AMC 8 · 2014 · #12

Easy mode Grade 7

Problem

A magazine prints $3$ photos of $3$ famous people, all grown up.

Next to them, the magazine prints $3$ baby photos of the same $3$ people. But the baby photos are not labeled, so you don't know which baby grew up to be which person.

A reader has to match each famous person to their own baby photo. Imagine the reader just guesses randomly.

What is the chance that the reader matches all $3$ baby photos correctly?

$\textbf{(A) }\frac{1}{9}\qquad\textbf{(B) }\frac{1}{6}\qquad\textbf{(C) }\frac{1}{4}\qquad\textbf{(D) }\frac{1}{3}\qquad\textbf{(E) }\frac{1}{2}$

Pick an answer.

(A)

$frac{1}{9}$

(B)

$frac{1}{6}$

(C)

$frac{1}{4}$

(D)

$frac{1}{3}$

(E)

$frac{1}{2}$

View mode:

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

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

#9 Solve an Easier Problem 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 $2 \times 1 = 2$.

$$2 \times 1 = 2 \text{ pairings}$$

💡 Trying the small version first checks that 'multiply the choices' gives the same answer as actually listing them.

#2 Make a Systematic List 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 $3 \times 2 \times 1$.

$$3 \times 2 \times 1 = 6 \text{ pairings}$$

💡 An organized list with a fixed ordering rule guarantees we hit every case once and only once.

#2 Make a Systematic List 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.

$$\text{favorable outcomes} = 1$$

💡 There is only one way to be 'all right,' but many ways to be partly wrong.

#2 Make a Systematic List 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 $=\dfrac{\text{favorable outcomes}}{\text{total outcomes}}$.

$$P(\text{all correct}) = \dfrac{1}{6} \;\Rightarrow\; \textbf{(B)}$$

💡 When every outcome is equally likely, probability is just 'good cases over all cases.'

[1] #9 4.OA.A.3 Warm up with the easier $2$-celebrity case. Label the celebrities $1, 2$ and the

[2] #2 7.SP.C.8 Now list every possible matching for $3$ celebrities $1, 2, 3$ and their correct

[3] #2 7.SP.C.8 Count the pairings that are fully correct. Only the first one in the list, $ABC$

[4] #2 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.3 Solve 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.8 Find 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.7 Develop 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 $\tfrac{1}{6}$.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Review

CCSS standards used (min grade 7)

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