2010 AMC 12A Problem 19

Question

Each of

2010

boxes in a line contains a single red marble, and for

1 \le k \le 2010,

the box in the

k

th 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) \lt \dfrac{1}{2010}?

45

63

64

201

1005

Answer: A

Concepts:telescoping basic probability inequality

Difficulty rating: 2240

Small hint:

The first

n-1

draws must be white, then the

n

th draw red

Big hint:

The white probabilities telescope:

\dfrac12\cdot\dfrac23\cdots\dfrac{n-1}{n}

Solution:

Since there are

k + 1

marbles in the

k

th box, there is a

\dfrac{k}{k + 1}

chance Isabella draws a white marble from it.

The probability of drawing a red marble is then

\dfrac{1}{k + 1}.

To stop after drawing the

n

th marble, the first

n - 1

marbles must have been white.

This happens with a probability of

\dfrac{1}{2} \cdot \dfrac{2}{3} \cdot \ldots \cdot \dfrac{n - 1}{n} \cdot \dfrac{1}{n + 1}.

Note that all the numerators cancel with the adjacent denominator, which means that this expression reduces to

\dfrac{1}{n(n + 1)}.

We have to find the smallest

n

such that

\dfrac{1}{n(n + 1)} \lt \dfrac{1}{2010}

n(n + 1) \gt 2010.

Since

44\cdot45=1980\lt2010

but

45\cdot46=2070\gt2010,

the smallest such

n

is

45.

Thus, A is the correct answer.