|
|
| (30 intermediate revisions by 4 users not shown) |
| Line 1: |
Line 1: |
| − | ==Problem==
| + | #redirect [[2022 AMC 10B Problems/Problem 23]] |
| − | Ant Amelia starts on the number line at <math>0</math> and crawls in the following manner. For <math>n=1,2,3,</math> Amelia chooses a time duration <math>t_n</math> and an increment <math>x_n</math> independently and uniformly at random from the interval <math>(0,1).</math> During the <math>n</math>th step of the process, Amelia moves <math>x_n</math> units in the positive direction, using up <math>t_n</math> minutes. If the total elapsed time has exceeded <math>1</math> minute during the <math>n</math>th step, she stops at the end of that step; otherwise, she continues with the next step, taking at most <math>3</math> steps in all. What is the probability that Amelia’s position when she stops will be greater than <math>1</math>?
| |
| − | | |
| − | <math>\textbf{(A) }\frac{1}{3} \qquad \textbf{(B) }\frac{1}{2} \qquad \textbf{(C) }\frac{2}{3} \qquad \textbf{(D) }\frac{3}{4} \qquad \textbf{(E) }\frac{5}{6}</math>
| |
| − | | |
| − | ==Solution==
| |
| − | Obviously the chance of Amelia stopping after only <math>1</math> step is <math>0</math>.
| |
| − | | |
| − | When Amelia takes <math>2</math> steps, then the sum of the time taken during the steps is greater than <math>1</math> minute. Let the time taken be <math>x</math> and <math>y</math> respectively, then we need <math>x+y>1</math> for <math>0<x<1, 0<y<1</math>, which has a chance of <math>\frac{1}{2}</math>. Let the lengths of steps be <math>a</math> and <math>b</math> respectively, then we need <math>a+b>1</math> for <math>0<a<1, 0<b<1</math>, which has a chance of <math>\frac{1}{2}</math>. Thus the total chance for this case is <math>\frac{1}{4}</math>.
| |
| − | | |
| − | When Amelia takes <math>3</math> steps, then by complementary counting the chance of taking <math>3</math> steps is <math>1-\frac{1}{2}=\frac{1}{2}</math>. Let the lengths of steps be <math>a</math>, <math>b</math> and <math>c</math> respectively, then we need <math>a+b+c>1</math> for <math>0<a<1, 0<b<1, 0<c<1</math>, which has a chance of <math>\frac{5}{6}</math>. Thus the total chance for this case is <math>\frac{5}{12}</math>.
| |
| − | | |
| − | Thus the answer is <math>\frac{1}{4}+\frac{5}{12}=\frac{2}{3}</math>.
| |
| − | | |
| − | ==Solution 2 (Clever)==
| |
| − | There are two cases: Amelia takes two steps or three steps.
| |
| − | | |
| − | The former case has a probability of <math>\frac{1}{2}</math>, as stated above, and thus the latter also has a probability of <math>\frac{1}{2}</math>.
| |
| − | | |
| − | The probability that Amelia passes <math>1</math> after two steps is also <math>\frac{1}{2}</math>, as it is symmetric to the probability above.
| |
| − | | |
| − | Thus, if the probability that Amelia passes <math>1</math> after three steps is <math>x</math>, our total probability is <math>\frac{1}{2} \cdot \frac{1}{2} + \frac{1}{2} \cdot x</math>. We know that <math>0 < x < 1</math>, and it is relatively obvious that <math>x > 1</math> (because the probability that <math>x > \frac{3}{2}</math> is \frac{1}{2}). This means that our total probability is between <math>\frac{1}{2}</math> and <math>\frac{3}{4}</math>, non-inclusive, so the only answer choice that fits is <math>\boxed{\textbf{(C) }\frac{2}{3}}</math>
| |
| − | | |
| − | ~mathboy100
| |
