Difference between revisions of "2021 AMC 10B Problems/Problem 18"
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− | Let <math>P_n</math> denote the | + | 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> | 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) = \frac {1}{ | + | 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> |
~ike.chen | ~ike.chen |
Revision as of 18:17, 12 February 2021
Contents
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 1
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
Solution 2
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 3
Note that the problem is basically asking us to find the probability that 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
Solution 4
Let denote the probability that the first odd number appears on roll and all our conditions are met. We now proceed with complementary counting.
For , it's impossible to have all evens appear before an odd. Note that for
Summing for all , we get our answer of
~ike.chen
Solution 5
Let be that probability that the condition in the problem is satisfied given that we need more distinct even numbers. Then, since there is a probability that we will roll an even number we already have rolled and will be in the same position again. Solving, we find that .
We can apply the same concept for and . We find that and so . Also, so . Since the problem is asking for , our answer is . -BorealBear
Video Solution by OmegaLearn (Conditional probability)
2021 AMC 10B (Problems • Answer Key • Resources) | ||
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