2019 AMC 12B Problems/Problem 16

Problem

There are lily pads in a row numbered $0$ to $11$ , in that order. There are predators on lily pads $3$ and $6$ , and a morsel of food on lily pad $10$ . Fiona the frog starts on pad $0$ , and from any given lily pad, has a $\frac{1}{2}$ chance to hop to the next pad, and an equal chance to jump $2$ pads. What is the probability that Fiona reaches pad $10$ without landing on either pad $3$ or pad $6$ ?

$\textbf{(A) } \frac{15}{256} \qquad \textbf{(B) } \frac{1}{16} \qquad \textbf{(C) } \frac{15}{128}\qquad \textbf{(D) } \frac{1}{8} \qquad \textbf{(E) } \frac14$

Solution 1

Firstly, notice that if Fiona jumps over the predator on pad $3$ , she must on pad $4$ . Similarly, she must land on $7$ if she makes it past $6$ . Thus, we can split the problem into $3$ smaller sub-problems, separately finding the probability Fiona skips $3$ , the probability she skips $6$ (starting at $4$ ) and the probability she doesn't skip $10$ (starting at $7$ ). Notice that by symmetry, the last of these three sub-prolems is the complement of the first sub-problem, so the probability will be $1 - \text{the probability obtained in the first sub-problem}$ .

In the analysis below, we call the larger jump a $2$ -jump, and the smaller a $1$ -jump.

For the first sub-problem, consider Fiona's options. She can either go $1, 1, 2$ , with probability $\frac{1}{8}$ , or she can go $2, 2$ , with probability $\frac{1}{4}$ . These are the only two options, so they together make the answer $\frac{1}{8}+\frac{1}{4}=\frac{3}{8}$ . We now also know the answer to the last sub-problem is $1-\frac{3}{8}=\frac{5}{8}$ .

For the second sub-problem, Fiona $\textit{must}$ go $1, 2$ , with probability $\frac{1}{4}$ , since any other option would result in her death to a predator.

Thus, since the three sub-problems are independent, the final answer is $\frac{3}{8} \cdot \frac{1}{4} \cdot \frac{5}{8} = \boxed{\textbf{(A) }\frac{15}{256}}$ .

Solution 2

Consider – independently – every lily pad that Fiona could reach.

Given that she can only jump at most $2$ places per move, and still wishes to avoid pads $3$ and $6$ , she must also land on numbers $2$ , $4$ , $5$ , and $7$ .

There are two ways to achieve this - one would be $(1,2)$ on her first move, and the other is just $(2)$ . The total sum is then $\frac{1}{2} \times \frac{1}{2} + \frac{1}{2} = \frac{3}{4}$ , which put into our first column and move on. The frog must subsequently go to space $4$ , again with probability $\frac{1}{2}$ . Thus, be sure to multiply by $\frac{1}{2}$ again, yielding the result of $\frac{3}{8}$ .

Similarly, multiply your product by $\frac{1}{2}$ once more, to arrive at spot $5$ : $\frac{3}{8} \times {1}{2} = \frac{3}{16}$ . For number $7$ , take another $\frac{1}{2}$ , giving us $\frac{3}{16} \times \frac{1}{2} = \frac{3}{32}$ .

Next, we must look at a number of options. For a fuller picture, it would be best to break down the choices. The only possibilities here are $(8,9,10)$ , $(8,10)$ , and $(9,10)$ , as the path straight to point $10$ is not available. That leaves us with a partial count of $\frac{1}{8} + \frac{1}{4} + \frac{1}{4} = frac{5}{8}$ . Multiply, to find that $\frac{3}{32} \times \frac{5}{8} = \boxed{\textbf{(A)} \frac{15}{256}}$ . $\square$

--anna0kear.

Solution 3 (Recursion)

Let $p_n$ be the probability of landing on lily pad $n$ . We immediately notice that, if there are no restrictions: $p_n = \frac{1}{2} \cdot p_{n-1} + \frac{1}{2} \cdot p_{n-2}$

This is because, given that we are at lily pad $n-2$ , there is a 50% chance that we will go to lily pad $n$ , and the same applies for lily pad $n-1$ . We will now compute the values of $p_n$ recursively, but we will skip over $3$ and $6$ . That is, we will not consider any jumps from lily pads 3 or 6 when considering the probabilities. We obtain the following chart, where an X represents an unused/uncomputed value:

$[asy] unitsize(40); string[] vals = {"1", "$1/2$", "$3/4$", "X", "$3/8$", "$3/16$", "X", "$3/32$", "$3/64$", "$9/128$", "$15/256$", "X"}; for(int i =0; i<= 11; ++i) { draw((i,0)--(i+1,0)--(i+1,1)--(i,1)--cycle); label((string) i, (i+0.5,0), S); label(vals[i], (i+0.5, 0.5)); } [/asy]$

As we can see, the answer is $\boxed{\textbf{(A) } \frac{15}{256}}$

(Solution by vedadehhc)

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

Solution 3 (Recursion)

See Also