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.
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
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:
.
See Also
VIDEO Solution: https://www.youtube.com/watch?v=fe5bdwszsos
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.