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 $\tfrac12$ and it disappears with probability $\tfrac12.$ For $n = 1, 2,$ or $3,$ 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 $4?$

$\dfrac{1}{101}$

$\dfrac{1}{100}$

$\dfrac{1}{99}$

$\dfrac{1}{98}$

$\dfrac{1}{97}$

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.

2025 AMC 10B Problem 24

Solution:

Problem 24 in Other Years