Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m
m (open-up. G'L)
Line 19: Line 19:
 
Consider – independently – every spot that the frog could attain.
 
Consider – independently – every spot that the frog could attain.
  
−
As long as it wishes to avoid pads <math>3</math> and <math>6</math> – but can only jump a maximum of <math>2 places at a time – it must determinablitlyes. land on numbers 2, 4, 5, and 7. Get it?
+
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>.
  
−
There's two ways to get there: either \tex(1,2) or just </math>2<math> for the first move. There's also a 1/4 chance that it skips. Thus, we put 3/4 into our column and then move on-fir now.
+
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}.
  
−
It must then make it to space #4, another 1/2 probability. So you can make sure to multiple by 1/2 again, giving 3/8.
+
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>.
  
 +
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
  
−
It's going to be getting a bit different now: for the next few choices, it also has to get to 5. Since our probability has also given and taken into account the factor that it's already landed on 4, multiple by #5 again to get to the spot-ment: 3/8 * 1/2 = 3/16.
+
--anna0kear.
−
 
 
−
Now there's another 1/2, giving us 3/16 * 1/2 = 3/32 for partie -8.
 
−
 
 
−
Here, we must look at possible options. There's only a 3/4 chance that it makes it directly. So, if we want a fuller picture then let's break it down. The possibilities are (8,9,10) (8,10) and (9,10) since going straight to #10 is not allowed. That leaves us with a partial count of 1/8 + 1/4 + 1/4 = 5/8 which we know works because the total sum can still add up to 1. Then, multiply, and that gives </math>3/32 * 5/8 = \boxed{\textbf{(A) }\frac{15}{256}}$.
 
  
 
==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}}
 
{{MAA Notice}}
 
{{MAA Notice}}
 +
−−−−−

Revision as of 22:33, 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$ (Error compiling LaTeX. Unknown error_msg)\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 $(8,9,10), (8,10), and (9,10)$, as the path straight to point $10$ is not available. That leaves us with a partial count of $\frac{1}{8} + \frac{1}{4} + \frac{1}{4} = frac{5}{8}. Multiply, to find the result of turning output's, answer$\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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

−−−−−

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_12B_Problems/Problem_16&oldid=102748"