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Difference between revisions of "2022 AMC 12B Problems/Problem 22"

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Thus the answer is <math>\frac{1}{4}+\frac{5}{12}=\frac{2}{3}</math>.
 
Thus the answer is <math>\frac{1}{4}+\frac{5}{12}=\frac{2}{3}</math>.
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==Solution 2 (Clever)==
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There are two cases: Amelia takes two steps or three steps.
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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>.
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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.
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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>
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~mathboy100

Revision as of 19:05, 18 November 2022

Problem

Ant Amelia starts on the number line at $0$ and crawls in the following manner. For $n=1,2,3,$ Amelia chooses a time duration $t_n$ and an increment $x_n$ independently and uniformly at random from the interval $(0,1).$ During the $n$th step of the process, Amelia moves $x_n$ units in the positive direction, using up $t_n$ minutes. If the total elapsed time has exceeded $1$ minute during the $n$th step, she stops at the end of that step; otherwise, she continues with the next step, taking at most $3$ steps in all. What is the probability that Amelia’s position when she stops will be greater than $1$?

$\textbf{(A) }\frac{1}{3} \qquad \textbf{(B) }\frac{1}{2} \qquad \textbf{(C) }\frac{2}{3} \qquad \textbf{(D) }\frac{3}{4} \qquad \textbf{(E) }\frac{5}{6}$

Solution

Obviously the chance of Amelia stopping after only $1$ step is $0$.

When Amelia takes $2$ steps, then the sum of the time taken during the steps is greater than $1$ minute. Let the time taken be $x$ and $y$ respectively, then we need $x+y>1$ for $0<x<1, 0<y<1$, which has a chance of $\frac{1}{2}$. Let the lengths of steps be $a$ and $b$ respectively, then we need $a+b>1$ for $0<a<1, 0<b<1$, which has a chance of $\frac{1}{2}$. Thus the total chance for this case is $\frac{1}{4}$.

When Amelia takes $3$ steps, then by complementary counting the chance of taking $3$ steps is $1-\frac{1}{2}=\frac{1}{2}$. Let the lengths of steps be $a$, $b$ and $c$ respectively, then we need $a+b+c>1$ for $0<a<1, 0<b<1, 0<c<1$, which has a chance of $\frac{5}{6}$. Thus the total chance for this case is $\frac{5}{12}$.

Thus the answer is $\frac{1}{4}+\frac{5}{12}=\frac{2}{3}$.

Solution 2 (Clever)

There are two cases: Amelia takes two steps or three steps.

The former case has a probability of $\frac{1}{2}$, as stated above, and thus the latter also has a probability of $\frac{1}{2}$.

The probability that Amelia passes $1$ after two steps is also $\frac{1}{2}$, as it is symmetric to the probability above.

Thus, if the probability that Amelia passes $1$ after three steps is $x$, our total probability is $\frac{1}{2} \cdot \frac{1}{2} + \frac{1}{2} \cdot x$. We know that $0 < x < 1$, and it is relatively obvious that $x > 1$ (because the probability that $x > \frac{3}{2}$ is \frac{1}{2}). This means that our total probability is between $\frac{1}{2}$ and $\frac{3}{4}$, non-inclusive, so the only answer choice that fits is $\boxed{\textbf{(C) }\frac{2}{3}}$

~mathboy100

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