2019 AMC 12B Problems/Problem 16

Problem

Lily pads numbered from $0$ to $11$ lie in a row on a pond. Fiona the frog sits on pad $0$ , a morsel of food sits on pad $10$ , and predators sit on pads $3$ and $6$ . At each unit of time the frog jumps either to the next higher numbered pad or to the pad after that, each with probability $\frac{1}{2}$ , independently from previous jumps. What is the probability that Fiona skips over pads $3$ and $6$ and lands on pad $10$ ?

$\textbf{(A) }\frac{15}{256}\qquad\textbf{(B) }\frac{1}{16}\qquad\textbf{(C) }\frac{15}{128}\qquad\textbf{(D) }\frac{1}{8}\qquad\textbf{(E) }\frac{1}{4}$

Solution 1

First, notice that Fiona, if she jumps over the predator on pad $3$ , \textbf{must} land on pad $4$ . Similarly, she must land on $7$ if she makes it past $6$ . Thus, we can split it into $3$ smaller problems counting the probability Fiona skips $3$ , Fiona skips $6$ (starting at $4$ ) and \textit{doesn't} skip $10$ (starting at $7$ ). Incidentally, the last one is equivalent to the first one minus $1$ .

Let's call the larger jump a $2$ -jump, and the smaller a $1$ -jump.

For the first mini-problem, let's see our options. Fiona can either go $1, 1, 2$ (probability of \frac{1}{8}), or she can go $2, 2$ (probability of \frac{1}{4}). These are the only two options, so they together make the answer $\frac{3}{8}$ . We now also know the answer to the last mini-problem ( $\frac{5}{8}$ ).

For the second mini-problem, Fiona \textit{must} go $1, 2$ (probability of \frac{1}{4}). Any other option results in her death to a predator.

Thus, the final answer is $\frac{3}{8} \cdot \frac{1}{4} \cdot \frac{5}{8} = \frac{15}{256} = \boxed{A}$

2019 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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2019 AMC 12B Problems/Problem 16

Contents

Problem

Solution 1

Solution 2

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