Difference between revisions of "2019 AMC 12B Problems/Problem 13"
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<cmath>\sum_{i=1}^{\infty} \frac{1}{4^i} = \boxed{\textbf{(C) } \frac{1}{3}}</cmath> | <cmath>\sum_{i=1}^{\infty} \frac{1}{4^i} = \boxed{\textbf{(C) } \frac{1}{3}}</cmath> | ||
− | -scrabbler94 | + | (Note: to find this sum, we use the formula <math>\sum_{i=1}^{\infty} r^i = \frac{r}{1-r}</math>. Since in this case, <math>r = \frac{1}{4}</math>, the answer is <math>\frac{\frac{1}{4}}{\frac{3}{4}} = \frac{1}{3}</math>. If you don't know this formula, you may instead note that if you multiply the sum by <math>4</math>, it is equivalent to adding <math>1</math>. Thus: <math>4n = n+1</math>, which clearly simplifies to <math>n = \frac{1}{3}</math>. |
+ | |||
+ | - scrabbler94 (explanation of infinite sum provided by Robin) | ||
==See Also== | ==See Also== |
Revision as of 15:15, 14 February 2019
- The following problem is from both the 2019 AMC 10B #17 and 2019 AMC 12B #13, so both problems redirect to this page.
Problem
A red ball and a green ball are randomly and independently tossed into bins numbered with the positive integers so that for each ball, the probability that it is tossed into bin is for What is the probability that the red ball is tossed into a higher-numbered bin than the green ball?
Solution
The probability of the red ball landing in a higher-numbered bin is the same as the probability of the green ball landing in a higher numbered bin. The probability of both landing in the same bin is . The sum is equal to . Therefore the other two probabilities have to both be .
Solution by a1b2
Solution 2 (variant)
Suppose the green ball goes in bin , for some . The probability of this occurring is . Given this occurs, the probability that the red ball goes in a higher-numbered bin is . Thus the probability that the green ball goes in bin , and the red ball goes in a bin greater than , is . Summing from to infinity, we get
(Note: to find this sum, we use the formula . Since in this case, , the answer is . If you don't know this formula, you may instead note that if you multiply the sum by , it is equivalent to adding . Thus: , which clearly simplifies to .
- scrabbler94 (explanation of infinite sum provided by Robin)
See Also
2019 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 16 |
Followed by Problem 18 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
2019 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 12 |
Followed by Problem 14 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.