🔥 New! 2025 AMC 10 B

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Problem 24 Easy

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 \frac { 1 } { 2 } and it disappears with probability \frac { 1 } { 2 }.

· For n = 1 , 2, or 3 , if it is at n, then it moves to n+1 with probability \frac { 1 } { 4 }, it moves to n-1 with probability \frac { 1 } { 4 }, and it disappears with probability \frac { 1 } { 2 }.

What is the probability that the frog reaches 4?

  • A.

    \frac { 1 } { 1 0 1 }

  • B.

    \frac { 1 } { 1 0 0 }

  • C.

    \frac { 1 } { 9 9 }

  • D.

    \frac { 1 } { 9 8 }

  • E.

    \frac { 1 } { 9 7 }

Answer:E
Solution

Let P(i) be the probability that the frog reaches 4 starting at i, where 0 \leq i \leq 3.

Then:

P(0) = \frac{1}{2} P(1)

P(1) = \frac{1}{4} P(2) + \frac{1}{4} P(0)

P(2) = \frac{1}{4} P(3) + \frac{1}{4} P(1)

P(3) = \frac{1}{4} + \frac{1}{4} P(2)

From the first equation: P(1) = 2P(0)

Plugging into the second equation: P(2) = 7P(0)

Solving the third equation: P(3) = 26P(0)

From the fourth equation: 4 \times 26 P(0) = 1 + 7 \cdot P(0), which gives P(0) = \frac{1}{97}.

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