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Difference between revisions of "2021 AMC 10B Problems/Problem 18"
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==Solution 4==
 
==Solution 4==
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Let <math>P_n</math> denote the probabilty that first odd number appears on roll <math>n</math>. We now proceed with complementary counting.  
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Let <math>P_n</math> denote the probabilty that the first odd number appears on roll <math>n</math> and our conditions are met. We now proceed with complementary counting.  
  
−
<math>n \le 3</math>, it's impossible to have all <math>3</math> evens appear before an odd. Note that for <math>n \ge 4,</math> there's a <math>\frac {1}{2^{n}} - \frac {1}{2^{n}} \cdot \frac {3(2^{n-1}-3)+3}{3^{n-1}}</math>
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For <math>n \le 3</math>, it's impossible to have all <math>3</math> evens appear before an odd. Note that for <math>n \ge 4,</math> <cmath>P_n = \frac {1}{2^{n}} - \frac {1}{2^{n}} \left(\frac{\binom{3}{2}(2^{n-1}-2)+\binom{3}{2}}{3^{n-1}}\right) = \frac {1}{2^{n}} - \left(\frac {3(2^{n-1})-3}{2^{n} \cdot 3^{n-1}}\right) = \frac {1}{2^{n}} - \left(\frac {1}{2 \cdot 3^{n-2}} - \frac{1}{2^{n} \cdot 3^{n-2}} \right).</cmath>
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 +
Summing for all <math>n</math>, we get our answer of <cmath>\left (\frac {1}{2^{4}} + \frac {1}{2^{5}} + ... \right) - \left (\frac {1}{2 \cdot 3^{2}} + \frac {1}{2 \cdot 3^{3}} + ... \right) + \left (\frac {1}{2^{4} \cdot 3^{2}} + \frac {1}{2^{5} \cdot 3^{3}} + ... \right) = \left (\frac {1}{8} \right) - \left(\frac {\frac {1}{18}}{ \frac{2}{3}} \right) + \left(\frac {\frac {1}{144}}{\frac {5}{6}} \right) = \left (\frac {1}{8} \right) - \left(\frac {1}{12} \right) + \left (\frac{1}{120} \right) = \frac {1}{120}.</cmath>
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~ike.chen
  
 
==Solution 5==
 
==Solution 5==

Revision as of 17:12, 12 February 2021

Problem

A fair $6$-sided die is repeatedly rolled until an odd number appears. What is the probability that every even number appears at least once before the first occurrence of an odd number?

$\textbf{(A)} ~\frac{1}{120} \qquad\textbf{(B)} ~\frac{1}{32} \qquad\textbf{(C)} ~\frac{1}{20} \qquad\textbf{(D)} ~\frac{3}{20} \qquad\textbf{(E)} ~\frac{1}{6}$


Solution

There is a $\frac{3}6$ chance that the first number we choose is even.

There is a $\frac{2}5$ chance that the next number that is distinct from the first is even.

There is a $\frac{1}4$ chance that the next number distinct from the first two is even.

$\frac{3}6 * \frac{2}5 * \frac{1}4 = \frac{1}{20}$, so the answer is $\boxed{ C) \frac{1}{20} }$

~Tucker

Solution 2

Every set of three numbers chosen from $\{1,2,3,4,5,6\}$ has an equal chance of being the first 3 distinct numbers rolled.

Therefore, the probability that the first 3 distinct numbers are $\{2,4,6\}$ is $\frac{1}{{6 \choose 3}}=\boxed{(C)~\frac{1}{20}}$

~kingofpineapplz

Solution 3

Note that the problem is basically asking us to find the probability that in some permutation of $1,2,3,4,5,6$ that we get the three even numbers in the first three spots.

There are $6!$ ways to order the $6$ numbers and $3!(3!)$ ways to order the evens in the first three spots and the odds in the next three spots.

Therefore the probability is $\frac{3!(3!)}{6!} = \frac{1}{20} = \boxed{\textbf{(C)}}$.


--abhinavg0627

Solution 4

Let $P_n$ denote the probabilty that the first odd number appears on roll $n$ and our conditions are met. We now proceed with complementary counting.

For $n \le 3$, it's impossible to have all $3$ evens appear before an odd. Note that for $n \ge 4,$ \[P_n = \frac {1}{2^{n}} - \frac {1}{2^{n}} \left(\frac{\binom{3}{2}(2^{n-1}-2)+\binom{3}{2}}{3^{n-1}}\right) = \frac {1}{2^{n}} - \left(\frac {3(2^{n-1})-3}{2^{n} \cdot 3^{n-1}}\right) = \frac {1}{2^{n}} - \left(\frac {1}{2 \cdot 3^{n-2}} - \frac{1}{2^{n} \cdot 3^{n-2}} \right).\]

Summing for all $n$, we get our answer of \[\left (\frac {1}{2^{4}} + \frac {1}{2^{5}} + ... \right) - \left (\frac {1}{2 \cdot 3^{2}} + \frac {1}{2 \cdot 3^{3}} + ... \right) + \left (\frac {1}{2^{4} \cdot 3^{2}} + \frac {1}{2^{5} \cdot 3^{3}} + ... \right) = \left (\frac {1}{8} \right) - \left(\frac {\frac {1}{18}}{ \frac{2}{3}} \right) + \left(\frac {\frac {1}{144}}{\frac {5}{6}} \right) = \left (\frac {1}{8} \right) - \left(\frac {1}{12} \right) + \left (\frac{1}{120} \right) = \frac {1}{120}.\]

~ike.chen

Solution 5

Let $E_n$ be that probability that the condition in the problem is satisfied given that we need $n$ more distinct even numbers. Then, \[E_1=\frac{1}{6}+\frac{1}{3}\cdot E_1+\frac{1}{2}\cdot 0,\] since there is a $\frac{1}{3}$ probability that we will roll an even number we already have rolled and will be in the same position again. Solving, we find that $E_1=\frac{1}{4}$.

We can apply the same concept for $E_2$ and $E_3$. We find that $E_2=\frac{1}{3}\cdot E_1+\frac{1}{6}\cdot E_2+\frac{1}{2}\cdot 0$, and so $E_2=\frac{1}{10}$. Also, $E_3=\frac{1}{2}\cdot E_2+\frac{1}{2}\cdot 0$, so $E_3=\frac{1}{20}$. Since the problem is asking for $E_3$, our answer is $\boxed{\textbf{(C) }\frac{1}{20}}$. -BorealBear

Video Solution by OmegaLearn (Conditional probability)

https://youtu.be/IX-Y38KPxqs

2021 AMC 10B (Problems • Answer Key • Resources)
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Problem 17 		Followed by
Problem 19
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All AMC 10 Problems and Solutions
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