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

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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>.
 
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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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}.
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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>.
  
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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}, giving us <math>{3}{16} \times \frac{1}{2} = \frac{3}{32}</math>.
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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}, giving us </math>{3}{16} \times \frac{1}{2} = \frac{3}{32}<math>.
  
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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 <math>(8,9,10), (8,10), and (9,10)</math>, as the path straight to point <math>10</math> is not available. That leaves us with a partial count of <math>\frac{1}{8} + \frac{1}{4} + \frac{1}{4} = frac{5}{8}. Multiply, to find the result of turning output's, answer </math>\frac{3}{32} \times {5}{8} = \boxed{\textbf{(A)} \frac{15}{256}}$. \square
+
Next, we must look at a number of options. For a fuller picture, it would be best to break down the choices. The only possibilities here are </math>(8,9,10)<math>, </math>(8,10)<math>, and </math>(9,10)<math>, as the path straight to point </math>10<math> is not available. That leaves us with a partial count of </math>\frac{1}{8} + \frac{1}{4} + \frac{1}{4} = frac{5}{8}<math>. Multiply, to find that </math>\frac{3}{32} \times {5}{8} = \boxed{\textbf{(A)} \frac{15}{256}}<math>. </math>\square$
  
 
--anna0kear.
 
--anna0kear.

Revision as of 21:35, 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 $\frac{1}{2}, giving us${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$ (Error compiling LaTeX. Unknown error_msg)(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$\frac{3}{32} \times {5}{8} = \boxed{\textbf{(A)} \frac{15}{256}}$.$\square$

--anna0kear.

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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