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2021 AMC 10B Problems/Problem 18

Revision as of 22:27, 28 October 2022 by Pi is 3.14 (talk | contribs)

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 $\frac36$ 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.

$\frac{3}{6} \cdot \frac{2}{5} \cdot \frac{1}{4} = \frac{1}{20}$, 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 $n \ge 4,$ \[P_n = \frac {1}{2^{n}} - \frac {1}{2^{n}} \left(\frac{\binom{3}{2}(2^{n-1}-2)+\binom{3}{1}}{3^{n-1}}\right)\] since there's a $\frac {1}{2^{n}}$ chance that the first odd appears on roll $n$ (disregarding the other conditions) and the other term is subtracting the probability that less than $3$ of the evens show up before the first odd roll. Simplifying, we arrive at \[P_n= \frac {1}{2^{n}} - \left(\frac {3(2^{n-1})-3}{2^{n} \cdot 3^{n-1}}\right) = \frac {1}{2^{n}} - \frac {1}{2 \cdot 3^{n-2}} + \frac{1}{2^{n} \cdot 3^{n-2}}.\]

Summing for all $n$, we get our answer of \[\left (\frac {1}{2^{4}} + \frac {1}{2^{5}} + ... \right) - \left (\frac {1}{2 \cdot 3^{2}} + \frac {1}{2 \cdot 3^{3}} + ... \right) + \left (\frac {1}{2^{4} \cdot 3^{2}} + \frac {1}{2^{5} \cdot 3^{3}} + ... \right) = \left (\frac {1}{8} \right) - \left(\frac {\frac {1}{18}}{ \frac{2}{3}} \right) + \left(\frac {\frac {1}{144}}{\frac {5}{6}} \right) = \frac {1}{8} - \frac {1}{12} + \frac{1}{120} = \boxed{\textbf{(C) }\frac{1}{20}.}\]

~ike.chen

Solution 5 (States)

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

Solution 1a (same as solution 1 but with a little more of explanation)

The probability of choosing an even number on the first turn is $\frac{1}{2}$, now since you already chose that number, it is irrelevant to the problem now, so, if you chose the number again, it doesn't really matter to our problem anymore. Now, with our remaining $5$ numbers, the probability of choosing another even number is $\frac{2}{5}$, and again, after you have chosen that number, it is out of our problem. Now, you just have $4$ numbers left and the probability of choosing the last even number is $\frac{1}{4}$, so the answer is $\frac{1}{2} \times \frac{2}{5} \times \frac{1}{4}$ $=$ $\frac{1}{20}$.

~math31415926535

Solution 6 (Infinite Geometric Sequence Method)

Let's say that our even integers are found in the first $n$ numbers where n must be greater than or equal to $3$. Then, we can form an argument based on this. There are $3^n$ total ways to assign even numbers being $($2$,$4$,$6$)$ to each space. Furthermore, we must subtract the cases where we have only a single even integer present in the total $n$ spaces and the case where there are only $2$ distinct even integers present. There is $1$ way we can have a single even integer in the entire $n$ spaces, therefore giving us just $1$ option for each of the three integers, so we have $3$ total cases for this. Moreover, the amount of cases with just $2$ distinct even integers is $2^n$ but subtract the cases where all of the n spaces is either a single integer giving us $2^n-2$, but we multiply by $3C2$ because of the ways to choose $2$ distinct even integers that are used in the sequence of $n$. Finally, we have $\sum_{n=3}^{\infty} \frac{(3^n-3-3(2^n-2))}{6^n}$ note: we must divide all of this by $6^n$ for probability. Additionally, over the entire summation, we multiply by $1/2$ because of the $1/2$ probability of selecting an odd at the end of all the evens. Therefore, if you compute this using infinite geometric sequences, you get $1/20$. $\boxed{\textbf{(C) }\frac{1}{20}.}$

~Jske25

Solution 2a (same as 2 but with more explanation)

The question basically asks for the probability that $2$, $4$, and $6$ are all rolled before $1$, $3$ or $5$. This probability is equal to the probability that any 3 number combination is rolled before the remaining numbers are rolled (for example, rolling $1, 2$ and $3$ before $4$, $5$ and $6$). There are $\binom{6}{3}$ combinations possible and summing them all up must result in 1 so the probability of the specific combination $2$ $4$ $6$ being chosen is$\boxed{\textbf{(C) }\frac{1}{20}.}$

~AwesomeK

Solution 8 (Complementary + PIE)

We know that the sequence with odd ending will eventually add up to probability 1.

For every even number appears at least once, let's look at the complementary, i.e. without some even numbers ${2, 4, 6}$. $A = \{ without 2\}, B = \{ without 4\}, C = \{ without 6\}$.

$1 - A\cup B\cup C = 1 - (A + B + C - A\cap B - B\cap C - A\cap C + A\cap B\cap C)$

Let $f(x) = \frac{1}{2} + x\frac{1}{2} + x^2\frac{1}{2} + \ldots = \frac{1}{2} \frac{1}{1-x}$

2 even numbers left $|A|=|B|=|C|: f(\frac{2}{6}) = \frac{1}{2} \frac{1}{1-\frac{1}{3}} = \frac{3}{4}$,

1 even number left $|A\cap B| = |B\cap C| = |A\cap C|: f(\frac{1}{6}) = \frac{1}{2} \frac{1}{1-\frac{1}{6}} = \frac{3}{5}$

0 even number left $|A\cap B\cap C|: f(0) = \frac{1}{2}$

ans $= 1 - A\cup B\cup C = 1- (\frac{3}{4} * 3 - \frac{3}{5} * 3 + \frac{1}{2}) = \frac{1}{20}$

~aliciawu

Solution 10

We don't care what happens after we pick the first three distinct even numbers. This can be done is $3! = 6$ ways. Totally there are $6 \cdot 5 \cdot 4$ ways to pick three distinct numbers. So the answer is $\frac{6}{6\cdot5\cdot4} = \frac{1}{20}$.

~coolmath_2018

Video Solution by OmegaLearn

https://youtu.be/IX-Y38KPxqs

~pi_is_3.14

Video Solution by hurdler (complementary probability)

https://www.youtube.com/watch?v=k2Jy4ni9tK8

Video Solution by TheBeautyofMath

https://youtu.be/FV9AnyERgJQ?t=480

~IceMatrix

Video Solution by Interstigation (Simple Bash With PIE)

(which stands for Principle of Inclusion and Exclusion) https://youtu.be/2SGmSYZ5bqU

~Interstigation

See Also

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

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