AMC 8 · 2017 · #10

Grade 7 probabilitycounting

Archetype Small-Domain Constrained Search

probability-basiccombinations-basicsystematic-enumeration systematic-enumerationcasework ↑ Prerequisites: fraction-arithmeticsystematic-enumeration

📏 Short solution 💡 2 insights

📘 View easy version →

Problem

A box contains five cards, numbered 1, 2, 3, 4, and 5. Three cards are selected randomly without replacement from the box. What is the probability that 4 is the largest value selected?

$\textbf{(A) }\frac{1}{10}\qquad\textbf{(B) }\frac{1}{5}\qquad\textbf{(C) }\frac{3}{10}\qquad\textbf{(D) }\frac{2}{5}\qquad\textbf{(E) }\frac{1}{2}$

Pick an answer.

(A)

$frac{1}{10}$

(B)

$frac{1}{5}$

(C)

$frac{3}{10}$

(D)

$frac{2}{5}$

(E)

$frac{1}{2}$

View mode:

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

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

#2 Make a Systematic List 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.

$$\{1,2,3\},\{1,2,4\},\{1,2,5\},\{1,3,4\},\{1,3,5\},\{1,4,5\},\{2,3,4\},\{2,3,5\},\{2,4,5\},\{3,4,5\}$$

💡 Writing out all the equally likely outcomes is exactly the "organized list" sample-space move.

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

Count the total.
The list has $10$ sets, so the sample space has $10$ equally likely outcomes.

$$\text{Total sets} = 10$$

💡 When every outcome is equally likely, the total number of outcomes is the denominator of the probability.

#3 Eliminate Possibilities 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 $5$.

$$\text{Survivors: } \{1,2,4\},\{1,3,4\},\{2,3,4\}$$

💡 Eliminating sets that break the "max $= 4$" rule leaves exactly the favorable ones.

#2 Make a Systematic List 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).

$$P(\text{max}=4) = \dfrac{3}{10} \;\Rightarrow\; \textbf{(C)}$$

💡 Counting favorable outcomes and dividing by total outcomes is the definition of probability for equally likely events.

[1] #2 7.SP.C.8 List every way to choose $3$ cards from $\{1, 2, 3, 4, 5\}$ in increasing order,

[2] #2 7.SP.C.7 Count the total. The list has $10$ sets, so the sample space has $10$ equally li

[3] #3 7.SP.C.8 Cross off sets where $4$ is NOT the largest. Any set containing $5$ has $5$ as t

[4] #2 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.7 Develop 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.8 Find 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!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 7)

More like this