Try this beautiful Problem on Probability in Marbles based on smallest value AMC 10 A, 2010. You may use sequential hints to solve the problem.
Each of 2010 boxes in a line contains a single red marble, and for $1 \leq k \leq 2010$, the box in the $k$ 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 $P(n)<\frac{1}{2010}$ ?
,
Probability
Combination
Marbles
Pre College Mathematics
AMC-10A, 2010 Problem-23
$45$
Given that Each of 2010 boxes in a line contains a single red marble, and for $1 \leq k \leq 2010$, the box in the $k$ th position also contains $k$ white marbles..
Therefore 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 $\frac{1}{k+1}$
Now can you finish the problem?
Also given that She stops when she first draws a red marble. Let $P(n)$ be the probability that Isabella stops after drawing exactly $n$ marbles.
Therefore we can say $P(n)=\left(\frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdots \frac{n-1}{n}\right) \cdot \frac{1}{n+1}=\frac{1}{n(n+1)}$
Now Can you finish the Problem?
Therefore the probability $\frac{1}{n(n+1)}<\frac{1}{2010}$ or $n(n+1)>2010$
Now $ n^2+n-2010>0$
Now to find out the factorization we see that $45 \times 46=2070$ and $ 44 \times 45 =1980$
As $n$ is smallest so $n=45$
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