smc

Question

Each of 2010 boxes in a line contains a single red marble, and for 1 k 2010, the box in the kth 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) < 12010?

Accepted Answer

The probability of drawing a white marble from box k is kk + 1, and the probability of drawing a red marble from box k is 1k+1. To stop after drawing n marbles, we must draw a white marble from boxes 1, 2, , n-1, and draw a red marble from box n. Thus, P(n) = (12 23 34 n - 1n) 1n +1 = 1n (n + 1). So, we must have 1n(n + 1) < 12010 or n(n+1) > 2010. Since n(n+1) increases as n increases, we can simply test values of n; after some trial and error, we get that the minimum value of n is 45, since 45(46) = 2070 but 44(45) = 1980.