AMC 10 · 2010 · #19

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

She draws from one box after another, stopping the first time a red marble appears. Find the earliest stop whose chance is tiny.

Pick an answer.

(A)

(B)

(C)

(D)

201

(E)

1005

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

One box gives its two probabilities directly.

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 late means missing every earlier box.

P(n)=(1/2·2/3·3/4…(n-1)/n)·1/(n+1)

3STEP 3

The product telescopes

The long product collapses to a single fraction.

1/2·2/3·3/4…(n-1)/n=1/n

4STEP 4

A clean formula for P(n)

That leaves one clean formula.

P(n)=1/n·1/(n+1)=1/(n(n+1))

5STEP 5

Turn the goal into an inequality

The goal becomes a simple product inequality.

1/(n(n+1)) ＜ 1/2010⇔ n(n+1) ＞ 2010

6STEP 6

Find the smallest n that works

Testing neighbours gives 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