2022 AMC 10B Problems/Problem 23
- The following problem is from both the 2022 AMC 10B #23 and 2022 AMC 12B #22, so both problems redirect to this page.
Contents
- 1 Problem
- 2 Solution 1 (Geometric Probability)
- 3 Solution 2 (Generalization and Induction)
- 4 Solution 3 (Observations)
- 5 Remark (Calculus)
- 6 Remark (Rigorous Calculus)
- 7 Video Solution by OmegaLearn Using Geometric Probability
- 8 Video Solution
- 9 Video Solution by The Power of Logic (#22 and #23)
- 10 Video Solution by Interstigation
- 11 See Also
Problem
Ant Amelia starts on the number line at and crawls in the following manner. For Amelia chooses a time duration and an increment independently and uniformly at random from the interval During the th step of the process, Amelia moves units in the positive direction, using up minutes. If the total elapsed time has exceeded minute during the th step, she stops at the end of that step; otherwise, she continues with the next step, taking at most steps in all. What is the probability that Amelia’s position when she stops will be greater than ?
Solution 1 (Geometric Probability)
Let and be random variables that are independently and uniformly distributed in the interval Note that as shown below: Let and be random variables that are independently and uniformly distributed in the interval Note that as shown below: We have two cases:
- Amelia takes exactly steps.
- Amelia takes exactly steps.
We need and So, the probability is
We need and So, the probability is
Together, the answer is
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
~MRENTHUSIASM
Solution 2 (Generalization and Induction)
We can in fact find the probability that any number of randomly distributed numbers on the interval sum to more than using geometric probability, as shown in the video below.
If we graph the points that satisfy , , we get the triangle with points , , and . If we graph the points that satisfy , , we get the tetrahedron with points , , , and .
Of course, the probability of either of these cases happening is simply the area/volume of the points we graphed divided by the total area of the graph, which is always (this would be much simpler than my calculus proof above).
Thus, we can now solve for the probability that the sum is less than one for numbers using induction.
The probability that the sum is less than one is .
For just number, the probability is .
Suppose that the probability for numbers is . We will prove that the probability for numbers is . To prove this, we consider that the area of an -dimensional tetrahedron is simply the area/volume of the base times the height divided by .
Of course, the area of the base is , and the height is , and thus, we obtain as our volume (this may be hard to visualize for higher dimensions). The induction step is complete.
The probability of the sum being less than is , and the probability of the sum being more than is . This trivializes the problem. The answer is
~mathboy100
Solution 3 (Observations)
There are two cases: Amelia takes two steps or three steps.
The former case has a probability of , as stated in Solution 1, and thus the latter also has a probability of .
The probability that Amelia passes after two steps is also , as it is symmetric to the probability above.
Thus, if the probability that Amelia passes after three steps is , our total probability is . We know that , and it is relatively obvious that (because the probability that is ). This means that our total probability is between and , non-inclusive, so the only answer choice that fits is .
~mathboy100
Remark (Calculus)
It is not immediately clear why three random numbers between and have a probability of of summing to more than . Here is a proof:
Let us start by finding the probability that two random numbers between and have a sum of more than , where .
Suppose that our two numbers are and . Then, the probability that (which means that ) is , and the probability that is .
If , the probability that is . This is because the probability that is equal to the probability that , which is , so our total probability is .
Let us now find the average of the probability that when . Since is a random number between and , its average is . Thus, our average is .
Hence, our total probability is equal to Now, let us find the probability that three numbers uniformly distributed between and sum to more than .
Let our three numbers be , , and . Then, the probability that is equal to the probability that is greater than , which is equal to .
To find the total probability, we must average over all values of . This average is simply equal to the area under the curve from to , all divided by . We can compute this value using integrals: For those who don't know calculus, is the area under the curve from to .
~mathboy100
Remark (Rigorous Calculus)
In the language of probability, we have three random variables , , , each independent and uniformly distributed over the interval . We are interested in the probability that , or .
It follows from the definition of the joint probability density function that (we assume that ): \begin{align} P(X+Y+Z>1)=\iiint_{x+y+z>1} f(x,y,z) \,dx\,dy\,dz. \end{align} Since are independent, we have , where represent the probability densities of respectively. Recall that are uniformly distributed over the interval . Hence, .
It remains to evaluate the following triple integral \begin{align} \iiint_{x+y+z>1} \,dx\,dy\,dz=\frac{5}{6}. \end{align} Therefore, .
~tsun26
Video Solution by OmegaLearn Using Geometric Probability
~ pi_is_3.14
Video Solution
~ThePuzzlr
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution by The Power of Logic (#22 and #23)
Video Solution by Interstigation
~Interstigation
See Also
2022 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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 |
2022 AMC 12B (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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.