Difference between revisions of "2022 AMC 10B Problems/Problem 23"
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+ | ==Problem== | ||
+ | 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== | ==Solution== | ||
Revision as of 00:54, 18 November 2022
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
We use the following lemma to solve this problem.
Let be independent random variables that are uniformly distributed on
. Then for
,
For ,
Now, we solve this problem.
We denote by the last step Amelia moves. Thus,
.
We have
where the second equation follows from the property that and
are independent sequences, the third equality follows from the lemma above.
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Solution 2 (Elimination)
There is a probability that Amelia is past
after
turn, so Amelia can only pass
after
turns or
turns. The probability of finishing in
turns is
(due to the fact that the probability of getting
is the same as the probability of getting
), and thus the probability of finishing in
turns is also
.
It is also clear that the probability of Amelia being past in
turns is equal to
.
Therefore, if is the probability that Amelia finishes if she takes three turns, our final probability is
.
must be a number between
and
(non-inclusive), and it is clearly greater than
, because the probability of getting more than
in
turns is
. Thus, the answer must be between
and
, non-inclusive, so the only answer that makes sense is
.
~mathboy100
Video Solution
~ThePuzzlr
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)