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

(This problem was equal to the 12B #16)
(Tag: New redirect)
(Undo revision 148929 by Sugar rush (talk)I have strong reason to believe that #18 is not a duplicating question (from many resources I found online. Please contact MAA directly if want change)
(Tags: Removed redirect, Undo)
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#redirect [[2021 AMC 12B Problems/Problem 16]]
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==Problem==
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A fair <math>6</math>-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?
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<math>\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}</math>
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==Solution 1==
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There is a <math>\frac36</math> chance that the first number we choose is even.
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There is a <math>\frac{2}{5}</math> chance that the next number that is distinct from the first is even.
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There is a <math>\frac{1}{4}</math> chance that the next number distinct from the first two is even.
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<math>\frac{3}{6} \cdot \frac{2}{5} \cdot \frac{1}{4} = \frac{1}{20}</math>, so the answer is <math>\boxed{\textbf{(C) }\frac{1}{20}}.</math>
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~Tucker
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==Solution 2==
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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.
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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>
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~kingofpineapplz
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==Solution 3 (Quicksolve) ==
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Note that the problem is basically asking us to find the probability that in some permutation of <math>1,2,3,4,5,6</math> that we get the three even numbers in the first three spots.
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There are <math>6!</math> ways to order the <math>6</math> numbers and <math>3!(3!)</math> ways to order the evens in the first three spots and the odds in the next three spots.
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Therefore the probability is <math>\frac{3!(3!)}{6!} = \frac{1}{20} = \boxed{\textbf{(C)}}</math>.
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--abhinavg0627
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==Solution 4==
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Let <math>P_n</math> denote the probability that the first odd number appears on roll <math>n</math> and all our conditions are met. We now proceed with complementary counting.
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For <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> <cmath>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)</cmath> since there's a <math>\frac {1}{2^{n}}</math> chance that the first odd appears on roll <math>n</math> (disregarding the other conditions) and the other term is subtracting the probability that less than <math>3</math> of the evens show up before the first odd roll. Simplifying, we arrive at <cmath>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}}.</cmath>
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Summing for all <math>n</math>, we get our answer of <cmath>\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}.}</cmath>
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~ike.chen
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==Solution 5==
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Let <math> E_n </math> be that probability that the condition in the problem is satisfied given that we need <math> n </math> more distinct even numbers. Then,
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<cmath> E_1=\frac{1}{6}+\frac{1}{3}\cdot E_1+\frac{1}{2}\cdot 0, </cmath>
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since there is a <math> \frac{1}{3} </math> probability that we will roll an even number we already have rolled and will be in the same position again. Solving, we find that <math> E_1=\frac{1}{4} </math>.
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We can apply the same concept for <math> E_2 </math> and <math> E_3 </math>. We find that <cmath> E_2=\frac{1}{3}\cdot E_1+\frac{1}{6}\cdot E_2+\frac{1}{2}\cdot 0, </cmath> and so <math> E_2=\frac{1}{10} </math>. Also, <cmath> E_3=\frac{1}{2}\cdot E_2+\frac{1}{2}\cdot 0, </cmath> so <math> E_3=\frac{1}{20} </math>. Since the problem is asking for <math> E_3 </math>, our answer is <math> \boxed{\textbf{(C) }\frac{1}{20}} </math>. -BorealBear
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==Solution 6 (same as solution 1 but with a little more of explanation)==
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The probability of choosing an even number on the first turn is <math>\frac{1}{2}</math>, 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 <math>5</math> numbers, the probability of choosing another even number is <math>\frac{2}{5}</math>, and again, after you have chosen that number, it is out of our problem. Now, you just have <math>4</math> numbers left and the probability of choosing the last even number is <math>\frac{1}{4}</math>, so the answer is <math>\frac{1}{2} \times \frac{2}{5} \times \frac{1}{4}</math> <math>=</math> <math>\frac{1}{20}</math>.
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~math31415926535
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== Video Solution by OmegaLearn (Conditional probability) ==
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https://youtu.be/IX-Y38KPxqs
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== Video Solution by hurdler (complementary probability) ==
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https://www.youtube.com/watch?v=k2Jy4ni9tK8
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==Video Solution by TheBeautyofMath==
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https://youtu.be/FV9AnyERgJQ?t=480
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~IceMatrix
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==See Also==
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{{AMC10 box|year=2021|ab=B|num-b=17|num-a=19}}
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{{MAA Notice}}

Revision as of 17:41, 8 March 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 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 (Quicksolve)

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

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 6 (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

Video Solution by OmegaLearn (Conditional probability)

https://youtu.be/IX-Y38KPxqs

Video Solution by hurdler (complementary probability)

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

Video Solution by TheBeautyofMath

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

~IceMatrix

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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