AMC 8 · 2015 · #11

Grade 7 probability

probability-basicpermutations-basicsystematic-enumeration systematic-enumeration ↑ Prerequisites: fraction-arithmeticmulti-digit-arithmetic

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Problem

In the small country of Mathland, all automobile license plates have four symbols. The first must be a vowel (A, E, I, O, or U), the second and third must be two different letters among the 21 non-vowels, and the fourth must be a digit (0 through 9). If the symbols are chosen at random subject to these conditions, what is the probability that the plate will read "AMC8"?

$\textbf{(A) } \frac{1}{22,050} \qquad \textbf{(B) } \frac{1}{21,000}\qquad \textbf{(C) } \frac{1}{10,500}\qquad \textbf{(D) } \frac{1}{2,100} \qquad \textbf{(E) } \frac{1}{1,050}$

Pick an answer.

(A)

$frac{1}{22,050}$

(B)

$frac{1}{21,000}$

(C)

$frac{1}{10,500}$

(D)

$frac{1}{2,100}$

(E)

$frac{1}{1,050}$

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Toolkit + CCSS Solution

Understand

Restated: A Mathland license plate has $4$ symbols: position $1$ is a vowel (A, E, I, O, U), positions $2$ and $3$ are two different non-vowel letters (from the $21$ consonants), and position $4$ is a digit ($0$-$9$). If every legal plate is equally likely, what is the probability that a random plate spells exactly "AMC8"?

Givens: Position $1$: one of $5$ vowels (A, E, I, O, U); Position $2$: one of $21$ non-vowel letters; Position $3$: one of the $21$ non-vowels, but different from position $2$; Position $4$: one of $10$ digits ($0$-$9$); All legal plates are equally likely; Answer choices: (A) $\tfrac{1}{22{,}050}$, (B) $\tfrac{1}{21{,}000}$, (C) $\tfrac{1}{10{,}500}$, (D) $\tfrac{1}{2{,}100}$, (E) $\tfrac{1}{1{,}050}$

Unknowns: The probability that a random legal plate is exactly "AMC8"

Understand

Plan

Primary tool: #7 Identify Subproblems

Secondary: #2 Make a Systematic List

Probability here is $\dfrac{1}{\text{total legal plates}}$, so the real work is counting the total. Tool #7 (Subproblems) splits one big count into $4$ tiny ones — "how many choices for each position?" — which then multiply. Tool #2 (Systematic List) is the bookkeeping behind each subproblem: list the vowels, the non-vowels, the leftover non-vowels, and the digits to make sure each count is honest. After multiplying, the probability falls out as $1$ over that total because "AMC8" is exactly one of the equally-likely plates.

Execute — Answer: B

#2 Make a Systematic List 7.SP.C.8 Step 1

Subproblem 1 — count the choices for position $1$.
The vowels are listed: A, E, I, O, U.
That is $5$ letters.

$$\text{choices}_1 = 5$$

💡 Listing the sample space for one slot is a Grade 7 "organized list" move for compound events.

#7 Identify Subproblems 4.OA.A.3 Step 2

Subproblem 2 — count the choices for position $2$.
The English alphabet has $26$ letters and $5$ of them are vowels, so the non-vowels number $26 - 5 = 21$.
Any of those $21$ may go in position $2$.

$$\text{choices}_2 = 26 - 5 = 21$$

💡 Subtracting the vowel count from $26$ is a Grade 4 multi-step subtraction inside a counting word problem.

#7 Identify Subproblems 7.SP.C.8 Step 3

Subproblem 3 — count the choices for position $3$.
It must be a non-vowel different from position $2$, so one of the $21$ non-vowels is already used up.

$$\text{choices}_3 = 21 - 1 = 20$$

💡 Treating position $3$ as a dependent slot (one option removed) is exactly the "compound event with the previous outcome subtracted" idea.

#2 Make a Systematic List 7.SP.C.8 Step 4

Subproblem 4 — count the choices for position $4$.
The digits $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$ give $10$ options.

$$\text{choices}_4 = 10$$

💡 Another organized-list count of the per-slot sample space.

