AMC 10 · 2002 · #24

Grade 7 probability

Archetype Overlap / Complement Counting └ Sub-archetype Parity Partition: Complement via Odd-Even Split

probability-basiccombinations-basicsystematic-enumeration casework ↑ Prerequisites: probability-basic

📏 Long solution 💡 3 insights

Problem

Tina randomly selects two distinct numbers from the set $\{ 1, 2, 3, 4, 5 \}$ , and Sergio randomly selects a number from the set $\{ 1, 2, ..., 10 \}$ . What is the probability that Sergio's number is larger than the sum of the two numbers chosen by Tina?

Solution 1

This is not too bad using casework.

Tina gets a sum of 3: This happens in only one way $(1,2)$ and Sergio can choose a number from 4 to 10, inclusive. There are 7 ways that Sergio gets a desirable number here.

Tina gets a sum of 4: This once again happens in only one way $(1,3)$ . Sergio can choose a number from 5 to 10, so 6 ways here.

Tina gets a sum of 5: This can happen in two ways $(1,4)$ and $(2,3)$ . Sergio can choose a number from 6 to 10, so $2\cdot5=10$ ways here.

Tina gets a sum of 6: Two ways here $(1,5)$ and $(2,4)$ . Sergio can choose a number from 7 to 10, so $2\cdot4=8$ here.

Tina gets a sum of 7: Two ways here $(2,5)$ and $(3,4)$ . Sergio can choose from 8 to 10, so $2\cdot3=6$ ways here.

Tina gets a sum of 8: Only one way possible $(3,5$ ). Sergio chooses 9 or 10, so 2 ways here.

Tina gets a sum of 9: Only one way $(4,5)$ . Sergio must choose 10, so 1 way.

In all, there are $7+6+10+8+6+2+1=40$ ways. Tina chooses two distinct numbers in $\binom{5}{2}=10$ ways while Sergio chooses a number in $10$ ways, so there are $10\cdot 10=100$ ways in all. Since $\frac{40}{100}=\frac{2}{5}$ , our answer is $\boxed{\text{(A)}\frac{2}{5}}$ .

Solution 2

We invoke some symmetry. Let $T$ denote Tina's sum, and let $S$ denote Sergio's number. Observe that, for $i = 2, 3, \ldots, 10$ , $\text{Pr}(T=i) = \text{Pr}(T=12-i)$ .

If Tina's sum is $i$ , then the probability that Sergio's number is larger than Tina's sum is $\frac{10-i}{10}$ . Thus, the probability $P$ is

$P = \text{Pr}(S>T) = \sum_{i=2}^{10} \text{Pr}(T=i) \times \frac{10-i}{10}$

Using the symmetry observation, we can also write the above sum as
$P = \sum_{i=2}^{10} \text{Pr}(T=12-i) \times \frac{10-i}{10} = \sum_{i=2}^{10} \text{Pr}(T=i) \times \frac{i-2}{10}$
where the last equality follows as we reversed the indices of the sum (by replacing $12-i$ with $i$ ). Thus, adding the two equivalent expressions for $P$ , we have

\begin{align*} 2P &= \sum_{i=2}^{10} \text{Pr}(T=i) \times \left(\frac{10-i}{10} + \frac{i-2}{10}\right) \ &= \sum_{i=2}^{10} \text{Pr}(T=i) \times \frac{4}{5} \ &= \frac{4}{5} \sum_{i=2}^{10} \text{Pr}(T=i) \ &= \frac{4}{5} \end{align*}

Since this represents twice the desired probability, the answer is $P = \boxed{\textbf{(A)} \frac{2}{5}}$ . -scrabbler94

Pick an answer.

(A)

$\ 2/5$

(B)

$\ 9/20$

(C)

$\ 1/2$

(D)

$\ 11/20$

(E)

$\ 24/25$

AMC 10 2002 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.

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Toolkit + CCSS Solution

Understand

Restated: Tina picks two different numbers from $\{1,2,3,4,5\}$ and adds them, getting a sum $T$. Separately, Sergio picks one number $S$ from $\{1,2,\dots,10\}$. Every choice is equally likely. Find the probability that Sergio's number beats Tina's sum, i.e. that $S > T$.

Givens: Tina chooses $2$ distinct numbers from $\{1,2,3,4,5\}$; her value is their sum $T$.; Sergio chooses $1$ number $S$ from $\{1,2,\dots,10\}$.; All of Tina's pairs are equally likely, and all of Sergio's $10$ numbers are equally likely, and the two choices are independent.; Answer choices: (A) $2/5$, (B) $9/20$, (C) $1/2$, (D) $11/20$, (E) $24/25$.

Unknowns: The probability that $S > T$.

Understand

Plan

Primary tool: #2 Make a Systematic List

Secondary: #7 Identify Subproblems, #16 Change Focus / Count the Complement

The two picks are independent, so the honest way to get the probability is to count outcomes: (number of winning $(T,S)$ situations) $\div$ (total situations). Because Tina's sum only takes a handful of values ($3$ through $9$), a systematic list organised by her sum (Tool #2) captures every case without missing or double-counting. For each fixed sum we split off a clean subproblem (Tool #7): how many pairs make that sum, and how many of Sergio's numbers exceed it. Grouping by the value $T$ (Tool #16 — pick the most useful thing to organise around) turns a messy $10 \times 10$ grid into seven tidy rows we can add up.

Execute — Answer: A

#7 Identify Subproblems 7.SP.C.8 Step 1

Count the total equally-likely outcomes

Tina picks $2$ distinct numbers from $5$, which can happen in $\binom{5}{2}=10$ ways (order does not matter for a sum).
Sergio independently picks $1$ of $10$ numbers.
Since the picks are independent and every combination is equally likely, the total number of equally-likely $(\text{Tina pair}, \text{Sergio number})$ outcomes is $10 \times 10 = 100$.
This $100$ will be the denominator of the probability.

$$\binom{5}{2}\times 10 = 10 \times 10 = 100$$

💡 Independent choices multiply, so the whole sample space is just Tina's count times Sergio's count.

