2022 AMC 8 Problems/Problem 25
Contents
Problem
A cricket randomly hops between leaves, on each turn hopping to one of the other leaves with equal probability. After hops what is the probability that the cricket has returned to the leaf where it started?
Solution 1 (Casework)
Let denote the leaf where the cricket starts and denote one of the other leaves. Note that:
- If the cricket is at then the probability that it hops to next is
- If the cricket is at then the probability that it hops to next is
- If the cricket is at then the probability that it hops to next is
We apply casework to the possible paths of the cricket:
-
The probability for this case is
-
The probability for this case is
Together, the probability that the cricket returns to after hops is
~MRENTHUSIASM
Solution 2 (Casework)
We can label the leaves as shown below:
Carefully counting cases, we see that there are ways for the cricket to return to leaf after four hops if its first hop was to leaf :
By symmetry, we know that there are ways if the cricket's first hop was to leaf , and there are ways if the cricket's first hop was to leaf . So, there are ways in total for the cricket to return to leaf after four hops.
Since there are possible ways altogether for the cricket to hop to any other leaf four times, the answer is .
~mahaler
Solution 3 (Recursion)
Denote to be the probability that the cricket would return back to the first point after hops. Then, we get the recursive formula because if the leaf is not on the target leaf, then there is a probability that it will make it back.
With this formula and the fact that (After one jump, the cricket can never be back to the target leaf.), we have so our answer is .
~wamofan
Video Solution
https://www.youtube.com/watch?v=85A6av3oqRo
~Mathematical Dexterity
Video Solution
https://youtu.be/Ij9pAy6tQSg?t=2588
~Interstigation
See Also
2022 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last Problem | |
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