#7 Identify Subproblems 4.OA.A.3 Step 5

Combine the four subproblems by the multiplication principle to get the total number of legal plates.
Each plate is one independent pick from each slot, so the counts multiply.

$$\text{total} = 5 \times 21 \times 20 \times 10 = (5 \times 20) \times (21 \times 10) = 100 \times 210 = 21{,}000$$

💡 Combining the per-slot counts by multiplication is the Grade 4 multi-step "how many in total" pattern.

#7 Identify Subproblems 7.SP.C.7 Step 6

Check that "AMC8" is itself a legal plate: A is a vowel (✓ pos $1$), M is a non-vowel (✓ pos $2$), C is a different non-vowel (✓ pos $3$), $8$ is a digit (✓ pos $4$).
So "AMC8" is one specific outcome out of the $21{,}000$ equally-likely legal plates, giving probability $\tfrac{1}{21{,}000}$.

$$P(\text{AMC8}) = \dfrac{1}{21{,}000} \;\Rightarrow\; \textbf{(B)}$$

💡 Forming probability as "favorable / total" on an equally-likely sample space is the Grade 7 probability-model definition.

[1] #2 7.SP.C.8 Subproblem 1 — count the choices for position $1$. The vowels are listed: A, E,

[2] #7 4.OA.A.3 Subproblem 2 — count the choices for position $2$. The English alphabet has $26$

[3] #7 7.SP.C.8 Subproblem 3 — count the choices for position $3$. It must be a non-vowel differ

[4] #2 7.SP.C.8 Subproblem 4 — count the choices for position $4$. The digits $0, 1, 2, 3, 4, 5,

[5] #7 4.OA.A.3 Combine the four subproblems by the multiplication principle to get the total nu

[6] #7 7.SP.C.7 Check that "AMC8" is itself a legal plate: A is a vowel (✓ pos $1$), M is a non-

Review

Reasonableness: The denominator must equal the total number of legal plates, and $5 \times 21 \times 20 \times 10$ visibly equals $21{,}000$. Choice (A) $22{,}050 = 5 \times 21 \times 21 \times 10$ would be the count if positions $2$ and $3$ were allowed to repeat — but the problem forbids repeats, so (A) is the classic trap. Choices (C), (D), (E) shrink the denominator, which would only happen if the favorable count were larger than $1$ — but "AMC8" is a single specific plate. So (B) is the only number consistent with both the constraints and "exactly one favorable plate".

Alternative: Tool #3 (Eliminate Possibilities) on the choices: rewrite each denominator as a product of factors that fit "vowels $\times$ consonants $\times$ consonants $\times$ digits". $22{,}050 = 5 \cdot 21 \cdot 21 \cdot 10$ assumes positions $2$ and $3$ may repeat — eliminated by the "different" constraint. $10{,}500 = 5 \cdot 21 \cdot 20 \cdot 5$ halves the digit count, which the problem does not. $2{,}100$ and $1{,}050$ drop another factor entirely. Only $21{,}000 = 5 \cdot 21 \cdot 20 \cdot 10$ matches every constraint, so (B) survives.

CCSS standards used (min grade 7)

4.OA.A.3 Solve multi-step word problems using four operations with whole numbers (Subtracting $26 - 5 = 21$ for the non-vowel count and multiplying the per-slot counts $5 \times 21 \times 20 \times 10 = 21{,}000$ for the total.)
7.SP.C.7 Develop probability models and use them to find probabilities of events (Treating every legal plate as equally likely and reading the probability of "AMC8" as $\tfrac{1}{\text{total plates}}$.)
7.SP.C.8 Find probabilities of compound events using organized lists, tables, and simulation (Counting the sample space slot-by-slot ($5$ vowels, $21$ non-vowels, $20$ remaining non-vowels, $10$ digits) and combining by the multiplication principle for compound events.)

⭐ Count each slot's options separately, multiply them all to get the total, then put $1$ over that — that's the Grade 7 way to handle a "random plate" probability.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Review

CCSS standards used (min grade 7)

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