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

(→‎Problem: Added problem statement.)
(→‎Solution: I pasted the solution here. It was originally on the 12B 19 (incorrect), so I just moved it here.)
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==Solution==
 
==Solution==
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First, notice that Fiona, if she jumps over the predator on pad <math>3</math>, \textbf{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 \textit{doesn't} skip <math>10</math> (starting at <math>7</math>). Incidentally, the last one is equivalent to the first one minus <math>1</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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For the first mini-problem, let's see our options. Fiona can either go <math>1, 1, 2</math> (probability of \frac{1}{8}), or she can go <math>2, 2</math> (probability of \frac{1}{4}). 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 mini-problem (<math>\frac{5}{8}</math>).
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For the second mini-problem, Fiona \textit{must} go <math>1, 2</math> (probability of \frac{1}{4}). Any other option results 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{A}</math>
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2019|ab=B|num-b=15|num-a=17}}
 
{{AMC12 box|year=2019|ab=B|num-b=15|num-a=17}}

Revision as of 15:59, 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

First, notice that Fiona, if she jumps over the predator on pad $3$, \textbf{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 $\frac{3}{8} \cdot \frac{1}{4} \cdot \frac{5}{8} = \frac{15}{256} = \boxed{A}$

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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