2025 AMC 12B Problem 20

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

\dfrac{1}{2}

and it disappears with probability

\dfrac{1}{2}.

• For

n = 1,

2,

3,

if it is at

n,

then it moves to

n+1

with probability

\dfrac{1}{4},

it moves to

n-1

with probability

\dfrac{1}{4},

and it disappears with probability

\dfrac{1}{2}.

What is the probability that the frog reaches

4?

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Solution

Let

f(n)

be the probability of reaching

4

from

n.

Then

f(0) = \tfrac{1}{2} f(1),

and for

n = 1, 2, 3,

f(n) = \tfrac{1}{4} f(n+1) + \tfrac{1}{4} f(n-1),

with

f(4) = 1.

Solving upward gives

f(2) = \tfrac{7}{2} f(1)

and

f(3) = 13 f(1);

then

4 \cdot 13 f(1) = 1 + \tfrac{7}{2} f(1)

yields

f(1) = \tfrac{2}{97},

f(0) = \tfrac{1}{97}.

Thus, the correct answer is E.