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AMC 10 · 2022 · #23

Grade 7 probability
Archetype Overlap / Complement Counting
probability-basicgeometric-probabilityexpected-valuecomplementary-countingfraction-arithmetic complementary-countingeasier-related-problemidentify-subproblems ↑ Prerequisites: probability-basicgeometric-probability
📏 Long solution 💡 4 insights

Problem

Ant Amelia starts on the number line at 00 and crawls in the following manner. For n=1,2,3,n=1,2,3, Amelia chooses a time duration tnt_n and an increment xnx_n independently and uniformly at random from the interval (0,1).(0,1). During the nnth step of the process, Amelia moves xnx_n units in the positive direction, using up tnt_n minutes. If the total elapsed time has exceeded 11 minute during the nnth step, she stops at the end of that step; otherwise, she continues with the next step, taking at most 33 steps in all. What is the probability that Amelia’s position when she stops will be greater than 11?

(A) 13(B) 12(C) 23(D) 34(E) 56\textbf{(A) }\frac{1}{3} \qquad \textbf{(B) }\frac{1}{2} \qquad \textbf{(C) }\frac{2}{3} \qquad \textbf{(D) }\frac{3}{4} \qquad \textbf{(E) }\frac{5}{6}

Pick an answer.

(A)
$\frac{1}{3}$
(B)
$\frac{1}{2}$
(C)
$\frac{2}{3}$
(D)
$\frac{3}{4}$
(E)
$\frac{5}{6}$
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Toolkit + CCSS Solution

Understand

Restated: Ant Amelia starts at $0$. For each step $n = 1, 2, 3$ she picks a time $t_n$ and a distance $x_n$, each independently and uniformly from $(0,1)$. On step $n$ she moves $x_n$ to the right and uses $t_n$ minutes. She stops at the end of a step whenever the total elapsed time first exceeds $1$ minute, but she takes at most $3$ steps. Find the probability that her final position exceeds $1$.

Givens: $t_1, t_2, t_3, x_1, x_2, x_3$ are mutually independent, each uniform on $(0,1)$; After step $n$, total time $T_n = t_1 + \dots + t_n$ and total position $X_n = x_1 + \dots + x_n$; Amelia stops as soon as $T_n > 1$, but no later than $n = 3$; Answer choices: (A) $\tfrac{1}{3}$, (B) $\tfrac{1}{2}$, (C) $\tfrac{2}{3}$, (D) $\tfrac{3}{4}$, (E) $\tfrac{5}{6}$

Unknowns: $P(\text{final position} > 1)$

Understand

Restated: Ant Amelia starts at $0$. For each step $n = 1, 2, 3$ she picks a time $t_n$ and a distance $x_n$, each independently and uniformly from $(0,1)$. On step $n$ she moves $x_n$ to the right and uses $t_n$ minutes. She stops at the end of a step whenever the total elapsed time first exceeds $1$ minute, but she takes at most $3$ steps. Find the probability that her final position exceeds $1$.

Givens: $t_1, t_2, t_3, x_1, x_2, x_3$ are mutually independent, each uniform on $(0,1)$; After step $n$, total time $T_n = t_1 + \dots + t_n$ and total position $X_n = x_1 + \dots + x_n$; Amelia stops as soon as $T_n > 1$, but no later than $n = 3$; Answer choices: (A) $\tfrac{1}{3}$, (B) $\tfrac{1}{2}$, (C) $\tfrac{2}{3}$, (D) $\tfrac{3}{4}$, (E) $\tfrac{5}{6}$

Plan

Primary tool: #7 Identify Subproblems

Secondary: #9 Solve an Easier Related Problem, #1 Draw a Diagram, #16 Change Focus / Complement

Tool #7 (Subproblems) — the stopping rule splits into exactly two mutually exclusive scenarios (stop at step 2 vs. step 3), so the answer is a sum of two probabilities. Tool #9 (Easier) — first answer the cleaner sub-question "what is $P(u_1 + u_2 > 1)$ for two uniform $(0,1)$ random variables?" with a small picture, before tackling three variables. Tool #1 (Diagram) — sketch the unit square (sum-of-two case) and unit cube with tetrahedron (sum-of-three case) to read the probabilities geometrically. Tool #16 (Complement) — for the sum-of-three, count the complement (sum $\le 1$, a tiny tetrahedron) rather than the bulky $5/6$ region directly.

