Each of 2010 boxes in a line contains one red marble and for `1lekle2010`, the box is the `k^(th)` position also contain `k` white marbles. A child begins at the fist box and successively drawn a single marble at random each box in order. The stops when be fist draws a red marble. Let `p(n)` be the probability that the stops after drawing exactly `n` marbles. The possible value (s) of `n` for which `p(n)lt 1/ 2010` is:

Options

A. 44
B. 45
C. 46
D. 47

Answer

To solve the problem, we need to find the probability \( p(n) \) that the child stops after drawing exactly \( n \) marbles, and determine the possible values of \( n \) for which \( p(n) < \frac{1}{2010} \).

### Step-by-Step Solution:

1. **Understanding the Problem**:
Each of the 2010 boxes contains one red marble and \( k \) white marbles in the \( k^{th} \) box. The child draws marbles from the boxes in order until they draw a red marble.

2. **Calculating Probabilities**:
- The probability of drawing a white marble from box \( k \) is:
\[ P(\text{White from box } k) = \frac{k}{k + 1} \]
- The probability of drawing a red marble from box \( k \) is:
\[ P(\text{Red from box } k) = \frac{1}{k + 1} \]

3. **Condition for Stopping**:
The child stops after drawing \( n \) marbles if they draw white marbles from boxes 1 to \( n-1 \) and a red marble from box \( n \). Therefore, the probability \( p(n) \) can be expressed as:
\[ p(n) = \left( \prod_{k=1}^{n-1} \frac{k}{k+1} \right) \cdot \frac{1}{n+1} \]

4. **Simplifying the Product**:
The product \( \prod_{k=1}^{n-1} \frac{k}{k+1} \) simplifies to:
\[ \prod_{k=1}^{n-1} \frac{k}{k+1} = \frac{1}{n} \]
Therefore, we have:
\[ p(n) = \frac{1}{n} \cdot \frac{1}{n+1} = \frac{1}{n(n+1)} \]

5. **Setting Up the Inequality**:
We need to find \( n \) such that:
\[ \frac{1}{n(n+1)} < \frac{1}{2010} \]
This leads to:
\[ n(n+1) > 2010 \]

6. **Finding Values of \( n \)**:
We need to find integer values of \( n \) that satisfy this inequality. We can test values of \( n \):
- For \( n = 44 \):
\[ 44 \times 45 = 1980 \quad (\text{not greater than } 2010) \]
- For \( n = 45 \):
\[ 45 \times 46 = 2070 \quad (\text{greater than } 2010) \]
- For \( n = 46 \):
\[ 46 \times 47 = 2162 \quad (\text{greater than } 2010) \]
- For \( n = 47 \):
\[ 47 \times 48 = 2256 \quad (\text{greater than } 2010) \]

7. **Conclusion**:
The possible values of \( n \) for which \( p(n) < \frac{1}{2010} \) are \( n = 45, 46, 47, \ldots \).

### Final Answer:
The possible values of \( n \) for which \( p(n) < \frac{1}{2010} \) are \( n \geq 45 \).