Difference between revisions of "2020 AMC 12B Problems/Problem 16"

Revision as of 01:39, 8 February 2020

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

Solution 3

Here X stands for R or B, and Y for the remaining color. After 3 rounds one can either have a 4+1 configuration (XXXXY), or 3+2 configuration (XXXYY). The probability of getting to XXXYYY from XXXYY is $\frac{2}{5}$ . Observe that the probability of arriving to 4+1 configuration is $\frac{2}{3} \cdot \frac{3}{4} = \frac{1}{2}$ ( $\frac{2}{3}$ to get from XXY to XXXY, $\frac{3}{4}$ to get from XXXY to XXXXY). Thus the probability of arriving to 3+2 configuration is also $\frac{1}{2}$ , and the answer is $\frac{1}{2} \cdot \frac{2}{5} = \boxed{\textbf{(B)}\ \frac{1}{5}}.$

@@ Line 18: / Line 18: @@
 ==Solution 3==
 Here X stands for R or B, and Y for the remaining color.
-After 3 rounds one can either have a 4+1 configuration (XXXXY), or 3+2 configuration (XXXYY). The probability of getting to XXXYYY from XXXYY is  <math>\frac{2}{5}</math>. Observe that the probability of arriving to 4+1 configuration is  <math>\frac{2}{3} \cdot \frac{3}{4} = \frac{1}{2}</math> ( <math>\frac{2}{3}</math> to get from XXY to XXXY, <math>\frac{3}{4}</math> to get from XXXY to XXXXY). Thus the probability of arriving to XXXYY is also <math>\frac{1}{2}</math>, and the answer is <cmath>\frac{1}{2} \cdot \frac{2}{5} = \boxed{\textbf{(B)}\ \frac{1}{5}}. </cmath>
+After 3 rounds one can either have a 4+1 configuration (XXXXY), or 3+2 configuration (XXXYY). The probability of getting to XXXYYY from XXXYY is  <math>\frac{2}{5}</math>. Observe that the probability of arriving to 4+1 configuration is  <math>\frac{2}{3} \cdot \frac{3}{4} = \frac{1}{2}</math> ( <math>\frac{2}{3}</math> to get from XXY to XXXY, <math>\frac{3}{4}</math> to get from XXXY to XXXXY). Thus the probability of arriving to 3+2 configuration is also <math>\frac{1}{2}</math>, and the answer is <cmath>\frac{1}{2} \cdot \frac{2}{5} = \boxed{\textbf{(B)}\ \frac{1}{5}}. </cmath>
 ==See Also==

2020 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
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