Difference between revisions of "2020 AMC 10B Problems/Problem 18"
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+ | {{duplicate|[[2020 AMC 10B Problems|2020 AMC 10B #18]] and [[2020 AMC 12B Problems|2020 AMC 12B #16]]}} | ||
+ | |||
==Problem== | ==Problem== | ||
− | An urn contains one red ball and one blue ball. A box of extra red and blue balls | + | An urn contains one red ball and one blue ball. A box of extra red and blue balls lies 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? |
<math>\textbf{(A) } \frac16 \qquad \textbf{(B) }\frac15 \qquad \textbf{(C) } \frac14 \qquad \textbf{(D) } \frac13 \qquad \textbf{(E) } \frac12</math> | <math>\textbf{(A) } \frac16 \qquad \textbf{(B) }\frac15 \qquad \textbf{(C) } \frac14 \qquad \textbf{(D) } \frac13 \qquad \textbf{(E) } \frac12</math> | ||
− | ==Solution== | + | ==Solution 1== |
Let <math>R</math> denote that George selects a red ball and <math>B</math> that he selects a blue one. Now, in order to get <math>3</math> balls of each color, he needs <math>2</math> more of both <math>R</math> and <math>B</math>. | Let <math>R</math> denote that George selects a red ball and <math>B</math> that he selects a blue one. Now, in order to get <math>3</math> balls of each color, he needs <math>2</math> more of both <math>R</math> and <math>B</math>. | ||
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Adding up the cases, we have <math>\frac{1}{15}+\frac{1}{15}+\frac{1}{15}=\boxed{\textbf{(B) }\frac15}</math> ~quacker88 | Adding up the cases, we have <math>\frac{1}{15}+\frac{1}{15}+\frac{1}{15}=\boxed{\textbf{(B) }\frac15}</math> ~quacker88 | ||
+ | |||
+ | ==Solution 2== | ||
+ | We know that we need to find the probability of adding 2 red and 2 blue balls in some order. | ||
+ | There are 6 ways to do this, since there are <math>\binom{4}{2}=6</math> ways to arrange <math>RRBB</math> in some order. | ||
+ | We will show that the probability for each of these 6 ways is the same. | ||
+ | |||
+ | We first note that the denominators should be counted by the same number. This number is <math>2 \cdot 3 \cdot 4 \cdot 5=120</math>. This is because 2, 3, 4, and 5 represent how many choices there are for the four steps. No matter what the <math>k-th</math> step involves <math>k+1</math> numbers to choose from. | ||
+ | |||
+ | The numerators are the number of successful operations. No matter the order, the first time a red is added will come from 1 choice and the second time will come from 2 choices, since that is how many reds are in the urn originally. | ||
+ | The same goes for the blue ones. The numerator must equal <math>(1 \cdot 2)^2</math>. | ||
+ | |||
+ | Therefore, the probability for each of the orderings of <math>RRBB</math> is <math>\frac{4}{120}=\frac{1}{30}</math>. There are 6 of these, so the total probability is <math>\boxed{\bf{(B)} \frac{1}{5}}</math>. | ||
+ | |||
+ | ==Solution 3== | ||
+ | First, notice that when George chooses a ball he just adds another ball of the same color. On George's first move, he either chooses the red or the blue with a <math>\frac{1}{2}</math> chance each. We can assume he chooses Red(chance <math>\frac{1}{2}</math>), and then multiply the final answer by two for symmetry. Now, there are two red balls and one blue ball in the urn. Then, he can either choose another Red(chance <math>\frac{2}{3}</math>), in which case he must choose two blues to get three of each, with probability <math>\frac{1}{4}\cdot\frac{2}{5}=\frac{1}{10}</math> or a blue for two blue and two red in the urn, with chance <math>\frac{1}{3}</math>. If he chooses blue next, he can either choose a red then a blue, or a blue then a red. Each of these has a <math>\frac{1}{2}\cdot\frac{2}{5}</math> for total of <math>2\cdot\frac{1}{5}=\frac{2}{5}</math>. The total probability that he ends up with three red and three blue is <math>2\cdot\frac{1}{2}(\frac{2}{3}\cdot\frac{1}{10}+\frac{1}{3}\cdot\frac{2}{5})=\frac{1}{15}+\frac{2}{15}=\boxed{\bf{(B)} \frac{1}{5}}</math>. ~aop2014 | ||
+ | |||
+ | ==Solution 4== | ||
+ | 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>. | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | ~JHawk0224 | ||
+ | |||
+ | ==Solution 5== | ||
+ | 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>. | ||
+ | |||
+ | ~FANYUCHEN20020715 | ||
+ | |||
+ | Edited by Kinglogic | ||
+ | |||
+ | ==Solution 6== | ||
+ | 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 <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> | ||
+ | |||
+ | ==Solution 7== | ||
+ | We can try to use dynamic programming to solve this problem. (Informatics Olympiad hahaha) | ||
+ | |||
+ | 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. | ||
+ | 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. | ||
+ | |||
+ | Then we have | ||
+ | <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> | ||
+ | |||
+ | Then we start can with <math>dp[1][1] = 1</math> and try to compute <math>dp[3][3]</math>. | ||
+ | |||
+ | <cmath>\begin{array}{|c || c | c | c | c | c |} | ||
+ | \hline | ||
+ | i \text{\ \textbackslash\ } j & 1 & 2 & 3\\ \hline\hline | ||
+ | 1 & 1 & 1/2 & 1/3\\ \hline | ||
+ | 2 & 1/2 & 1/3 & 1/4\\ \hline | ||
+ | 3 & 1/3 & 1/4 & 1/5\\ \hline | ||
+ | \end{array}</cmath> | ||
+ | The answer is <math>\boxed{\textbf{(B)}\ \frac{1}{5}}</math>. | ||
+ | |||
+ | (Solution by CircleOO) | ||
+ | |||
+ | ==Solution 8 (Brute force)== | ||
+ | Since there are only <math>4</math> operations, and each operation only has <math>2</math> possibilities, this problem is simple enough to brute force. So we draw a probability tree containing every possible outcome after <math>4</math> moves, add the probabilities of every outcome with <math>3</math> red balls and <math>3</math> blue balls, and find that the answer is <math>\boxed{\textbf{(B)}\ \frac{1}{5}}</math>. | ||
+ | |||
+ | -SmileKat32 | ||
==Video Solution== | ==Video Solution== | ||
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{{AMC10 box|year=2020|ab=B|num-b=17|num-a=19}} | {{AMC10 box|year=2020|ab=B|num-b=17|num-a=19}} | ||
+ | {{AMC12 box|year=2020|ab=B|num-b=15|num-a=17}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 02:51, 15 May 2021
- The following problem is from both the 2020 AMC 10B #18 and 2020 AMC 12B #16, so both problems redirect to this page.
