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

Revision as of 23:35, 14 February 2019

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

Consider – independently – every spot that the frog could attain.

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

There are two ways to get to that point – one would be $(1,2)$ on the 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)

@@ Line 34: / Line 34: @@
 <cmath>p_n = \frac{1}{2} \cdot p_{n-1} + \frac{1}{2} \cdot p_{n-2}</cmath>
-This is because, given that we are at lily pad <math>n-2</math>, there is a 50% chance that we will go to lily pad <math>n</math>, and the same applies for lily pad <math>n-1</math>. We will now compute the values of <math>p_n</math> recursively, but we will skip over <math>3</math> and <math>6</math>. That is, we will not consider any jumps from lily pads 3 or 6 when considering the probabilities. We obtain the following chart:
+This is because, given that we are at lily pad <math>n-2</math>, there is a 50% chance that we will go to lily pad <math>n</math>, and the same applies for lily pad <math>n-1</math>. We will now compute the values of <math>p_n</math> recursively, but we will skip over <math>3</math> and <math>6</math>. 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", "$1/2$", "$3/4$", "$X$", "$3/8$", "$3/16$", "$X$", "$3/32$", "$3/64$", "$9/128$", "$15/256$", "$X$"};
+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);

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