Difference between revisions of "2021 AMC 10B Problems/Problem 18"
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− | ==Solution== | + | ==Solution 1== |
− | There is a <math>\ | + | There is a <math>\frac36</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{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. | + | There is a <math>\frac{1}{4}</math> chance that the next number distinct from the first two is even. |
− | <math>\frac{3}6 | + | <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> |
~Tucker | ~Tucker | ||
+ | ==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. | 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. | ||
Line 25: | Line 26: | ||
~kingofpineapplz | ~kingofpineapplz | ||
− | + | ==Solution 3 == | |
− | ==Solution | ||
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. | 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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--abhinavg0627 | --abhinavg0627 | ||
+ | |||
+ | ==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}{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> | ||
+ | |||
+ | 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> | ||
+ | |||
+ | ~ike.chen | ||
+ | |||
+ | ==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 | ||
+ | |||
+ | ==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 <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>. | ||
+ | |||
+ | ~math31415926535 | ||
+ | |||
+ | ==Solution 6 (Infinite Geometric Sequence Method)== | ||
+ | Let's say that our even integers are found in the first <math>n</math> numbers where n must be greater than or equal to <math>3</math>. Then, we can form an argument based on this. There are <math>3^n</math> total ways to assign even numbers being <math>(</math>2<math>, </math>4<math>, </math>6<math>)</math> to each space. Furthermore, we must subtract the cases where we have only a single even integer present in the total <math>n</math> spaces and the case where there are only <math>2</math> distinct even integers present. There is <math>1</math> way we can have a single even integer in the entire <math>n</math> spaces, therefore giving us just <math>1</math> option for each of the three integers, so we have <math>3</math> total cases for this. Moreover, the amount of cases with just <math>2</math> distinct even integers is <math>2^n</math> but subtract the cases where all of the n spaces is either a single integer giving us <math>2^n-2</math>, but we multiply by <math>3C2</math> because of the ways to choose <math>2</math> distinct even integers that are used in the sequence of <math>n</math>. Finally, we have <math>\sum_{n=3}^{\infty} \frac{(3^n-3-3(2^n-2))}{6^n}</math> note: we must divide all of this by <math>6^n</math> for probability. Additionally, over the entire summation, we multiply by <math>1/2</math> because of the <math>1/2</math> probability of selecting an odd at the end of all the evens. Therefore, if you compute this using infinite geometric sequences, you get <math>1/20</math>. <math>\boxed{\textbf{(C) }\frac{1}{20}.}</math> | ||
+ | |||
+ | ~Jske25 | ||
+ | |||
+ | https://www.youtube.com/watch?v=T-kdDFeHzns&t=17s | ||
+ | |||
+ | ==Solution 2a (same as 2 but with more explanation)== | ||
+ | The question basically asks for the probability that <math>2</math>, <math>4</math>, and <math>6</math> are all rolled before <math>1</math>, <math>3</math> or <math>5</math>. This probability is equal to the probability that any 3 number combination is rolled before the remaining numbers are rolled (for example, rolling <math>1, 2</math> and <math>3</math> before <math>4</math>, <math>5</math> and <math>6</math>). There are <math>\binom{6}{3}</math> combinations possible and summing them all up must result in 1 so the probability of the specific combination <math>2</math> <math>4</math> <math>6</math> being chosen is<math>\boxed{\textbf{(C) }\frac{1}{20}.}</math> | ||
+ | |||
+ | ~AwesomeK | ||
+ | |||
+ | == 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== | ||
+ | {{AMC10 box|year=2021|ab=B|num-b=17|num-a=19}} | ||
+ | {{MAA Notice}} |
Latest revision as of 07:22, 17 July 2021
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2
- 4 Solution 3
- 5 Solution 4
- 6 Solution 5
- 7 Solution 1a (same as solution 1 but with a little more of explanation)
- 8 Solution 6 (Infinite Geometric Sequence Method)
- 9 Solution 2a (same as 2 but with more explanation)
- 10 Video Solution by hurdler (complementary probability)
- 11 Video Solution by TheBeautyofMath
- 12 Video Solution by Interstigation (Simple Bash With PIE)
- 13 See Also
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 since there's a chance that the first odd appears on roll (disregarding the other conditions) and the other term is subtracting the probability that less than of the evens show up before the first odd roll. Simplifying, we arrive at
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
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 , 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 numbers, the probability of choosing another even number is , and again, after you have chosen that number, it is out of our problem. Now, you just have numbers left and the probability of choosing the last even number is , so the answer is .
~math31415926535
Solution 6 (Infinite Geometric Sequence Method)
Let's say that our even integers are found in the first numbers where n must be greater than or equal to . Then, we can form an argument based on this. There are total ways to assign even numbers being 246 to each space. Furthermore, we must subtract the cases where we have only a single even integer present in the total spaces and the case where there are only distinct even integers present. There is way we can have a single even integer in the entire spaces, therefore giving us just option for each of the three integers, so we have total cases for this. Moreover, the amount of cases with just distinct even integers is but subtract the cases where all of the n spaces is either a single integer giving us , but we multiply by because of the ways to choose distinct even integers that are used in the sequence of . Finally, we have note: we must divide all of this by for probability. Additionally, over the entire summation, we multiply by because of the probability of selecting an odd at the end of all the evens. Therefore, if you compute this using infinite geometric sequences, you get .
~Jske25
https://www.youtube.com/watch?v=T-kdDFeHzns&t=17s
Solution 2a (same as 2 but with more explanation)
The question basically asks for the probability that , , and are all rolled before , or . This probability is equal to the probability that any 3 number combination is rolled before the remaining numbers are rolled (for example, rolling and before , and ). There are combinations possible and summing them all up must result in 1 so the probability of the specific combination being chosen is
~AwesomeK
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.