AMC 8 · 2017 · #10
Easy mode Grade 7Problem
Imagine a box with 5 cards inside. The cards are labeled 1, 2, 3, 4, and 5.
You reach in and pull out 3 cards at random, all at once. (You don't put any card back.)
Now look at the 3 numbers you pulled. We want the largest one to be exactly 4.
What is the probability of that?
Pick an answer.
AMC 8 2017 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 box has five cards numbered $1, 2, 3, 4, 5$. We pull out $3$ cards at the same time (no putting back). What is the probability that the biggest number on the cards we pulled is exactly $4$?
Givens: There are $5$ cards numbered $1, 2, 3, 4, 5$; We pick $3$ of the $5$ cards without replacement; Every set of $3$ cards is equally likely; Answer choices: (A) $\tfrac{1}{10}$, (B) $\tfrac{1}{5}$, (C) $\tfrac{3}{10}$, (D) $\tfrac{2}{5}$, (E) $\tfrac{1}{2}$
Unknowns: The probability that the largest of the three drawn cards equals $4$
Understand
Restated: A box has five cards numbered $1, 2, 3, 4, 5$. We pull out $3$ cards at the same time (no putting back). What is the probability that the biggest number on the cards we pulled is exactly $4$?
Givens: There are $5$ cards numbered $1, 2, 3, 4, 5$; We pick $3$ of the $5$ cards without replacement; Every set of $3$ cards is equally likely; Answer choices: (A) $\tfrac{1}{10}$, (B) $\tfrac{1}{5}$, (C) $\tfrac{3}{10}$, (D) $\tfrac{2}{5}$, (E) $\tfrac{1}{2}$
Plan
Primary tool: #2 Make a Systematic List
Secondary: #3 Eliminate Possibilities
Only $\binom{5}{3} = 10$ possible 3-card sets exist, so we can literally write every one of them down (Tool #2). Once the list is in front of us, Tool #3 (Eliminate) makes it easy to mark which sets have $4$ as the largest: any set containing card $5$ is out, and any set not containing card $4$ is out. Counting what survives gives the probability directly — no combinatorial formulas needed.
Execute — Answer: C
7.SP.C.8 Step 1 - List every way to choose $3$ cards from $\{1, 2, 3, 4, 5\}$ in increasing order, starting with the smallest card.
- Picking an ordering rule first guarantees no duplicates and no missing cases.
💡 Writing out all the equally likely outcomes is exactly the "organized list" sample-space move.
7.SP.C.7 Step 2 - Count the total.
- The list has $10$ sets, so the sample space has $10$ equally likely outcomes.
💡 When every outcome is equally likely, the total number of outcomes is the denominator of the probability.
7.SP.C.8 Step 3 - Cross off sets where $4$ is NOT the largest.
- Any set containing $5$ has $5$ as the largest, and any set without $4$ has a largest of $3$ or less.
- The survivors must contain $4$ and not contain $5$.
💡 Eliminating sets that break the "max $= 4$" rule leaves exactly the favorable ones.
7.SP.C.7 Step 4 - Form the probability as favorable $\div$ total.
- There are $3$ favorable sets out of $10$, so the probability is $\tfrac{3}{10}$, which is choice (C).
💡 Counting favorable outcomes and dividing by total outcomes is the definition of probability for equally likely events.
7.SP.C.8 List every way to choose $3$ cards from $\{1, 2, 3, 4, 5\}$ in increasing order, 7.SP.C.7 Count the total. The list has $10$ sets, so the sample space has $10$ equally li 7.SP.C.8 Cross off sets where $4$ is NOT the largest. Any set containing $5$ has $5$ as t 7.SP.C.7 Form the probability as favorable $\div$ total. There are $3$ favorable sets out Review
Reasonableness: By symmetry, the largest card in a random $3$-card draw from $\{1,\dots,5\}$ is $5$ with probability $\tfrac{\binom{4}{2}}{10}=\tfrac{6}{10}$, is $4$ with probability $\tfrac{\binom{3}{2}}{10}=\tfrac{3}{10}$, and is $3$ with probability $\tfrac{\binom{2}{2}}{10}=\tfrac{1}{10}$. These sum to $\tfrac{6+3+1}{10}=1$, confirming our $\tfrac{3}{10}$ for max $=4$ is consistent with all possibilities.
Alternative: Tool #16 (Change Focus / Complement) is not the cleanest here, but Tool #9 (Easier Related Problem) is a nice alternative: think of it as "first pick card $4$, then pick the other $2$ from the smaller cards $\{1,2,3\}$." There are $\binom{3}{2}=3$ ways to do that, out of $\binom{5}{3}=10$ total, giving $\tfrac{3}{10}$ — same answer without writing the full list.
CCSS standards used (min grade 7)
7.SP.C.7Develop probability models and use them to find probabilities of events (Using the uniform probability model (every $3$-card set equally likely) to set probability = favorable $\div$ total $= \tfrac{3}{10}$.)7.SP.C.8Find probabilities of compound events using organized lists, tables, and simulation (Writing out the organized list of all $10$ three-card subsets, then identifying the $3$ subsets that satisfy "largest card $= 4$.")
⭐ This AMC 8 problem only needs Grade 7 probability with organized lists you already know — write out every way, count the ones that fit, divide!
⭐ This AMC 8 problem only needs Grade 7 probability with organized lists you already know — write out every way, count the ones that fit, divide!
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