Question

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 $1?$

$\dfrac 13$

$\dfrac 12$

$\dfrac 23$

$\dfrac 34$

$\dfrac 56$

Solution:

The stopping time depends only on the time variables, while the final position depends only on the distance variables, so the corresponding probabilities multiply.

For two independent numbers in $(0,1),$ the probability that their sum is less than $1$ is the area of a right triangle, namely $\frac12.$

For three independent numbers in $(0,1),$ the probability that their sum is less than $1$ is the volume of a tetrahedron with side intercepts $1,$ namely $\frac16.$

If $t_1+t_2>1,$ Amelia stops after two steps. This has probability $\frac12,$ and independently $x_1+x_2>1$ has probability $\frac12,$ contributing $\frac14.$

If $t_1+t_2<1,$ Amelia takes the third step. This has probability $\frac12,$ and independently $x_1+x_2+x_3>1$ has probability $1-\frac16=\frac56,$ contributing $\frac5{12}.$

The total probability is $\frac14+\frac5{12}=\frac23.$

Thus, the answer is C .

2022 AMC 10B Problem 23

Solution:

Problem 23 in Other Years