Difference between revisions of "2020 AMC 10B Problems/Problem 18"

(Solution 2)
(Solution 2)
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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.
 
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 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.
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The same goes for the blue ones. The numerator must equal <math>(1 \cdot 2)^2</math>.
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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{(B) \frac{1}{5}}</math>.
  
 
==Video Solution==
 
==Video Solution==

Revision as of 20:48, 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?

$\textbf{(A) } \frac16 \qquad \textbf{(B) }\frac15 \qquad \textbf{(C) } \frac14 \qquad \textbf{(D) } \frac13 \qquad \textbf{(E) } \frac12$

Solution

Let $R$ denote that George selects a red ball and $B$ that he selects a blue one. Now, in order to get $3$ balls of each color, he needs $2$ more of both $R$ and $B$.

There are 6 cases: $RRBB, RBRB, RBBR, BBRR, BRBR, BRRB$ (we can confirm that there are only $6$ since $\binom{4}{2}=6$). However we can clump $RRBB + BBRR$, $RBRB + BRBR$, and $RBBR + BRRB$ together since they are equivalent by symmetry.


$\textbf{CASE 1: }$ $RRBB$ and $BBRR$

Let's find the probability that he picks the balls in the order of $RRBB$.

The probability that the first ball he picks is red is $\frac{1}{2}$.

Now there are $2$ reds and $1$ blue in the urn. The probability that he picks red again is now $\frac{2}{3}$.

There are $3$ reds and $1$ blue now. The probability that he picks a blue is $\frac{1}{4}$.

Finally, there are $3$ reds and $2$ blues. The probability that he picks a blue is $\frac{2}{5}$.

So the probability that the $RRBB$ case happens is $\frac{1}{2} \cdot \frac{2}{3} \cdot \frac{1}{4} \cdot \frac{2}{5} = \frac{1}{30}$. However, since the $BBRR$ case is the exact same by symmetry, case 1 has a probability of $\frac{1}{30} \cdot 2 = \frac{1}{15}$ chance of happening.


$\textbf{CASE 2: }$ $RBRB$ and $BRBR$

Let's find the probability that he picks the balls in the order of $RBRB$.

The probability that the first ball he picks is red is $\frac{1}{2}$.

Now there are $2$ reds and $1$ blue in the urn. The probability that he picks blue is $\frac{1}{3}$.

There are $2$ reds and $2$ blues now. The probability that he picks a red is $\frac{1}{2}$.

Finally, there are $3$ reds and $2$ blues. The probability that he picks a blue is $\frac{2}{5}$.

So the probability that the $RBRB$ case happens is $\frac{1}{2} \cdot \frac{1}{3} \cdot \frac{1}{2} \cdot \frac{2}{5} = \frac{1}{30}$. However, since the $BRBR$ case is the exact same by symmetry, case 2 has a probability of $\frac{1}{30} \cdot 2 = \frac{1}{15}$ chance of happening.


$\textbf{CASE 3: }$ $RBBR$ and $BRRB$

Let's find the probability that he picks the balls in the order of $RBBR$.

The probability that the first ball he picks is red is $\frac{1}{2}$.

Now there are $2$ reds and $1$ blue in the urn. The probability that he picks blue is $\frac{1}{3}$.

There are $2$ reds and $2$ blues now. The probability that he picks a blue is $\frac{1}{2}$.

Finally, there are $2$ reds and $3$ blues. The probability that he picks a red is $\frac{2}{5}$.

So the probability that the $RBBR$ case happens is $\frac{1}{2} \cdot \frac{1}{3} \cdot \frac{1}{2} \cdot \frac{2}{5} = \frac{1}{30}$. However, since the $BRBR$ case is the exact same by symmetry, case 3 has a probability of $\frac{1}{30} \cdot 2 = \frac{1}{15}$ chance of happening.

Adding up the cases, we have $\frac{1}{15}+\frac{1}{15}+\frac{1}{15}=\boxed{\textbf{(B) }\frac15}$ ~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 $\binom{4}{2}=6$ ways to arrange $RRBB$ 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 $2 \cdot 3 \cdot 4 \cdot 5=120$. This is because 2, 3, 4, and 5 represent how many choices there are for the four steps. No matter what the $k-th$ step involves $k+1$ 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 $(1 \cdot 2)^2$.

Therefore, the probability for each of the orderings of $RRBB$ is $\frac{4}{120}=\frac{1}{30}$. There are 6 of these, so the total probability is $\boxed{(B) \frac{1}{5}}$.

Video Solution

https://youtu.be/3BvJeZU3T-M

~IceMatrix

See Also

2020 AMC 10B (ProblemsAnswer KeyResources)
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

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