AMC 8 · 2014 · #12
Grade 7 probabilityProblem
A magazine printed photos of three celebrities along with three photos of the celebrities as babies. The baby pictures did not identify the celebrities. Readers were asked to match each celebrity with the correct baby pictures. What is the probability that a reader guessing at random will match all three correctly?
Pick an answer.
AMC 8 2014 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 magazine shows $3$ celebrities and $3$ unlabeled baby photos. A reader pairs each celebrity with one baby photo by guessing. What is the probability that all three pairings are correct?
Givens: $3$ celebrities and $3$ baby photos, each photo belongs to exactly one celebrity; The reader matches them by random guessing (every pairing is equally likely); Answer choices: (A) $\tfrac{1}{9}$, (B) $\tfrac{1}{6}$, (C) $\tfrac{1}{4}$, (D) $\tfrac{1}{3}$, (E) $\tfrac{1}{2}$
Unknowns: The probability that the random guess matches all $3$ celebrity–baby pairs correctly
Understand
Restated: A magazine shows $3$ celebrities and $3$ unlabeled baby photos. A reader pairs each celebrity with one baby photo by guessing. What is the probability that all three pairings are correct?
Givens: $3$ celebrities and $3$ baby photos, each photo belongs to exactly one celebrity; The reader matches them by random guessing (every pairing is equally likely); Answer choices: (A) $\tfrac{1}{9}$, (B) $\tfrac{1}{6}$, (C) $\tfrac{1}{4}$, (D) $\tfrac{1}{3}$, (E) $\tfrac{1}{2}$
Plan
Primary tool: #2 Make a Systematic List
Secondary: #9 Solve an Easier Problem
There are only a handful of possible matchings, so Tool #2 (Make a Systematic List) lets us write out every possible guess in order and just count. To make sure the counting rule is right, we first try Tool #9 (Solve an Easier Problem) on the $2$-celebrity version — small enough to list by hand — and notice that the count is $2 \times 1 = 2$. The same idea extends to $3 \times 2 \times 1 = 6$ for the real problem. Once we have the $6$ orderings, exactly one is the fully-correct match, so the probability is $\tfrac{1}{6}$.
Execute — Answer: B
4.OA.A.3 Step 1 - Warm up with the easier $2$-celebrity case.
- Label the celebrities $1, 2$ and the correct baby photos $A, B$ (so $1 \leftrightarrow A$, $2 \leftrightarrow B$ is the right answer).
- List every way to pair them: $(1\!\to\!A,\, 2\!\to\!B)$ and $(1\!\to\!B,\, 2\!\to\!A)$.
- That is $2$ orderings, matching $2 \times 1 = 2$.
💡 Trying the small version first checks that 'multiply the choices' gives the same answer as actually listing them.
7.SP.C.8 Step 2 - Now list every possible matching for $3$ celebrities $1, 2, 3$ and their correct baby photos $A, B, C$.
- Order the list by what celebrity $1$ gets, then by what celebrity $2$ gets: $ABC,\ ACB,\ BAC,\ BCA,\ CAB,\ CBA$.
- That is $6$ different pairings — exactly $3 \times 2 \times 1$.
💡 An organized list with a fixed ordering rule guarantees we hit every case once and only once.
7.SP.C.8 Step 3 - Count the pairings that are fully correct.
- Only the first one in the list, $ABC$, matches every celebrity to their own baby photo.
- Every other ordering swaps at least one pair, so it cannot be all-correct.
💡 There is only one way to be 'all right,' but many ways to be partly wrong.
7.SP.C.7 Step 4 - Since the reader is guessing at random, every one of the $6$ pairings is equally likely.
- Apply the basic probability rule: probability $=\dfrac{\text{favorable outcomes}}{\text{total outcomes}}$.
💡 When every outcome is equally likely, probability is just 'good cases over all cases.'
4.OA.A.3 Warm up with the easier $2$-celebrity case. Label the celebrities $1, 2$ and the 7.SP.C.8 Now list every possible matching for $3$ celebrities $1, 2, 3$ and their correct 7.SP.C.8 Count the pairings that are fully correct. Only the first one in the list, $ABC$ 7.SP.C.7 Since the reader is guessing at random, every one of the $6$ pairings is equally Review
Reasonableness: A probability of $\tfrac{1}{6} \approx 16.7\%$ feels right: the reader has to nail three guesses in a row with no information, and there are $6$ ways to scramble three items. Compared to $\tfrac{1}{2}$ (a coin flip) it is much smaller, and compared to $\tfrac{1}{9}$ it is bigger — which makes sense because $\tfrac{1}{9}$ would assume the three guesses are independent ($\tfrac{1}{3} \times \tfrac{1}{3} \times \tfrac{1}{3}$), but they are not: once celebrity $1$ is matched, only $2$ photos are left for celebrity $2$, and then only $1$ for celebrity $3$.
Alternative: Tool #6 (Guess and Check) on the answer choices: independent-guesses logic would give $\tfrac{1}{3} \cdot \tfrac{1}{3} \cdot \tfrac{1}{3} = \tfrac{1}{27}$ (not listed) — a clue that the guesses are not independent. Picking-without-replacement logic gives $\tfrac{1}{3} \cdot \tfrac{1}{2} \cdot \tfrac{1}{1} = \tfrac{1}{6}$, which is exactly choice (B). The other choices ($\tfrac{1}{9}, \tfrac{1}{4}, \tfrac{1}{3}, \tfrac{1}{2}$) do not match the $3! = 6$ denominator.
CCSS standards used (min grade 7)
4.OA.A.3Solve multistep word problems using the four operations (Confirming on the easier $2$-celebrity warm-up that 'multiply the choices' ($2 \times 1 = 2$) gives the same total as listing the pairings by hand.)7.SP.C.8Find probabilities of compound events using organized lists, tables, and tree diagrams (Listing all $6$ celebrity-to-baby pairings in order and counting both the total ($6$) and the favorable ($1$) outcomes.)7.SP.C.7Develop a probability model and use it to find probabilities of events (Applying the uniform-probability rule $P = \tfrac{\text{favorable}}{\text{total}} = \tfrac{1}{6}$ once every pairing is equally likely.)
⭐ List all $6$ ways to match $3$ celebrities to $3$ baby photos — only $1$ is fully right, so the probability is $\tfrac{1}{6}$.
⭐ List all $6$ ways to match $3$ celebrities to $3$ baby photos — only $1$ is fully right, so the probability is $\tfrac{1}{6}$.
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