AMC 10 · 2010 · #23
Grade 7 probabilityEach of 2010 boxes in a line contains a single red marble, and for 1≤k≤2010, the box in the kth position also contains k white marbles. Isabella begins at the first box and successively draws a single marble at random from each box, in order. She stops when she first draws a red marble. Let P(n) be the probability that Isabella stops after drawing exactly n marbles. What is the smallest value of n for which P(n)<20101?
Pick an answer.
AMC 10 2010 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: There are 2010 boxes in a row. Box $k$ holds 1 red marble and $k$ white marbles. Isabella draws one random marble from box 1, then box 2, and so on, stopping the moment she first draws red. If $P(n)$ is the probability she stops right after her $n$th draw, find the smallest $n$ with $P(n)<\frac{1}{2010}$.
Givens: Box $k$ contains exactly 1 red marble and $k$ white marbles, so $k+1$ marbles total; She draws once from each box in order, box 1 first; She stops the instant she draws a red marble; $P(n)$ is the probability she stops after exactly $n$ draws; Answer choices: (A) $45$, (B) $63$, (C) $64$, (D) $201$, (E) $1005$
Unknowns: The smallest $n$ for which $P(n)<\frac{1}{2010}$
Understand
Restated: There are 2010 boxes in a row. Box $k$ holds 1 red marble and $k$ white marbles. Isabella draws one random marble from box 1, then box 2, and so on, stopping the moment she first draws red. If $P(n)$ is the probability she stops right after her $n$th draw, find the smallest $n$ with $P(n)<\frac{1}{2010}$.
Givens: Box $k$ contains exactly 1 red marble and $k$ white marbles, so $k+1$ marbles total; She draws once from each box in order, box 1 first; She stops the instant she draws a red marble; $P(n)$ is the probability she stops after exactly $n$ draws; Answer choices: (A) $45$, (B) $63$, (C) $64$, (D) $201$, (E) $1005$
Plan
Primary tool: #5 Look for a Pattern
Secondary: #7 Identify Subproblems, #6 Guess and Check
Tool #5 (Look for a Pattern): the whole problem cracks open when you write $P(n)$ as a product of fractions and notice the numerators and denominators cancel in a chain, collapsing a scary product into the tiny formula $\frac{1}{n(n+1)}$. Tool #7 (Identify Subproblems): stopping at box $n$ is really one requirement per box — white, white, ..., white, then red — so I find each box's probability separately and multiply. Tool #6 (Guess and Check): once $P(n)=\frac{1}{n(n+1)}$, the condition becomes $n(n+1)>2010$, and the fastest finish is to test $n$ values near $\sqrt{2010}$ until the inequality first holds.
Execute — Answer: A
7.SP.C.5 Step 1 One box's two probabilities
- Box $k$ has 1 red marble and $k$ white marbles, so $k+1$ marbles in all.
- The chance of drawing white from box $k$ is $\frac{k}{k+1}$, and the chance of drawing red is $\frac{1}{k+1}$.
- These two add to 1, as they must.
💡 More white marbles in a box makes white more likely and red less likely, exactly as the fractions show.
7.SP.C.8 Step 2 What stopping at box n requires
- To stop after exactly $n$ draws, Isabella must draw white from every box $1,2,\ldots,n-1$ (so she keeps going) and then red from box $n$ (so she stops).
- Because the boxes are separate draws, multiply the probabilities together.
💡 A run of independent draws happens with the product of their chances, one factor per box.
5.NF.B.4 Step 3 The product telescopes
- Look at the white-marble product $\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdots\frac{n-1}{n}$.
- Each numerator cancels the denominator just before it: the 2 on top cancels the 2 below, the 3 cancels the 3, and so on down the chain.
- All that survives is the first numerator $1$ and the last denominator $n$, leaving $\frac{1}{n}$.
💡 In a chain of fractions where each top matches the next bottom, everything in the middle cancels and only the ends remain.
5.NF.B.4 Step 4 A clean formula for P(n)
- Now multiply the collapsed product by the red-draw factor $\frac{1}{n+1}$: $P(n)=\frac{1}{n}\cdot\frac{1}{n+1}=\frac{1}{n(n+1)}$.
