2019 AMC 12B Problems/Problem 16: Difference between revisions

Revision as of 15:44, 14 February 2019

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

@@ Line 2: / Line 2: @@
 ==Solution==
-We solve for the probability by doing <math>\frac{1-(Probability of Equality)}{2}</math>.
-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>
-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==
 {{AMC12 box|year=2019|ab=B|num-b=15|num-a=17}}