Probability
Geometric probability — marble draws
MJAT_TS1_P2
Grade 12
Question:
Each of 2010 boxes in a line contains one red marble, and for $1 \leq k \leq 2010$, the $k$-th box also contains $k$ white marbles. A child begins at the first box and successively draws a single marble at random from each box in order. The child stops when he first draws a red marble. Let $p(n)$ be the probability that he stops after drawing exactly $n$ marbles. The possible values of $n$ for which $p(n) < \dfrac{1}{2010}$ are
A) $n = 44$
B) $n = 45$
C) $n = 46$
D) $n = 47$
Step-by-Step Solution
Key Concept: $p(n) = \dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdots\dfrac{n-1}{n}\cdot\dfrac{1}{n+1} = \dfrac{1}{n(n+1)}$. We need $\dfrac{1}{n(n+1)} < \dfrac{1}{2010}$, i.e., $n(n+1) > 2010$.
$p(n) = \frac{1}{n(n+1)}$. Check: $n=44$: $\frac{1}{1980} > \frac{1}{2010}$ — NO. $n=45$: $\frac{1}{2070} < \frac{1}{2010}$ ✓. $n=46$: $\frac{1}{2162} < \frac{1}{2010}$ ✓. $n=47$: $\frac{1}{2256} < \frac{1}{2010}$ ✓. Answer: B, C, D.
Correct Answer: BCD