AMC 10 · 2019 · #17

Grade 8 probability

probability-basicsymmetry-argumentsequences-geometriccomplementary-counting complementary-countingpattern-recognition ↑ Prerequisites: probability-basicsequences-geometric

📏 Short solution 💡 3 insights

Problem

A red ball and a green ball are randomly and independently tossed into bins numbered with the positive integers so that for each ball, the probability that it is tossed into bin $k$ is $2^{-k}$ for $k = 1,2,3....$ What is the probability that the red ball is tossed into a higher-numbered bin than the green ball?
$\textbf{(A) } \frac{1}{4} \qquad\textbf{(B) } \frac{2}{7} \qquad\textbf{(C) } \frac{1}{3} \qquad\textbf{(D) } \frac{3}{8} \qquad\textbf{(E) } \frac{3}{7}$

Pick an answer.

(A)

$\frac{1}{4}$

(B)

$\frac{2}{7}$

(C)

$\frac{1}{3}$

(D)

$\frac{3}{8}$

(E)

$\frac{3}{7}$

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Toolkit + CCSS Solution

Understand

Restated: A red ball and a green ball are tossed independently into bins numbered $1, 2, 3, \ldots$. For each ball, the probability of landing in bin $k$ is $2^{-k}$. Find the probability that the red ball lands in a higher-numbered bin than the green ball.

Givens: $P(\text{red in bin } k) = 2^{-k}$ for $k = 1, 2, 3, \ldots$; Same distribution for green; the two tosses are independent; Verify the distribution is valid: $\sum_{k=1}^{\infty} 2^{-k} = 1$; Choices: (A) $1/4$, (B) $2/7$, (C) $1/3$, (D) $3/8$, (E) $3/7$

Unknowns: $P(R > G)$, where $R, G$ are the bin numbers of the red and green balls

Understand

Plan

Primary tool: #16 Change Focus / Count the Complement

Secondary: #15 Organize Information in More Ways, #5 Look for a Pattern

Tool #16 (Complement): direct summation $\sum_{g} \sum_{r > g} 2^{-g} \cdot 2^{-r}$ is messy. Instead, count the tie probability $P(R = G)$ and use symmetry. Tool #15 (Reorganize): the three events $R > G$, $R < G$, $R = G$ partition the sample space — that's the right way to slice the problem. Tool #5 (Pattern): the tie probability sums a geometric series in $4^{-k}$, an arithmetic move kids see early.

Execute — Answer: C

#15 Organize Information in More Ways 7.SP.C.7 Step 1

By the symmetry of the setup, swapping the colors red $\leftrightarrow$ green doesn't change the probabilities.
So $P(R > G) = P(R < G)$.
Call this common value $p$.

$$P(R > G) = P(R < G) = p$$

💡 Red beating green and green beating red are mirror images — same chance.

#15 Organize Information in More Ways 7.SP.C.7 Step 2

The three events $R > G$, $R < G$, $R = G$ exhaust all possibilities and don't overlap, so their probabilities sum to $1$.

$$P(R > G) + P(R < G) + P(R = G) = 1$$

💡 Exactly one of three things happens — split, win, or lose for red.

#16 Change Focus / Count the Complement 7.SP.C.8 Step 3

Compute the tie probability $P(R = G)$ by summing over each bin $k$: both balls land in bin $k$ with probability $2^{-k} \cdot 2^{-k} = 4^{-k}$ (independence).

$$P(R = G) = \sum_{k=1}^{\infty} 4^{-k}$$

💡 Tie cases are the easy ones to count — pick a bin, both balls land there.

#5 Look for a Pattern 8.EE.A.1 Step 4

Sum the geometric series with first term $\frac{1}{4}$ and common ratio $\frac{1}{4}$: $\sum_{k=1}^{\infty} 4^{-k} = \dfrac{1/4}{1 - 1/4} = \dfrac{1/4}{3/4} = \dfrac{1}{3}$.

$$P(R = G) = \dfrac{1/4}{1 - 1/4} = \dfrac{1}{3}$$

💡 Geometric sum with ratio $1/4$ — first-term-over-$(1-\text{ratio})$ rule.

#16 Change Focus / Count the Complement 6.EE.B.7 Step 5

Plug back: $2p + \dfrac{1}{3} = 1 \Rightarrow 2p = \dfrac{2}{3} \Rightarrow p = \dfrac{1}{3}$.
So $P(R > G) = \dfrac{1}{3}$.
Answer: choice (C).

$$2p + \dfrac{1}{3} = 1 \Rightarrow p = \dfrac{1}{3}$$

💡 Solve a one-step equation for $p$ — the symmetry trick paid off.

[1] #15 7.SP.C.7 By the symmetry of the setup, swapping the colors red $\leftrightarrow$ green do

[2] #15 7.SP.C.7 The three events $R > G$, $R < G$, $R = G$ exhaust all possibilities and don't o

[3] #16 7.SP.C.8 Compute the tie probability $P(R = G)$ by summing over each bin $k$: both balls

[4] #5 8.EE.A.1 Sum the geometric series with first term $\frac{1}{4}$ and common ratio $\frac{1

[5] #16 6.EE.B.7 Plug back: $2p + \dfrac{1}{3} = 1 \Rightarrow 2p = \dfrac{2}{3} \Rightarrow p =

Review

Reasonableness: Three outcomes (red > green, red < green, tie) each carry probability $1/3$ — pleasingly symmetric. Cross-check with the first few bins: ties at $(1,1), (2,2), (3,3), \ldots$ have total probability $1/4 + 1/16 + 1/64 + \ldots = 1/3$. ✓. And the red-wins events $(R, G) = (2, 1), (3, 1), (3, 2), (4, 1), \ldots$ summed directly $\sum_{g=1}^{\infty} 2^{-g} \sum_{r=g+1}^{\infty} 2^{-r} = \sum_{g=1}^{\infty} 2^{-g} \cdot 2^{-g} = \sum 4^{-g} = 1/3$. ✓.

Alternative: Tool #2 (Systematic List) + Tool #5 (Pattern): condition on the green ball landing in bin $g$. Given $g$, $P(R > g) = \sum_{r=g+1}^{\infty} 2^{-r} = 2^{-g}$. So $P(R > G) = \sum_{g=1}^{\infty} 2^{-g} \cdot 2^{-g} = \sum_{g=1}^{\infty} 4^{-g} = 1/3$ — same answer, longer path.

CCSS standards used (min grade 8)

6.EE.B.7 Solve real-world problems by writing and solving equations of the form px = q (Solving $2p + 1/3 = 1$ for $p$.)
7.SP.C.7 Develop probability models and use them to find probabilities of events (Using symmetry $P(R > G) = P(R < G)$ and the partition of the sample space into three events.)
7.SP.C.8 Find probabilities of compound events using organized lists, tables, and simulation (Computing the tie probability as $\sum_k P(R=k) P(G=k) = \sum_k 4^{-k}$ using independence.)
8.EE.A.1 Know and apply the properties of integer exponents (Summing the infinite geometric series $\sum_{k=1}^{\infty} 4^{-k} = 1/3$.)

⭐ This AMC 10 problem only needs Grade 8 exponent reasoning you already know — by symmetry, red beating green and green beating red are equally likely, and ties happen $1/3$ of the time (geometric series). So $P(R > G) = (1 - 1/3)/2 = 1/3$. The answer is $\textbf{(C)}$.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 8)

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