2022 AMC 12B Problems/Problem 22
Contents
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)
Solution 2 (Calculus)
We use the following lemma to solve this problem.
Let be independent random variables that are uniformly distributed on
. Then for
,
For
,
Check the Remark section for proof.
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 3 (Casework)
There are two cases: Amelia takes two steps or three steps.
The former case has a probability of , as stated above, 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
Solution 4 (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
Remark
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
Video Solution by OmegaLearn Using Geometric Probability
~ pi_is_3.14
Video Solution
~ThePuzzlr
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
See Also
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 |
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 |
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