Difference between revisions of "2022 AMC 12B Problems/Problem 22"
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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. | 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> | + | 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 <math>\frac{1}{2}</math>). 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 | ~mathboy100 |
Revision as of 20:06, 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
Obviously the chance of Amelia stopping after only step is
.
When Amelia takes steps, then the sum of the time taken during the steps is greater than
minute. Let the time taken be
and
respectively, then we need
for
, which has a chance of
. Let the lengths of steps be
and
respectively, then we need
for
, which has a chance of
. Thus the total chance for this case is
.
When Amelia takes steps, then by complementary counting the chance of taking
steps is
. Let the lengths of steps be
,
and
respectively, then we need
for
, which has a chance of
. Thus the total chance for this case is
.
Thus the answer is .
Solution 2 (Clever)
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