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Difference between revisions of "2019 AMC 12B Problems/Problem 16"

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(Fixed the problem statement, and improved the formatting and clarity of Solution 1)
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==Problem==
 
==Problem==
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Lily pads numbered from <math>0</math> to <math>11</math> lie in a row on a pond. Fiona the frog sits on pad <math>0</math>, a morsel of food sits on pad <math>10</math>, and predators sit on pads <math>3</math> and <math>6</math>. At each unit of time the frog jumps either to the next higher numbered pad or to the pad after that, each with probability <math>\frac{1}{2}</math>, independently from previous jumps. What is the probability that Fiona skips over pads <math>3</math> and <math>6</math> and lands on pad <math>10</math>?
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There are lily pads in a row numbered <math>0</math> to <math>11</math>, in that order. There are predators on lily pads <math>3</math> and <math>6</math>, and a morsel of food on lily pad <math>10</math>. Fiona the frog starts on pad <math>0</math>, and from any given lily pad, has a <math>\frac{1}{2}</math> chance to hop to the next pad, and an equal chance to jump <math>2</math> pads. What is the probability that Fiona reaches pad <math>10</math> without landing on either pad <math>3</math> or pad <math>6</math>?
  
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<math>\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}</math>
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<math>\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</math>
  
 
==Solution 1==
 
==Solution 1==
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First, notice that Fiona, if she jumps over the predator on pad <math>3</math>, must land on pad <math>4</math>. Similarly, she must land on <math>7</math> if she makes it past <math>6</math>. Thus, we can split it into <math>3</math> smaller problems counting the probability Fiona skips <math>3</math>, Fiona skips <math>6</math> (starting at <math>4</math>) and <math>\textit{doesn't}</math> skip <math>10</math> (starting at <math>7</math>). Incidentally, the last one is the complement of the first sub-problem (i.e. equivalent to <math>1</math> - probability of first sub-problem).
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Firstly, notice that if Fiona jumps over the predator on pad <math>3</math>, she must on pad <math>4</math>. Similarly, she must land on <math>7</math> if she makes it past <math>6</math>. Thus, we can split the problem into <math>3</math> smaller sub-problems, separately finding the probability Fiona skips <math>3</math>, the probability she skips <math>6</math> (starting at <math>4</math>) and the probability she ''doesn't'' skip <math>10</math> (starting at <math>7</math>). Notice that by symmetry, the last of these three sub-prolems is the complement of the first sub-problem, so the probability will be <math>1 - \text{the probability obtained in the first sub-problem}</math>.
  
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Let's call the larger jump a <math>2</math>-jump, and the smaller a <math>1</math>-jump.  
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In the analysis below, we call the larger jump a <math>2</math>-jump, and the smaller a <math>1</math>-jump.  
  
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For the first sub-problem, let's see our options. Fiona can either go <math>1, 1, 2</math> <math>\left(\text{probability of } \frac{1}{8} \right)</math>, or she can go <math>2, 2</math> <math>\left(\text{probability of } \frac{1}{4} \right)</math>. These are the only two options, so they together make the answer <math>\frac{3}{8}</math>. We now also know the answer to the last sub-problem is <math>\frac{5}{8}</math>.
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For the first sub-problem, consider Fiona's options. She can either go <math>1, 1, 2</math>, with probability <math>\frac{1}{8}</math>, or she can go <math>2, 2</math>, with probability <math>\frac{1}{4}</math>. These are the only two options, so they together make the answer <math>\frac{1}{8}+\frac{1}{4}=\frac{3}{8}</math>. We now also know the answer to the last sub-problem is <math>1-\frac{3}{8}=\frac{5}{8}</math>.
  
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For the second sub-problem, Fiona <math>\textit{must}</math> go <math>1, 2</math> <math>\left (\text{probability of } \frac{1}{4} \right)</math>. Any other option results in her death to a predator.
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For the second sub-problem, Fiona <math>\textit{must}</math> go <math>1, 2</math>, with probability <math>\frac{1}{4}</math>, since any other option would result in her death to a predator.
  
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Thus, the final answer is <math>\frac{3}{8} \cdot \frac{1}{4} \cdot \frac{5}{8} = \frac{15}{256} = \boxed{\textbf{(A) }\frac{15}{256}}</math>.
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Thus, since the three sub-problems are independent, the final answer is <math>\frac{3}{8} \cdot \frac{1}{4} \cdot \frac{5}{8} = \boxed{\textbf{(A) }\frac{15}{256}}</math>.
  
 
==Solution 2==
 
==Solution 2==
  
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Consider – independently – every spot that the frog could attain.
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Consider – independently – every lily pad that Fiona could reach.
  
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Given that it can only jump at most <math>2</math> places per move, and still wishes to avoid pads <math>3</math> and <math>6</math>, it must also land on numbers <math>2</math>, <math>4</math>, <math>5</math>, and <math>7</math>.
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Given that she can only jump at most <math>2</math> places per move, and still wishes to avoid pads <math>3</math> and <math>6</math>, she must also land on numbers <math>2</math>, <math>4</math>, <math>5</math>, and <math>7</math>.
  
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There are two ways to get to that point – one would be <math>(1,2)</math> on the first move, and the other is just <math>(2)</math>. The total sum is then <math>\frac{1}{2} \times \frac{1}{2} + \frac{1}{2} = \frac{3}{4}</math>, which put into our first column and move on. The frog must subsequently go to space <math>4</math>, again with probability <math>\frac{1}{2}</math>. Thus, be sure to multiply by <math>\frac{1}{2}</math> again, yielding the result of <math>\frac{3}{8}</math>.
+
There are two ways to achieve this - one would be <math>(1,2)</math> on her first move, and the other is just <math>(2)</math>. The total sum is then <math>\frac{1}{2} \times \frac{1}{2} + \frac{1}{2} = \frac{3}{4}</math>, which put into our first column and move on. The frog must subsequently go to space <math>4</math>, again with probability <math>\frac{1}{2}</math>. Thus, be sure to multiply by <math>\frac{1}{2}</math> again, yielding the result of <math>\frac{3}{8}</math>.
  
 
Similarly, multiply your product by <math>\frac{1}{2}</math> once more, to arrive at spot <math>5</math>: <math>\frac{3}{8} \times {1}{2} = \frac{3}{16}</math>. For number <math>7</math>, take another <math>\frac{1}{2}</math>, giving us <math>\frac{3}{16} \times \frac{1}{2} = \frac{3}{32}</math>.
 
Similarly, multiply your product by <math>\frac{1}{2}</math> once more, to arrive at spot <math>5</math>: <math>\frac{3}{8} \times {1}{2} = \frac{3}{16}</math>. For number <math>7</math>, take another <math>\frac{1}{2}</math>, giving us <math>\frac{3}{16} \times \frac{1}{2} = \frac{3}{32}</math>.

Revision as of 22:01, 18 February 2019

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)

See Also

2019 AMC 12B (Problems • Answer Key • Resources)
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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