Difference between revisions of "2010 AMC 10A Problems/Problem 23"

Revision as of 20:41, 24 January 2020

Problem

Each of $2010$ boxes in a line contains a single red marble, and for $1 \le k \le 2010$ , the box in the $k\text{th}$ 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 $P(n) < \frac{1}{2010}$ ?

$\textbf{(A)}\ 45 \qquad \textbf{(B)}\ 63 \qquad \textbf{(C)}\ 64 \qquad \textbf{(D)}\ 201 \qquad \textbf{(E)}\ 1005$

Solutions

Solution 1

The probability of drawing a white marble from box $k$ is $\frac{k}{k+1}$ . The probability of drawing a red marble from box $n$ is $\frac{1}{n+1}$ .

The probability of drawing a red marble at box $n$ is therefore

$\frac{1}{n+1} \left( \prod_{k=1}^{n-1}\frac{k}{k+1} \right) < \frac{1}{2010}$

$\frac{1}{n+1} \left( \frac{1}{n} \right) < \frac{1}{2010}$

$(n+1)n > 2010$

It is then easy to see that the lowest integer value of $n$ that satisfies the inequality is $\boxed{45\ \textbf{(A)}}$ .

An easy way to know that $45$ is the answer is that $50*51=2550$ , so you know $n<50$ - the only solution for n under $50$ is $45$ .

Solution 2

Using the first few values of $n$ , it is easy to derive a formula for $P(n)$ . The chance that she stops on the second box ( $n=2$ ) is the chance of drawing a white marble then a red marble: $\frac{1}2 \cdot \frac{1}3$ . The chance that she stops on the third box ( $n=3$ ) is the chance of drawing two white marbles then a red marble: $\frac{1}2 \cdot \frac{2}3 \cdot \frac{1}4$ . If $n=4$ , $P(n) = \frac{1}2 \cdot \frac{2}3 \cdot \frac{3}4 \cdot \frac{1}5$ .

Cross-cancelling in the fractions gives $P(2) =\frac{1}{2\cdot3}$ , $P(3) = \frac{1}{3\cdot4}$ , and $P(4) = \frac{1}{4\cdot5}$ . From this, it is clear that $P(n) = \frac{1}{(n)(n+1)}$ . (Alternatively, $P(n) = \frac{(n-1)!}{(n+1)!}$ .)

$\frac{1}{(n+1)(n)} < \frac{1}{2010}$

The lowest integer that satisfies the above inequality is $\boxed{(A) 45}$ .

Solution 3

We can think about it like this. For Isabella to stop, this just means she didn't get a red the previous $n-1$ times. So we can write a fraction for each probability that she doesn't get a red. For the first move it is just $\frac{1}{2}$ , then it is $\frac{2}{3}$ , and so on until $\frac{n-1}{n}$ . The last fraction is the probability she does get the red, which is $\frac{1}{n+1}$ . In the fraction set $\frac{1}{2}\frac{2}{3}...\frac{n-1}{n}$ . All the terms cancel except the 1, and the n, so we have $\frac{1}{n}\frac{1}{n+1}<\frac{1}{2010}$ , and by some estimating and guess and check, we get $\boxed{(\text{B}) 63}$

@@ Line 31: / Line 31: @@
 The lowest integer that satisfies the above inequality is <math>\boxed{(A) 45}</math>.
+=== Solution 3 ===
+We can think about it like this. For Isabella to stop, this just means she didn't get a red the previous <math>n-1</math> times. So we can write a fraction for each probability that she doesn't get a red. For the first move it is just <math>\frac{1}{2}</math>, then it is <math>\frac{2}{3}</math>, and so on until <math>\frac{n-1}{n}</math>. The last fraction is the probability she does get the red, which is <math>\frac{1}{n+1}</math>. In the fraction set <math>\frac{1}{2}\frac{2}{3}...\frac{n-1}{n}</math>. All the terms cancel except the 1, and the n, so we have <math>\frac{1}{n}\frac{1}{n+1}<\frac{1}{2010}</math>, and by some estimating and guess and check, we get <math>\boxed{(\text{B}) 63}</math>
 == See also ==

2010 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
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