AMC 10 · 2010 · #23

Grade 7 probability

Archetype Small-Domain Constrained Search └ Sub-archetype Sum-Target Search

probability-basictelescoping-sumconditional-probability bound-inequality-then-enumerate ↑ Prerequisites: probability-basic

📏 Long solution 💡 3 insights

Problem

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 nth draw, find the smallest n with P(n) ＜ 1/2010.

Pick an answer.

(A)

(B)

(C)

(D)

201

(E)

1005

AMC 10 2010 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Look for a Pattern

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 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)=1/(n(n+1)), the condition becomes n(n+1) ＞ 2010, and the fastest finish is to test n values near √(2010) until the inequality first holds.

1STEP 1

One box's two probabilities

Box k holds 1 red and k white marbles, k+1 in all, so white comes up with chance k/(k+1) and red with 1/(k+1).

P(white from box k)=k/(k+1), P(red from box k)=1/(k+1)

2STEP 2

What stopping at box n requires

Stopping after exactly n draws means white from every box 1,2,…,n-1 and then red from box n; the draws are independent, so multiply.

P(n)=(1/2·2/3·3/4…(n-1)/n)·1/(n+1)

3STEP 3

The product telescopes

In 1/2·2/3·3/4…(n-1)/n every numerator cancels the denominator before it, so only the first 1 and the last n survive, leaving 1/n.

1/2·2/3·3/4…(n-1)/n=1/n

4STEP 4

A clean formula for P(n)

Multiply that 1/n by the red-draw factor 1/(n+1) and the messy product becomes one neat fraction, P(n)=1/(n(n+1)).

P(n)=1/n·1/(n+1)=1/(n(n+1))

5STEP 5

Turn the goal into an inequality

We want 1/(n(n+1)) ＜ 1/2010, and a smaller unit fraction just means a bigger bottom, so the goal becomes n(n+1) ＞ 2010.

1/(n(n+1)) ＜ 1/2010⇔ n(n+1) ＞ 2010

6STEP 6

Find the smallest n that works

Since n(n+1) only grows, test near √(2010)≈44.8: 44·45=1980 falls short, 45·46=2070 clears it, so the smallest n is 45, choice (A).

44 · 45=1980 ≯ 2010, 45 · 46=2070 ＞ 2010 → n=45 (A)

Answer

The formula P(n)=1/(n(n+1)) passes quick tests: P(1)=1/2 (box 1 is half red, correct), and the probabilities Σ_n 1/(n(n+1))=Σ_n(1/n-1/(n+1)) telescope toward 1, as a full probability distribution should. The threshold n(n+1) ＞ 2010 sits right at √(2010)≈ 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.

💡Key takeaway

Write the stopping chance as a product of fractions, watch the middle cancel down to 1/(n(n+1)), and the whole question becomes 'when does n(n+1) pass 2010?'

One box's two probabilities
What stopping at box n requires
The product telescopes
A clean formula for P(n)
Turn the goal into an inequality
Find the smallest n that works