2010 AMC 10A Problem 23

Box $k$ contains $1$ red and $k$ white marbles, $k + 1$ in all, so the chance of drawing white there is $\dfrac{k}{k+1}$ and the chance of red is $\dfrac{1}{k+1}$.

Isabella stops after exactly $n$ draws when she draws white from boxes $1, 2, \dots, n-1$ and red from box $n$. The draws are independent, so
$$P(n) = \frac12 \cdot \frac23 \cdot \frac34 \cdots \frac{n-1}{n} \cdot \frac{1}{n+1}.$$
Every numerator cancels the previous denominator, leaving
$$P(n) = \frac{1}{n} \cdot \frac{1}{n+1} = \frac{1}{n(n+1)}.$$

We need the smallest $n$ with $n(n+1) > 2010$. Since $44 \cdot 45 = 1980 < 2010$ and $45 \cdot 46 = 2070 > 2010$, the answer is $n = 45$.

The answer is $\boxed{\textbf{(A)}\ 45}$.

"Stops at step $n$" always means "survives steps $1$ through $n - 1$, then fails at $n$," a product of sequential probabilities. When the survival factors have the form $\frac{k}{k+1}$, the product telescopes to a single fraction, and the final inequality is just estimating $\sqrt{2010} \approx 44.8$.

• Using only the red probability at box n, P(n) = 1/(n+1), and ignoring that she must have drawn white from every earlier box.
• Solving $n^2 < 2010$ instead of $n(n+1) > 2010$ and landing on $44$, or misreading the direction of the inequality.