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Question

Each of 2010 boxes in a line contains one red marble, and for 1k2010, the box in the kth 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(n) be the probability that he stops after drawing exactly n marbles. Then the possible value(s) of n for which P(n)<12010, is

A
47
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B
46
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C
44
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D
45
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Solution

The correct option is D 45
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 1,2,3,,(n1) and draw a red marble from box n.
P(n)=(122334n1n)1n+1
P(n)=1n(n+1)<12010
n(n+1)>2010
Since 44(45)=1980 and 45(46)=2070
n45

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