2019 AMC 10A Problems/Problem 22
- The following problem is from both the 2019 AMC 10A #22 and 2019 AMC 12A #20, so both problems redirect to this page.
Contents
Problem
Real numbers between 0 and 1, inclusive, are chosen in the following manner. A fair coin is flipped. If it lands heads, then it is flipped again and the chosen number is 0 if the second flip is heads, and 1 if the second flip is tails. On the other hand, if the first coin flip is tails, then the number is chosen uniformly at random from the closed interval . Two random numbers and are chosen independently in this manner. What is the probability that ?
Solution 1
There are several cases depending on what the first coin flip is when determining and what the first coin flip is when determining .
The four cases are:
Case 1: is either or , and is either or .
Case 2: is either or , and is chosen from the interval .
Case 3: is is chosen from the interval , and is either or .
Case 4: is is chosen from the interval , and is also chosen from the interval .
Each case has a chance of occurring (as it requires two coin flips).
For Case 1, we need and to be different. Therefore, the probability for success in Case 1 is .
For Case 2, if is 0, we need to be in the interval . If is 1, we need to be in the interval . Regardless of what is, the probability for success for Case 2 is .
By symmetry, Case 3 has the same success rate as Case 2.
For Case 4, we must use geometric probability because there are an infinite number of pairs that can be selected, whether they satisfy the inequality or not. Graphing gives us the following picture where the shaded area is the set of all the points that fulfill the inequality:
The shaded area is , which means the probability for success for case 4 is (since the total area of the bounding square, containing all possible pairs, is ).
Adding up the success rates from each case, we get:
.
Solution 2
Taking into account that there are two options for the result of the first coin flip, there are four possible combinations with equal possibility of initial coins flips.
(1) x: heads, y: heads
(2) x: heads, y: tails
(3) x: tails, y: heads
(4) x: tails, y: tails
(1) Because x and y both must be integers, so it's simple to see that the probability that is .
(2) Since x has to be an integer it is also easy to see that the probability is again .
(3) This is the same as case 2 so the probability is .
(4) This one is a little bit tricky. You could use geometric probability shown above, but if you don't understand it or happened to not think of it, there is another solution that involves using the multiple choice answers.
First, the probability we have so far is .This is greater than so is the not the answer. It is either or . Let's draw out the probability with two parallel lines on a paper with represent a number line from . The concept is to compare one point and find the fraction of the other line that contains points that are at least away from it. If the point we choose is on the endpoints, then the fraction of the other line that works is exactly . But as we move this point closer to the middle, the deadzone (area where the other point cannot be) grows, diminishing the probability. Finally, when it is directly in the middle, there are no points that pass the requirements except at and .
So, looking at the choices again, we have , , , and .
is exactly more than the probability we had before. Notice that this is impossible because we proved that the average probability of the fourth case is lower than , so the answer is .
-jackshi2006
Video Solution 1
Education, the Study of Everything
Video Solution 2
https://www.youtube.com/watch?v=fe5bdwszsos
Video Solution 3
https://youtu.be/ZhAZ1oPe5Ds?t=263
~ pi_is_3.14
Video Solution by Richard Rusczyk
https://www.youtube.com/watch?v=Fe5bDwSzsOs
See Also
2019 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
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 |
2019 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 19 |
Followed by Problem 21 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.