- The messy product became one neat fraction.
💡 Telescoping turns a long product into a single small fraction you can actually work with.
6.EE.B.5 Step 5 Turn the goal into an inequality
- We want $P(n)<\frac{1}{2010}$, that is $\frac{1}{n(n+1)}<\frac{1}{2010}$.
- For positive numbers, a smaller fraction means a larger denominator, so this is the same as $n(n+1)>2010$.
💡 Among unit fractions, the one with the bigger bottom is the smaller number.
6.EE.B.5 Step 6 Find the smallest n that works
- Now test values of $n$; since $n(n+1)$ grows as $n$ grows, I look near $\sqrt{2010}\approx 44.8$.
- Trying $n=44$ gives $44\cdot 45=1980$, which is not more than $2010$.
- Trying $n=45$ gives $45\cdot 46=2070$, which is more than $2010$.
- So $n=44$ fails but $n=45$ works, and $45$ is the smallest such $n$.
- The answer is (A).
💡 Because the product only increases, the first $n$ that clears the bar is the smallest answer.
7.SP.C.5 Box $k$ has 1 red marble and $k$ white marbles, so $k+1$ marbles in all. The cha 7.SP.C.8 To stop after exactly $n$ draws, Isabella must draw white from every box $1,2,\l 5.NF.B.4 Look at the white-marble product $\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cd 5.NF.B.4 Now multiply the collapsed product by the red-draw factor $\frac{1}{n+1}$: $P(n) 6.EE.B.5 We want $P(n)<\frac{1}{2010}$, that is $\frac{1}{n(n+1)}<\frac{1}{2010}$. For po 6.EE.B.5 Now test values of $n$; since $n(n+1)$ grows as $n$ grows, I look near $\sqrt{20 Review
Reasonableness: The formula $P(n)=\frac{1}{n(n+1)}$ passes quick tests: $P(1)=\frac{1}{2}$ (box 1 is half red, correct), and the probabilities $\sum_n \frac{1}{n(n+1)}=\sum_n\left(\frac{1}{n}-\frac{1}{n+1}\right)$ telescope toward 1, as a full probability distribution should. The threshold $n(n+1)>2010$ sits right at $\sqrt{2010}\approx 44.8$, so an answer near 45 is exactly what we expect. The larger choices 63, 64, 201, 1005 would each demand $n(n+1)$ far above 2010, overshooting badly; 44 just misses at 1980. So 45 is right.
Alternative: Skip building the general formula and reason directly about the boundary. We need $n(n+1)>2010$, and $2010\approx 45^2$ since $45^2=2025$. Because $n(n+1)$ straddles $n^2$, test the two integers around $\sqrt{2010}$: $44\cdot45=1980$ (too small) and $45\cdot46=2070$ (big enough). This confirms $n=45$ without expanding the telescoping product, using only the estimate that the answer must live next to $\sqrt{2010}$.
CCSS standards used (min grade 7)
7.SP.C.5Understand that the probability of a chance event is between 0 and 1 (Reading each box as a probability model — white with chance k/(k+1) and red with chance 1/(k+1), which sum to 1.)7.SP.C.8Find probabilities of compound events using organized lists, tables, and simulation (Multiplying the per-box chances for the run 'white, white, ..., white, red' to get P(n) as a compound-event probability.)5.NF.B.4Apply and extend understanding of multiplication to multiply a fraction by a fraction (Multiplying the chain of fractions and cancelling to telescope the product down to 1/(n(n+1)).)6.EE.B.5Understand solving an equation or inequality as a process of finding values (Rewriting the goal as n(n+1)>2010 and testing values of n to find the smallest one that makes it true.)
⭐ Write the stopping chance as a product of fractions, watch the middle cancel down to $\frac{1}{n(n+1)}$, and the whole question becomes 'when does $n(n+1)$ pass 2010?'
⭐ Write the stopping chance as a product of fractions, watch the middle cancel down to $\frac{1}{n(n+1)}$, and the whole question becomes 'when does $n(n+1)$ pass 2010?'
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