Difference between revisions of "2019 AMC 12B Problems/Problem 16"
m |
|||
| Line 13: | Line 13: | ||
For the second mini-problem, Fiona <math>\textit{must}</math> go <math>1, 2</math> (probability of <math>\frac{1}{4}</math>). Any other option results in her death to a predator. | For the second mini-problem, Fiona <math>\textit{must}</math> go <math>1, 2</math> (probability of <math>\frac{1}{4}</math>). Any other option results in her death to a predator. | ||
| − | Thus, the final answer is <math>\frac{3}{8} \cdot \frac{1}{4} \cdot \frac{5}{8} = \frac{15}{256} = | + | Thus, the final answer is <math>\frac{3}{8} \cdot \frac{1}{4} \cdot \frac{5}{8} = \frac{15}{256} = </math>\boxed{\textbf{(A) }\frac{15}{256}}$. |
==Solution 2== | ==Solution 2== | ||
| − | + | ||
| − | + | ||
| − | + | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2019|ab=B|num-b=15|num-a=17}} | {{AMC12 box|year=2019|ab=B|num-b=15|num-a=17}} | ||
| − | {{MAA Notice} | + | {{MAA Notice}} |
Revision as of 19:50, 14 February 2019
Contents
Problem
Lily pads numbered from 
to 
lie in a row on a pond. Fiona the frog sits on pad , a morsel of food sits on pad 
, and predators sit on pads 
and 
. At each unit of time the frog jumps either to the next higher numbered pad or to the pad after that, each with probability 
, independently from previous jumps. What is the probability that Fiona skips over pads and and lands on pad ?
Solution 1
First, notice that Fiona, if she jumps over the predator on pad , must land on pad 
. Similarly, she must land on 
if she makes it past . Thus, we can split it into smaller problems counting the probability Fiona skips , Fiona skips (starting at ) and 
skip (starting at ). Incidentally, the last one is equivalent to the first one minus 
.
Let's call the larger jump a 
-jump, and the smaller a -jump.
For the first mini-problem, let's see our options. Fiona can either go 
(probability of 
), or she can go 
(probability of 
). These are the only two options, so they together make the answer 
. We now also know the answer to the last mini-problem (
).
For the second mini-problem, Fiona 
go 
(probability of ). Any other option results in her death to a predator.
Thus, the final answer is 
\boxed{\textbf{(A) }\frac{15}{256}}$.
Solution 2
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. 
