2021 AMC 10B Problems/Problem 18

Revision as of 19:51, 11 February 2021 by Abhinavg0627 (talk | contribs) (Solution 2 (If you don't have much time))

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


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 $1,2,3,4,5,6$ 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

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