2019 AMC 12B Problems/Problem 16
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 
.
Solution 2
Consider – independently – every spot that the frog could attain.
Given that it can only jump at most places per move, and still wishes to avoid pads and , it must also land on numbers , , 
, and .
There are two ways to get to that point – one would be 
on the first move, and the other is just 
. The total sum is then 
, which put into our first column and move on. The frog must subsequently go to space , again with probability . Thus, be sure to multiply by again, yielding the result of $\frac{3}{8}.
Similarly, multiply your product by$ (Error compiling LaTeX. Unknown error_msg)\frac{1}{2}
5
\frac{3}{8} \times {1}{2} = \frac{3}{16}
7
\frac{1}{2}, giving us 
.
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 
, as the path straight to point is not available. That leaves us with a partial count of 
\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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