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 $4?$

$\dfrac{1}{101}$

$\dfrac{1}{100}$

$\dfrac{1}{99}$

$\dfrac{1}{98}$

$\dfrac{1}{97}$

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 $f(1) = \tfrac{2}{97},$ so $f(0) = \tfrac{1}{97}.$

Thus, the correct answer is E.

2025 AMC 12B Problem 20

Solution:

Problem 20 in Other Years