Difference between revisions of "2019 AMC 12B Problems/Problem 16"
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+ | ==Solution 3 (Recursion)== | ||
+ | |||
+ | Let <math>p_n</math> be the probability of landing on lily pad <math>n</math>. We immediately notice that, if there are no restrictions: | ||
+ | <cmath>p_n = \frac{1}{2} \cdot p_{n-1} + \frac{1}{2} \cdot p_{n-2}</cmath> | ||
+ | |||
+ | This is because, given that we are at lily pad <math>n-2</math>, there is a 50% chance that we will go to lilypad <math>n</math>, and the same applies for lily pad <math>n-1</math>. We will now compute the values of <math>p_n</math> recursively, but we will skip over <math>3</math> and <math>6</math>. That is, we will not consider any jumps from lily pads 3 or 6 when considering the probabilities. We obtain the following chart: | ||
+ | |||
+ | <asy> | ||
+ | for(int i =0; i<= 11; ++i) { | ||
+ | draw((i,0)--(i+1,0)--(i+1,1)--(0,i)--cycle); | ||
+ | } | ||
+ | </asy> | ||
==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 00:14, 15 February 2019
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
.
Similarly, multiply your product by once more, to arrive at spot
:
. For number
, take another
, 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 ,
, and
, as the path straight to point
is not available. That leaves us with a partial count of
. Multiply, to find that
.
--anna0kear.
Solution 3 (Recursion)
Let be the probability of landing on lily pad
. We immediately notice that, if there are no restrictions:
This is because, given that we are at lily pad , there is a 50% chance that we will go to lilypad
, and the same applies for lily pad
. We will now compute the values of
recursively, but we will skip over
and
. That is, we will not consider any jumps from lily pads 3 or 6 when considering the probabilities. We obtain the following chart:
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.