Execute — Answer: C

#7 Identify Subproblems 7.SP.C.8 Step 1
  • Split into two mutually exclusive scenarios based on when Amelia stops.
  • (S1) She stops after step 2: $T_2 > 1$ and the final position $X_2 = x_1 + x_2 > 1$.
  • (S2) She continues past step 2 and stops after step 3: $T_2 \le 1$ and final position $X_3 = x_1 + x_2 + x_3 > 1$.
  • Because all $t$'s are independent of all $x$'s, each scenario's joint probability factors as $P(\text{time event}) \cdot P(\text{position event})$.
$$P(\text{pos} > 1) = P(T_2 > 1)\,P(X_2 > 1) + P(T_2 \le 1)\,P(X_3 > 1)$$

💡 Grade 7 — the stop rule cleanly partitions the sample space; independence lets each scenario factor.

#9 Solve an Easier Related Problem 7.SP.C.7 Step 2
  • Easier sub-problem: with $u_1, u_2$ uniform $(0,1)$ independent, find $P(u_1 + u_2 > 1)$.
  • Draw the unit square in the $(u_1, u_2)$-plane: total area $1$.
  • The region $u_1 + u_2 \le 1$ is the lower-left triangle with legs $1$, area $\tfrac{1}{2}$.
  • So $P(u_1+u_2 > 1) = 1 - \tfrac{1}{2} = \tfrac{1}{2}$.
$$P(u_1+u_2 > 1) = \tfrac{1}{2}$$

💡 Grade 7 — geometric probability on a unit square: shaded area $=$ probability.

#1 Draw a Diagram 7.SP.C.7 Step 3
  • Same picture gives $P(T_2 > 1) = \tfrac{1}{2}$ (replace $u_i$ with $t_i$) and $P(X_2 > 1) = \tfrac{1}{2}$ (replace $u_i$ with $x_i$).
  • So scenario S1 contributes $P(\text{S1}) = \tfrac{1}{2} \cdot \tfrac{1}{2} = \tfrac{1}{4}$.
  • And $P(T_2 \le 1) = \tfrac{1}{2}$.
$$P(\text{S1}) = \tfrac{1}{2} \cdot \tfrac{1}{2} = \tfrac{1}{4}$$

💡 Grade 7 — multiply two independent probabilities.

#16 Change Focus / Complement 7.SP.C.8 Step 4
  • For S2 we need $P(X_3 > 1) = P(x_1 + x_2 + x_3 > 1)$.
  • Use the complement: count $P(x_1 + x_2 + x_3 \le 1)$.
  • With $x_1, x_2, x_3$ uniform $(0,1)$ independent, the sample space is the unit cube (volume $1$), and the region $x_1 + x_2 + x_3 \le 1$ is the tetrahedron with vertices $(0,0,0), (1,0,0), (0,1,0), (0,0,1)$ — volume $\tfrac{1}{3} \cdot \tfrac{1}{2} \cdot 1 = \tfrac{1}{6}$.
  • So $P(X_3 > 1) = 1 - \tfrac{1}{6} = \tfrac{5}{6}$.
$$P(X_3 > 1) = 1 - \tfrac{1}{6} = \tfrac{5}{6}$$

💡 Grade 7 — count the small tetrahedron (complement) instead of the larger region; tetrahedron volume $= \tfrac{1}{3} \cdot \text{base} \cdot \text{height}$.

#7 Identify Subproblems 7.SP.C.8 Step 5

Scenario S2 contributes $P(\text{S2}) = P(T_2 \le 1) \cdot P(X_3 > 1) = \tfrac{1}{2} \cdot \tfrac{5}{6} = \tfrac{5}{12}$.

$$P(\text{S2}) = \tfrac{1}{2} \cdot \tfrac{5}{6} = \tfrac{5}{12}$$

💡 Grade 7 — multiply two independent probabilities again.

#7 Identify Subproblems 5.NF.A.1 Step 6
  • Add the two mutually exclusive scenarios: $P(\text{pos} > 1) = \tfrac{1}{4} + \tfrac{5}{12} = \tfrac{3}{12} + \tfrac{5}{12} = \tfrac{8}{12} = \tfrac{2}{3}$.
  • Answer (C).
$$\tfrac{1}{4} + \tfrac{5}{12} = \tfrac{8}{12} = \tfrac{2}{3} \;\Rightarrow\; \textbf{(C)}$$

💡 Grade 5 — add fractions with a common denominator.

