2010 AMC 10A Problems/Problem 23
Problem
Each of boxes in a line contains a single red marble, and for , the box in the position also contains 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 be the probability that Isabella stops after drawing exactly marbles. What is the smallest value of for which ?
Solutions
Solution 1
The probability of drawing a white marble from box is . The probability of drawing a red marble from box is .
The probability of drawing a red marble at box is therefore
It is then easy to see that the lowest integer value of that satisfies the inequality is .
An easy way to know that is the answer is that , so you know - the only solution for n under is .
Solution 2
Using the first few values of , it is easy to derive a formula for . The chance that she stops on the second box () is the chance of drawing a white marble then a red marble: . The chance that she stops on the third box () is the chance of drawing two white marbles then a red marble:. If , .
Cross-cancelling in the fractions gives , , and . From this, it is clear that . (Alternatively, .)
The lowest integer that satisfies the above inequality is .
Solution 3
We can think about it like this. For Isabella to stop, this just means she didn't get a red the previous times. So we can write a fraction for each probability that she doesn't get a red. For the first move it is just , then it is , and so on until . The last fraction is the probability she does get the red, which is . In the fraction set . All the terms cancel except the 1, and the n, so we have , and by some estimating and guess and check, we get ~awsomek
See also
2010 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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