Question

Each of $2010$ boxes in a line contains one red marble, and for $1\le k\le 2010,$ the box in the $k^{\text{th}}$ position also contains $k$ white marbles. Ram begins at the first box and successively draws a single marble at random from each box, in order. He stops when he first draws a red marble. Let $P\left(n\right)$ be the probability that he stops after drawing exactly $n$ marbles. Then the possible value(s) of $n$ for which $P\left(n\right)<\frac{1}{2010},$ is

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

Answer 1

Answer: (A), (B), (D)

$45$

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

To stop after drawing $n$ marbles, we must draw a white marble from boxes $1,2,3,\cdots ,(n-1)$ and draw a red marble from box $n.$
$\therefore P\left(n\right)=(\frac{1}{2}\cdot \frac{2}{3}\cdot \frac{3}{4}\cdots \frac{n-1}{n})\cdot \frac{1}{n+1}$
$\Rightarrow P\left(n\right)=\frac{1}{n(n+1)}<\frac{1}{2010}$
$\Rightarrow n(n+1)>2010$
Since $44\left(45\right)=1980$ and $45\left(46\right)=2070$
$\therefore n\ge 45$