so probability is just a count of favorable sequences over the total.
7.SP.C.8 Step 2 List the head-counts that can match
- Coin $A$ can only show $0$ to $3$ heads, while coin $B$ can show $0$ to $4$.
- For the counts to be equal, the shared value must be reachable by both coins, so the only matches are $0$, $1$, $2$, and $3$ heads.
- Coin $B$ showing $4$ heads can never be matched, because coin $A$ has only $3$ flips.
- So there are exactly four separate cases to handle.
💡 Two amounts can only be equal at a value both are able to reach, so the extra fourth head on coin $B$ is automatically out.
7.SP.C.8 Step 3 Count matching sequences in each case
- For a target of $k$ heads, count how many of $A$'s sequences have exactly $k$ heads and how many of $B$'s do, then multiply because the coins are independent.
- The number of ways to place $k$ heads among $3$ flips is $\binom{3}{k}$, and among $4$ flips is $\binom{4}{k}$.
- Sweeping $k = 0,1,2,3$: $k=0$ gives $1\times1 = 1$; $k=1$ gives $3\times4 = 12$; $k=2$ gives $3\times6 = 18$; $k=3$ gives $1\times4 = 4$.
💡 For independent coins the favorable pairs are found by multiplying each coin's own count of ways.
7.SP.C.8 Step 4 Add the cases and form the probability
- The four cases cannot overlap — the coins match at exactly one head-count — so add their sequence counts: $1 + 12 + 18 + 4 = 35$ favorable combined outcomes.
- Divide by the $128$ equally likely total to get $\frac{35}{128}$.
- Since $35 = 5\times7$ shares no factor with $128 = 2^7$, the fraction is already in lowest terms.
- This is choice (D).
💡 Cases that cannot happen together add up, so the favorable counts stack into one numerator over the shared total.
7.SP.C.7 Coin $A$'s $3$ flips give $2\times2\times2 = 8$ equally likely head/tail sequenc 7.SP.C.8 Coin $A$ can only show $0$ to $3$ heads, while coin $B$ can show $0$ to $4$. For 7.SP.C.8 For a target of $k$ heads, count how many of $A$'s sequences have exactly $k$ he 7.SP.C.8 The four cases cannot overlap — the coins match at exactly one head-count — so a Review
Reasonableness: The value $\frac{35}{128} \approx 0.27$ is a little above $\frac14$, which fits the picture: matching is fairly common but far from guaranteed, so an answer near a quarter is believable while $\frac12$ (E) would be far too high. The choices also expose the traps — rounding to the clean-looking $\frac14$ gives (C), and slips in the case counts land on the near-miss numerators $19$ and $23$ over $128$. Only the exact count $1+12+18+4 = 35$ produces (D).
Alternative: Count the favorable sequences with a symmetry trick instead of case-by-case. Because $\binom{4}{k} = \binom{4}{4-k}$, the sum $\sum_k \binom{3}{k}\binom{4}{k}$ equals $\sum_k \binom{3}{k}\binom{4}{4-k}$. That last sum is Vandermonde's identity: it counts the ways to choose $4$ heads out of all $3+4 = 7$ coins, namely $\binom{7}{4} = 35$. The same $35$ over $128$ confirms $\frac{35}{128}$ and choice (D).
CCSS standards used (min grade 7)
7.SP.C.7Develop probability models and use them to find probabilities of events (Treating every flip-sequence as equally likely so probability becomes favorable outcomes over the total $128$.)7.SP.C.8Find probabilities of compound events using organized lists, tables, and simulation (Splitting "same number of heads" into the cases $k=0,1,2,3$, counting each with $\binom{3}{k}\binom{4}{k}$, and adding them.)
⭐ When two coin batches must match, handle one shared head-count at a time: multiply each batch's ways within a case, add the cases, and put it over all the equally likely outcomes.
⭐ When two coin batches must match, handle one shared head-count at a time: multiply each batch's ways within a case, add the cases, and put it over all the equally likely outcomes.
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