2025 AMC 12B Problem 20

Question

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,

or

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?

\dfrac{1}{101}

\dfrac{1}{100}

\dfrac{1}{99}

\dfrac{1}{98}

\dfrac{1}{97}

Let

f(n)

be the probability of eventually reaching

4

starting from

n;

write a first-step equation at each state

Solve the linear system with

f(4) = 1

and value

0

whenever the frog disappears

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},

so

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

Thus, the correct answer is E.

Problem 20 in other years