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| − | ==Problem==
| + | #REDIRECT [[2020 AMC 10B Problems/Problem 18]] |
| − | 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.
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| − | After the four iterations the urn contains six balls. What is the probability that the urn contains three balls of each color?
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| − | <math>\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}</math>
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| − | ==Solution==
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| − | Let the probability that the urn ends up with more red balls be denoted <math>P(R)</math>. Since this is equal to the probability there are more blue balls, the probability there are equal amounts is <math>1-2P(R)</math>. <math>P(R) =</math> the probability no more blues are chosen plus the probability only 1 more blue is chosen. The first case, <math>P(\text{no more blues}) = \frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}=\frac{1}{5}</math>.
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| − | The second case, <math>P(\text{1 more blue}) = 4\cdot\frac{1\cdot1\cdot2\cdot3}{2\cdot3\cdot4\cdot5} = \frac{1}{5}</math>. Thus, the answer is <math>1-2\left(\frac{1}{5}+\frac{1}{5}\right)=1-\frac{4}{5}=\boxed{\textbf{(B)}\ \frac{1}{5}}</math>.
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| − | ~JHawk0224
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| − | ==Solution 2==
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| − | By conditional probability after 4 rounds we have 5 cases: RRRBBB, RRRRBB, RRBBBB, RRRRRB and RBBBBB. Thus the probability is <math>\frac{1}{5}</math>. Put <math>\boxed{B}</math>.
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| − | ~FANYUCHEN20020715
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| − | Edited by Kinglogic
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| − | ==Solution 3==
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| − | Here X stands for R or B, and Y for the remaining color.
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| − | 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 <cmath>\frac{2}{3} \cdot \frac{3}{4} = \frac{1}{2}</cmath> (<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>
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| − | ==Solution 4==
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| − | We can try to use dynamic programming to solve this problem. (Informatics Olympiad hahaha)
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| − | We let <math>dp[i][j]</math> be the probability that we end up with <math>i</math> red balls and <math>j</math> blue balls.
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| − | Notice that there are only two ways that we can end up with <math>i</math> red balls and <math>j</math> blue balls: one is by fetching a red ball from the urn when we have <math>i - 1</math> red balls and <math>j</math> blue balls and the other is by fetching a blue ball from the urn when we have <math>i</math> red balls and <math>j - 1</math> blue balls.
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| − | Then we have
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| − | <math>dp[i][j] = \frac{i - 1}{i - 1 + j} dp[i - 1][j] + \frac{j - 1}{i - 1 + j} dp[i][j - 1]</math>
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| − | Then we start can with <math>dp[1][1] = 1</math> and try to compute <math>dp[3][3]</math>.
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| − | <cmath>\begin{array}{|c || c | c | c | c | c |}
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| − | \hline
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| − | i \text{\ \textbackslash\ } j & 1 & 2 & 3\\ \hline\hline
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| − | 1 & 1 & \frac{1}{2} & \frac{1}{3}\\ \hline
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| − | 2 & \frac{1}{2} & \frac{1}{3} & \frac{1}{4}\\ \hline
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| − | 3 & \frac{1}{3} & \frac{1}{4} & \frac{1}{5}\\ \hline
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| − | \end{array}</cmath>
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| − | The answer is <math>\boxed{\textbf{(B)}\ \frac{1}{5}}</math>.
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| − | (Solution by CircleOO)
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| − | ==See Also==
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| − | {{AMC12 box|year=2020|ab=B|num-b=15|num-a=17}}
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| − | {{MAA Notice}}
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