Difference between revisions of "2019 AMC 12B Problems/Problem 16"
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==Solution== | ==Solution== | ||
We solve for the probability by doing <math>\frac{1-(Probability of Equality)}{2}</math>. | We solve for the probability by doing <math>\frac{1-(Probability of Equality)}{2}</math>. | ||
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| + | We see that the probability of equality is the summation of all the probabilities that the balls land in the same container. Thus we have the probability of equality being equal to <math>(\frac{1}{2})(\frac{1}{2})+(\frac{1}{4})(\frac{1}{4})+(\frac{1}{16})(\frac{1}{16})...</math> | ||
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| + | The summation of this expression is equal to <cmath>\sum_{n=0}^{\infty} (1)(\frac{1}{4})^{n}-1</cmath>. Using the geometric sum formula, we obtain the summation of this expression to be <math>\frac{1}{\frac{3}{4}}-1</math> or <math>\frac{1}{3}</math>. | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2019|ab=B|num-b=15|num-a=17}} | {{AMC12 box|year=2019|ab=B|num-b=15|num-a=17}} | ||
Revision as of 13:12, 14 February 2019
Problem
Solution
We solve for the probability by doing 
.
We see that the probability of equality is the summation of all the probabilities that the balls land in the same container. Thus we have the probability of equality being equal to 
The summation of this expression is equal to ![\[\sum_{n=0}^{\infty} (1)(\frac{1}{4})^{n}-1\]](http://latex.artofproblemsolving.com/2/e/9/2e95b48186156ef9a8c13788cceb500d09f5345b.png)
. Using the geometric sum formula, we obtain the summation of this expression to be 
or 
.
See Also
| 2019 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 15 |
Followed by Problem 17 |
| 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 | |
