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 $3$ , 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 $\frac{3}{8} \cdot \frac{1}{4} \cdot \frac{5}{8} = \frac{15}{256} = \boxed{\textbf{(A) }\frac{15}{256}}$ .

Solution 2

A different, but longer (and more tedious) answer-checking approach would involve considering each spot that the frog could get to separately.

Since we know it can only jump a maximum of 2 places at the time: as long as it wishes to avoid pads 3 and 6, it MUST land on numbers 2, 4, 5, and 7. Get it?

There's two ways to get there: either \tex(1,2) or just $2$ for the first move. There's also a 1/4 chance that it skips. Thus, we put 3/4 into our column and then move on-fir now.

It must then make it to space #4, another 1/2 probability. So you can make sure to multiple by 1/2 again, giving 3/8.

It's going to be getting a bit different now: for the next few choices, it also has to get to 5. Since our probability has also given and taken into account the factor that it's already landed on 4, multiple by #5 again to get to the spot-ment: 3/8 * 1/2 = 3/16.

Now there's another 1/2, giving us 3/16 * 1/2 = 3/32 for partie -8.

Here, we must look at possible options. There's only a 3/4 chance that it makes it directly. So, if we want a fuller picture then let's break it down. The possibilities are (8,9,10) (8,10) and (9,10) since going straight to #10 is not allowed. That leaves us with a partial count of 1/8 + 1/4 + 1/4 = 5/8 which we know works because the total sum can still add up to 1. Then, multiply, and that gives $3/32 * 5/8 = \boxed{\textbf{(A) }\frac{15}{256}}$ .

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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