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

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#redirect [[2021 AMC 12B Problems/Problem 16]]
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?
<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>
==Solution 1==
There is a <math>\frac{3}{6}</math> chance that the first number we choose is even.
There is a <math>\frac{2}{5}</math> chance that the next number that is distinct from the first is even.
There is a <math>\frac{1}{4}</math> chance that the next number distinct from the first two is even.
<math>\frac{3}{6} * \frac{2}{5} * \frac{1}{4} = \frac{1}{20}</math>, so the answer is <math>\boxed{\textbf{(C) }\frac{1}{20}}.</math>
==Solution 2==
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.
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>
==Solution 3 (Quicksolve) ==
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.
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.
Therefore the probability is <math>\frac{3!(3!)}{6!} = \frac{1}{20} = \boxed{\textbf{(C)}}</math>.
==Solution 4==
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.
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}{2}}{3^{n-1}}\right) = \frac {1}{2^{n}} - \left(\frac {3(2^{n-1})-3}{2^{n} \cdot 3^{n-1}}\right) = \frac {1}{2^{n}} - \left(\frac {1}{2 \cdot 3^{n-2}} - \frac{1}{2^{n} \cdot 3^{n-2}} \right).</cmath>
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) = \left (\frac {1}{8} \right) - \left(\frac {1}{12} \right) + \left (\frac{1}{120} \right) = \boxed{\textbf{(C) }\frac{1}{20}.}</cmath>
==Solution 5==
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,
<cmath> E_1=\frac{1}{6}+\frac{1}{3}\cdot E_1+\frac{1}{2}\cdot 0, </cmath>
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>.
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
== Video Solution by OmegaLearn (Conditional probability) ==
{{AMC10 box|year=2021|ab=B|num-b=17|num-a=19}}

Revision as of 22:06, 13 February 2021

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