Difference between revisions of "2021 AMC 10B Problems/Problem 18"

Revision as of 18:46, 11 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

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

@@ Line 17: / Line 17: @@
 ~Tucker
+Every set of three numbers chosen from <math>\{1,2,3,4,5,6\}</math> has an equal chance of being the first 3 distinct numbers rolled.
+Therefore, the probability that the first 3 distinct numbers are <math>\{2,4,6\}</math> is <math>\frac{1}{{6 \choose 3}}=\boxed{(C)~\frac{1}{20}}</math>
+~kingofpineapplz

Page

Toolbox

Search

Difference between revisions of "2021 AMC 10B Problems/Problem 18"

Revision as of 18:46, 11 February 2021

Problem

Solution