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

Revision as of 16:14, 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 first odd number appears on roll $n$ . We now proceed with complementary counting.

$n \le 3$ , it's impossible to have all $3$ evens appear before an odd. Note that for $n \ge 4,$ there's a $\frac {1}{2^{n}} - \frac {1}{2^{n}} \cdot \frac {3(2^{n-1}-3)+3}{3^{n-1}}$

Video Solution by OmegaLearn (Conditional probability)

https://youtu.be/IX-Y38KPxqs

2021 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

@@ Line 35: / Line 35: @@
 --abhinavg0627
+==Solution 4==
+Let <math>P_n</math> denote the probabilty that first odd number appears on roll <math>n</math>. 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>
 == Video Solution by OmegaLearn (Conditional probability) ==

Page

Toolbox

Search