2020 AMC 12B Problems/Problem 16

Problem

An urn contains one red ball and one blue ball. A box of extra red and blue balls lie nearby. George performs the following operation four times: he draws a ball from the urn at random and then takes a ball of the same color from the box and returns those two matching balls to the urn. After the four iterations the urn contains six balls. What is the probability that the urn contains three balls of each color? $\textbf{(A)}\ \frac{1}{6} \qquad\textbf{(B)}\ \frac{1}{5} \qquad\textbf{(C)}\ \frac{1}{4} \qquad\textbf{(D)}\ \frac{1}{3} \qquad\textbf{(E)}\ \frac{1}{2}$

Solution

Let the probability that the urn ends up with more red balls be denoted $P(R)$ . Since this is equal to the probability there are more blue balls, the probability there are equal amounts is $1-2P(R)$ . $P(R) =$ the probability no more blues are chosen plus the probability only 1 more blue is chosen. The first case, $P(\text{no more blues}) = \frac{1}{2}*\frac{2}{3}*\frac{3}{4}*\frac{4}{5}=\frac{1}{5}$ .

The second case, $P(\text{1 more blue}) = 4*\frac{1*1*2*3}{2*3*4*5} = \frac{1}{5}$ . Thus, the answer is $1-2(\frac{1}{5}+\frac{1}{5})=1-\frac{4}{5}=\boxed{\textbf{(B)}\ \frac{1}{5}}$ .

~JHawk0224

Solution 2

By conditional probability after 4 rounds we have 5 cases: RRRBBB, RRRRBB, RRBBBB, RRRRRB and BRRRRR. Thus the probability is $\frac{1}{5}$ . Put $\boxed{B}$ .

~FANYUCHEN20020715

2020 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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2020 AMC 12B Problems/Problem 16

Contents

Problem

Solution

Solution 2

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