Contents
Problem
An urn contains one red ball and one blue ball. A box of extra red and blue balls lies 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?
Solution 1
Let denote that George selects a red ball and that he selects a blue one. Now, in order to get balls of each color, he needs more of both and .
There are 6 cases: (we can confirm that there are only since ). However we can clump , , and together since they are equivalent by symmetry.
and
Let's find the probability that he picks the balls in the order of .
The probability that the first ball he picks is red is .
Now there are reds and blue in the urn. The probability that he picks red again is now .
There are reds and blue now. The probability that he picks a blue is .
Finally, there are reds and blues. The probability that he picks a blue is .
So the probability that the case happens is . However, since the case is the exact same by symmetry, case 1 has a probability of chance of happening.
and
Let's find the probability that he picks the balls in the order of .
The probability that the first ball he picks is red is .
Now there are reds and blue in the urn. The probability that he picks blue is .
There are reds and blues now. The probability that he picks a red is .
Finally, there are reds and blues. The probability that he picks a blue is .
So the probability that the case happens is . However, since the case is the exact same by symmetry, case 2 has a probability of chance of happening.
and
Let's find the probability that he picks the balls in the order of .
The probability that the first ball he picks is red is .
Now there are reds and blue in the urn. The probability that he picks blue is .
There are reds and blues now. The probability that he picks a blue is .
Finally, there are reds and blues. The probability that he picks a red is .
So the probability that the case happens is . However, since the case is the exact same by symmetry, case 3 has a probability of chance of happening.
Adding up the cases, we have ~quacker88
Solution 2
We know that we need to find the probability of adding 2 red and 2 blue balls in some order. There are 6 ways to do this, since there are ways to arrange in some order. We will show that the probability for each of these 6 ways is the same.
We first note that the denominators should be counted by the same number. This number is . This is because 2, 3, 4, and 5 represent how many choices there are for the four steps. No matter what the step involves numbers to choose from.
The numerators are the number of successful operations. No matter the order, the first time a red is added will come from 1 choice and the second time will come from 2 choices, since that is how many reds are in the urn originally. The same goes for the blue ones. The numerator must equal .
Therefore, the probability for each of the orderings of is . There are 6 of these, so the total probability is .
Solution 3
First, notice that when George chooses a ball he just adds another ball of the same color. On George's first move, he either chooses the red or the blue with a chance each. We can assume he chooses Red(chance ), and then multiply the final answer by two for symmetry. Now, there are two red balls and one blue ball in the urn. Then, he can either choose another Red(chance ), in which case he must choose two blues to get three of each, with probability or a blue for two blue and two red in the urn, with chance . If he chooses blue next, he can either choose a red then a blue, or a blue then a red. Each of these has a for total of . The total probability that he ends up with three red and three blue is . ~aop2014
Solution 4
Let the probability that the urn ends up with more red balls be denoted . Since this is equal to the probability there are more blue balls, the probability there are equal amounts is . the probability no more blues are chosen plus the probability only 1 more blue is chosen. The first case, .
The second case, . Thus, the answer is .
~JHawk0224
Solution 5
By conditional probability after 4 rounds we have 5 cases: RRRBBB, RRRRBB, RRBBBB, RRRRRB and RBBBBB. Thus the probability is . Put .
~FANYUCHEN20020715
Edited by Kinglogic
Solution 6
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 . Observe that the probability of arriving to 4+1 configuration is ( to get from XXY to XXXY, to get from XXXY to XXXXY). Thus the probability of arriving to 3+2 configuration is also , and the answer is
Solution 7
We can try to use dynamic programming to solve this problem. (Informatics Olympiad hahaha)
We let be the probability that we end up with red balls and blue balls. Notice that there are only two ways that we can end up with red balls and blue balls: one is by fetching a red ball from the urn when we have red balls and blue balls and the other is by fetching a blue ball from the urn when we have red balls and blue balls.
Then we have
Then we start can with and try to compute .
The answer is .
(Solution by CircleOO)
Solution 8 (Brute force)
Since there are only operations, and each operation only has possibilities, this problem is simple enough to brute force. So we draw a probability tree containing every possible outcome after moves, add the probabilities of every outcome with red balls and blue balls, and find that the answer is .
-SmileKat32
Video Solution
~IceMatrix
See Also
2020 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
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 10 Problems and Solutions |
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 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.