2019 AMC 12B Problems/Problem 16

Problem

Solution

We solve for the probability by doing $\frac{1-(Probability of Equality)}{2}$ .

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 $(\frac{1}{2})(\frac{1}{2})+(\frac{1}{4})(\frac{1}{4})+(\frac{1}{16})(\frac{1}{16})...$

The summation of this expression is equal to $\sum_{n=0}^{\infty} (1)(\frac{1}{4})^{n}-1$ . Using the geometric sum formula, we obtain the summation of this expression to be $\frac{1}{\frac{3}{4}}-1$ or $\frac{1}{3}$ .

2019 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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2019 AMC 12B Problems/Problem 16

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