# 2021 AMC 10B Problems/Problem 18

## 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 1

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 $\boxed{\textbf{(C) }\frac{1}{20}}.$

~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 probability that the first odd number appears on roll $n$ and all 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 $$\left (\frac {1}{2^{4}} + \frac {1}{2^{5}} + ... \right) - \left (\frac {1}{2 \cdot 3^{2}} + \frac {1}{2 \cdot 3^{3}} + ... \right) + \left (\frac {1}{2^{4} \cdot 3^{2}} + \frac {1}{2^{5} \cdot 3^{3}} + ... \right) = \left (\frac {1}{8} \right) - \left(\frac {\frac {1}{18}}{ \frac{2}{3}} \right) + \left(\frac {\frac {1}{144}}{\frac {5}{6}} \right) = \left (\frac {1}{8} \right) - \left(\frac {1}{12} \right) + \left (\frac{1}{120} \right) = \boxed{\textbf{(C) }\frac{1}{20}.}$$

~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)

 2021 AMC 10B (Problems • Answer Key • Resources) Preceded byProblem 17 Followed byProblem 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