[1] #7 7.SP.C.8 Split into two mutually exclusive scenarios based on when Amelia stops. (S1) She
[2] #9 7.SP.C.7 Easier sub-problem: with $u_1, u_2$ uniform $(0,1)$ independent, find $P(u_1 + u
[3] #1 7.SP.C.7 Same picture gives $P(T_2 > 1) = \tfrac{1}{2}$ (replace $u_i$ with $t_i$) and $P
[4] #16 7.SP.C.8 For S2 we need $P(X_3 > 1) = P(x_1 + x_2 + x_3 > 1)$. Use the complement: count
[5] #7 7.SP.C.8 Scenario S2 contributes $P(\text{S2}) = P(T_2 \le 1) \cdot P(X_3 > 1) = \tfrac{1
[6] #7 5.NF.A.1 Add the two mutually exclusive scenarios: $P(\text{pos} > 1) = \tfrac{1}{4} + \t

Review

Reasonableness: Sanity. The two scenarios partition the event "Amelia stops eventually" — and they must, because Amelia always stops by step 3 (the rules force it). Each scenario probability is non-negative and at most the marginal $P(\text{stop after that step})$. We computed $\tfrac{1}{4} + \tfrac{5}{12} = \tfrac{2}{3} \approx 0.667$ — squarely in $(0,1)$, larger than $\tfrac{1}{2}$ (consistent with the fact that taking three steps usually pushes Amelia past $1$). Choice (C) matches. The other choices map to common miscounts: (B) $\tfrac{1}{2}$ ignores that step 3 is more likely to land past $1$; (E) $\tfrac{5}{6}$ takes $P(X_3 > 1)$ without the time condition; (A) $\tfrac{1}{3}$ is the complement guess.

Alternative: Tool #5 (Pattern) — directly compute $P(\text{pos} > 1)$ as $P(X_N > 1)$ where $N$ is the (random) stopping step. Condition on $N$: $P(N=2) = P(T_2 > 1) = \tfrac{1}{2}$ and $P(N=3) = \tfrac{1}{2}$. Since $\{N=2\}$ depends only on $t$'s while $X_n$ depends only on $x$'s, we factor each term cleanly: $\tfrac{1}{2}\cdot P(X_2 > 1) + \tfrac{1}{2}\cdot P(X_3 > 1) = \tfrac{1}{2}\cdot \tfrac{1}{2} + \tfrac{1}{2}\cdot \tfrac{5}{6} = \tfrac{2}{3}$.

CCSS standards used (min grade 7)

  • 5.NF.A.1 Add and subtract fractions with unlike denominators (Final addition $\tfrac{1}{4} + \tfrac{5}{12} = \tfrac{3}{12} + \tfrac{5}{12} = \tfrac{2}{3}$.)
  • 7.SP.C.7 Develop probability models and use them to find probabilities of events (Modeling the two-variable sum $P(u_1 + u_2 > 1) = \tfrac{1}{2}$ geometrically on the unit square.)
  • 7.SP.C.8 Find probabilities of compound events using organized lists, tables, tree diagrams, and simulation (Splitting into the two mutually exclusive stopping scenarios and combining each as a product of independent events.)

⭐ This AMC 10 problem only needs Grade 7 probability you already know — the stop rule cleanly splits the sample space into "stop at step 2" and "stop at step 3". For the first, both the time-sum and the position-sum live on a unit square, so each event has probability $\tfrac{1}{2}$. For the second, the position-sum of three uniforms exceeds $1$ with probability $1 - \tfrac{1}{6} = \tfrac{5}{6}$ (complement of a tetrahedron). Add: $\tfrac{1}{4} + \tfrac{5}{12} = \tfrac{2}{3}$.

⭐ This AMC 10 problem only needs Grade 7 probability you already know — the stop rule cleanly splits the sample space into "stop at step 2" and "stop at step 3". For the first, both the time-sum and the position-sum live on a unit square, so each event has probability $\tfrac{1}{2}$. For the second, the position-sum of three uniforms exceeds $1$ with probability $1 - \tfrac{1}{6} = \tfrac{5}{6}$ (complement of a tetrahedron). Add: $\tfrac{1}{4} + \tfrac{5}{12} = \tfrac{2}{3}$.

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