2025 AMC 10B Problem 24

A frog hops along the number line according to the following rules. It starts at

0.

If it is at

0,

then it moves to

1

with probability

\tfrac12

and it disappears with probability

\tfrac12.

For

n = 1,

2,

3,

if it is at

n,

then it moves to

n + 1

with probability

\tfrac14,

it moves to

n - 1

with probability

\tfrac14,

and it disappears with probability

\tfrac12.

What is the probability that the frog reaches

4?

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Solution

Let

p_n

be the probability of reaching

4

from position

n,

with

p_4 = 1.

The rules give

p_0 = \tfrac12 p_1,

p_1 = \tfrac14 p_0 + \tfrac14 p_2,

p_2 = \tfrac14 p_1 + \tfrac14 p_3,

and

p_3 = \tfrac14 p_2 + \tfrac14.

Work upward:

p_1 = \tfrac27 p_2

and

p_2 = \tfrac{7}{26} p_3.

These unwind to

p_3 = \tfrac{26}{97},

p_2 = \tfrac{7}{97},

p_1 = \tfrac{2}{97},

and finally

p_0 = \tfrac{1}{97}.

Therefore, the answer is E.