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

List Tina's pairs by their sum

Organise the $10$ pairs by the sum $T$ they produce.
Sum $3$: $(1,2)$ — $1$ pair.
Sum $4$: $(1,3)$ — $1$ pair.
Sum $5$: $(1,4),(2,3)$ — $2$ pairs.
Sum $6$: $(1,5),(2,4)$ — $2$ pairs.
Sum $7$: $(2,5),(3,4)$ — $2$ pairs.
Sum $8$: $(3,5)$ — $1$ pair.
Sum $9$: $(4,5)$ — $1$ pair.
These counts add to $1+1+2+2+2+1+1=10$, so every pair is accounted for exactly once.

$$T:\ 3,4,5,6,7,8,9 \ \to\ \text{pairs } 1,1,2,2,2,1,1\ (\text{total }10)$$

💡 Sorting the outcomes by their sum guarantees you see each pair once and never twice.

#2 Make a Systematic List 7.SP.C.7 Step 3

Count Sergio's winning numbers per sum

For a fixed sum $T$, Sergio wins when $S > T$, i.e.
$S \in \{T+1, \dots, 10\}$, which is $10 - T$ numbers.
Now weight each sum by how many pairs make it.
Sum $3$: $1$ pair $\times (10-3)=7$.
Sum $4$: $1\times 6 = 6$.
Sum $5$: $2\times 5 = 10$.
Sum $6$: $2\times 4 = 8$.
Sum $7$: $2\times 3 = 6$.
Sum $8$: $1\times 2 = 2$.
Sum $9$: $1\times 1 = 1$.

$$\#\{S>T\} = 10 - T$$

💡 Once the sum is set, the numbers bigger than it are simply everything above it up to $10$.

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

Add the winning outcomes

Add the row totals to get the number of $(T,S)$ outcomes where Sergio wins: $7+6+10+8+6+2+1 = 40$.
So $40$ of the $100$ equally-likely outcomes are wins for Sergio.

$$7+6+10+8+6+2+1 = 40$$

💡 The favorable count is just the sum of the per-sum winning counts.

#2 Make a Systematic List 7.SP.C.7 Step 5

Form the probability

Divide favorable by total: $\dfrac{40}{100} = \dfrac{2}{5}$.
Comparing with the choices, $\dfrac{2}{5}$ is option (A).

$$P(S>T) = \dfrac{40}{100} = \dfrac{2}{5} \;\Rightarrow\; \textbf{(A)}$$

💡 Probability of an equally-likely event is favorable outcomes over total outcomes.

[1] #7 7.SP.C.8 Tina picks $2$ distinct numbers from $5$, which can happen in $\binom{5}{2}=10$

[2] #2 7.SP.C.8 Organise the $10$ pairs by the sum $T$ they produce. Sum $3$: $(1,2)$ — $1$ pair

[3] #2 7.SP.C.7 For a fixed sum $T$, Sergio wins when $S > T$, i.e. $S \in \{T+1, \dots, 10\}$,

[4] #7 4.OA.A.3 Add the row totals to get the number of $(T,S)$ outcomes where Sergio wins: $7+6

[5] #2 7.SP.C.7 Divide favorable by total: $\dfrac{40}{100} = \dfrac{2}{5}$. Comparing with the

Review

Reasonableness: The answer $2/5 = 0.4$ should sit a little below one-half, and that makes sense: Tina's smallest possible sum is $3$, so Sergio (who can be as low as $1$) is often already behind before he even ties. A quick bound check: Tina's sums average $(3+9)/2 = 6$, and against a sum of $6$ Sergio wins with $4$ of his $10$ numbers, i.e. probability $0.4$ — consistent with $2/5$. Also $40/100$ reduces cleanly to $2/5$, one of the listed choices, and the pair counts summed to $10 = \binom{5}{2}$, confirming no case was missed or double-counted.

Alternative: Use symmetry (Tool #5, Look for a Pattern). Tina's sum distribution is symmetric about $6$: $P(T=i)=P(T=12-i)$. For a given sum $i$, $P(S>i)=\tfrac{10-i}{10}$, and by the symmetry this pairs with $P(S<i')$ for the mirrored sum. Adding the probability of $S>T$ to the mirror-image expression collapses every term to $\tfrac{10-i}{10}+\tfrac{i-2}{10}=\tfrac{4}{5}$, so $2P = \tfrac{4}{5}$ and $P = \tfrac{2}{5}$, again (A) — no case-by-case counting needed.

CCSS standards used (min grade 7)

7.SP.C.8 Find probabilities of compound events using organized lists, tables, and tree diagrams (Counting the $10 \times 10 = 100$ total outcomes and organizing Tina's pairs by their sum so each pair is counted exactly once.)
7.SP.C.7 Develop probability models and use them to find probabilities of events (Counting Sergio's winning numbers ($10-T$) for each sum and forming the probability $40/100 = 2/5$ as favorable over total.)
4.OA.A.3 Solve multistep word problems using the four operations (Weighting each sum by its pair count and adding the seven row totals $7+6+10+8+6+2+1 = 40$.)

⭐ Group Tina's pairs by their sum; for each sum, count how many of Sergio's numbers are bigger, add those up, and divide by all $100$ outcomes.

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: A

Count the total equally-likely outcomes

List Tina's pairs by their sum

Count Sergio's winning numbers per sum

Add the winning outcomes

Form the probability

Review

CCSS standards used (min grade 7)

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