Difference between revisions of "2021 AMC 10B Problems/Problem 18"
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Therefore the probability is <math>\frac{3!(3!)}{6!} = \frac{1}{20} = \boxed{\textbf{(C)}}</math>. | Therefore the probability is <math>\frac{3!(3!)}{6!} = \frac{1}{20} = \boxed{\textbf{(C)}}</math>. | ||
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Revision as of 18:51, 11 February 2021
Problem
A fair 
-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?
Solution
There is a 
chance that the first number we choose is even.
There is a 
chance that the next number that is distinct from the first is even.
There is a 
chance that the next number distinct from the first two is even.
, so the answer is 
~Tucker
Every set of three numbers chosen from 
has an equal chance of being the first 3 distinct numbers rolled.
Therefore, the probability that the first 3 distinct numbers are 
is 
~kingofpineapplz
Solution 2 (If you don't have much time)
Note that the problem is basically asking us to find the probability in some permutation of 
that we get the three even numbers in the first three spots.
There are 
ways to order the numbers and 
ways to order the evens in the first three spots and the odds in the next three spots.
Therefore the probability is 
.
--abhinavg0627
