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-problems 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 $1$-jump, $1$-jump, $2$-jump, with probability $\frac{1}{8}$, or she can go $2$-jump, $2$-jump, 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 must go $1$-jump, $2$-jump, 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

Observe that since Fiona 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 reach lily pad $2$, namely $1$-jump, $1$-jump, with probability $\frac{1}{4}$, or just a $2$-jump, with probability $\frac{1}{2}$. The total is thus $\frac{1}{4} + \frac{1}{2} = \frac{3}{4}$. Fiona must now make a $2$-jump to lily pad $4$, again with probability $\frac{1}{2}$, giving $\frac{3}{4} \cdot \frac{1}{2} = \frac{3}{8}$.

Similarly, Fiona must now make a $1$-jump to reach lily pad $5$, again with probability $\frac{1}{2}$, giving $\frac{3}{8} \cdot \frac{1}{2} = \frac{3}{16}$. Then she must make a $2$-jump to reach lily pad $7$, with probability $\frac{1}{2}$, yielding $\frac{3}{16} \cdot \frac{1}{2} = \frac{3}{32}$.

Finally, to reach lily pad $10$, Fiona has a few options - she can make $3$ consecutive $1$-jumps, with probability $\frac{1}{8}$, or $1$-jump, $2$-jump, with probability $\frac{1}{4}$, or $2$-jump, $1$-jump, again with probability $\frac{1}{4}$. The final answer is thus $\frac{3}{32} \cdot \left(\frac{1}{8} + \frac{1}{4} + \frac{1}{4}\right) = \frac{3}{32} \cdot \frac{5}{8} = \boxed{\textbf{(A) } \frac{15}{256}}$.

Solution 3 (recursion)

Let $p_n$ be the probability of landing on lily pad $n$. Observe that if there are no restrictions, we would have \[p_n = \frac{1}{2} \cdot p_{n-1} + \frac{1}{2} \cdot p_{n-2}\]

This is because, given that Fiona is at lily pad $n-2$, there is a $\frac{1}{2}$ probability that she will make a $2$-jump to reach lily pad $n$, and the same applies for a $1$-jump to reach 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]

Hence the answer is $\boxed{\textbf{(A) } \frac{15}{256}}$.

Note: If we let $p_n$ be the probability of surviving if the frog is on lily pad $n$, using $p_{10}$ = 1, we can solve backwards and obtain the following chart:

[asy]  unitsize(40); string[] vals = {"$15/256$", "$5/128$", "$5/64$", "X", "$5/32$", "$5/16$", "X", "$5/8$", "$3/4$", "$1/2$", "1", "X"}; for(int i =0; i<= 11; ++i) { draw((i,0)--(i+1,0)--(i+1,1)--(i,1)--cycle); label((string) (11-i), (i+0.5,0), S); label(vals[(11-i)], (i+0.5, 0.5)); } [/asy]

Solution 4 (simple casework bash)

This is equivalent to finding the probability for each of the valid ways of tiling a $1$-by-$11$ rectangular grid (with one end being lilypad $0$ and the other being lilypad $11$) with tiles of size $1 \cdot 1$ and $1 \cdot 2$ that satisfies the conditions mentioned in the problem. Obviously, for Fiona to skip pads $3$ and $6$, a $1 \cdot 2$ tile must be placed with one end at lilypad $2$ and the other at lilypad $4$, and another $1 \cdot 2$ must be placed with one end at lilypad $5$ and the other at lilypad $7$. Thus, since only a $1 \cdot 1$ tile can fit between the two aforementioned $1 \cdot 2$ tiles, we will place it there. Now, we can solve this problem with simple casework.

Case 1: Two $1 \cdot 1$ tiles fill the space between lilypads $0$ and $2$.

There are two ways to permute a placement of a $1 \cdot 1$ tile and a $1 \cdot 2$ tile between lilypads $7$ and $10$, so our probability for this sub-case is $\frac{2}{2^7} = \frac{1}{64}$. In the other subcase where the space between lilypads $7$ and $10$ is completely filled with $1 \cdot 1$ tiles, there is trivially only one tiling, thus the probability for this sub-case is $\frac{1}{2^8} = \frac{1}{256}.$ The total probability for this case is $\frac{1}{64} + \frac{1}{256} = \frac{5}{256}.$

Case 2: A single $1 \cdot 2$ tile fills the space between lilypads $0$ and $2$.

Note that the combined probability for this case will be double that of Case $1$, since a single $1 \cdot 2$ tile takes up one less tile than two $1 \cdot 1$ tiles. Thus, the probability for this case is $2 \cdot \frac{5}{256} = \frac{10}{256}.$

Summing our cases up, we obtain $\boxed{\textbf{(A) } \frac{15}{256}}$.

-fidgetboss_4000

See Also

2019 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

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