Difference between revisions of "2019 AMC 12B Problems/Problem 16"
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+ | ==Solution 4 (simple casework bash)== | ||
+ | This is equivalent to finding the probability for each of the valid ways of tiling a <math>1</math>-by-<math>11</math> rectangular grid (with one end being lilypad <math>0</math> and the other being lilypad <math>11</math>) with tiles of size <math>1 \cdot 1</math> and <math>1 \cdot 2</math> that satisfies the conditions mentioned in the problem. Obviously, for Fiona to skip pads <math>3</math> and <math>6</math>, a <math>1 \cdot 2</math> tile must be placed with one end at lilypad <math>2</math> and the other at lilypad <math>4</math>, and another <math>1 \cdot 2</math> must be placed with one end at lilypad <math>5</math> and the other at lilypad <math>7</math>. Thus, since only a <math>1 \cdot 1</math> tile can fit between the two aforementioned <math>1 \cdot 2</math> tiles, we will place it there. Now, we can solve this problem with simple casework. | ||
+ | |||
+ | Case 1: Two <math>1 \cdot 1</math> tiles fill the space between lilypads <math>0</math> and <math>2</math>. | ||
+ | |||
+ | There are two ways to permute a placement of a <math>1 \cdot 1</math> tile and a <math>1 \cdot 2</math> tile between lilypads <math>7</math> and <math>10</math>, so our probability for this sub-case is <math>\frac{2}{2^7} = \frac{1}{64}</math>. In the other subcase where the space between lilypads <math>7</math> and <math>10</math> is completely filled with <math>1 \cdot 1</math> tiles, there is trivially only one tiling, thus the probability for this sub-case is <math>\frac{1}{2^8} = \frac{1}{256}.</math> The total probability for this case is <math>\frac{1}{64} + \frac{1}{256} = \frac{5}{256}.</math> | ||
+ | |||
+ | Case 2: A single <math>1 \cdot 2</math> tile fills the space between lilypads <math>0</math> and <math>2</math>. | ||
+ | |||
+ | Note that the combined probability for this case will be double that of Case <math>1</math>, since a single <math>1 \cdot 2</math> tile takes up one less tile than two <math>1 \cdot 1</math> tiles. Thus, the probability for this case is <math>2 \cdot \frac{5}{256} = \frac{10}{256}.</math> | ||
+ | |||
+ | Summing our cases up, we obtain <math>\boxed{\textbf{(A) } \frac{15}{256}}</math>. | ||
+ | |||
+ | -fidgetboss_4000 | ||
==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 23:05, 2 April 2020
Contents
Problem
There are lily pads in a row numbered to
, in that order. There are predators on lily pads
and
, and a morsel of food on lily pad
. Fiona the frog starts on pad
, and from any given lily pad, has a
chance to hop to the next pad, and an equal chance to jump
pads. What is the probability that Fiona reaches pad
without landing on either pad
or pad
?
Solution 1
Firstly, notice that if Fiona jumps over the predator on pad , she must on pad
. Similarly, she must land on
if she makes it past
. Thus, we can split the problem into
smaller sub-problems, separately finding the probability Fiona skips
, the probability she skips
(starting at
) and the probability she doesn't skip
(starting at
). Notice that by symmetry, the last of these three sub-problems is the complement of the first sub-problem, so the probability will be
.
In the analysis below, we call the larger jump a -jump, and the smaller a
-jump.
For the first sub-problem, consider Fiona's options. She can either go -jump,
-jump,
-jump, with probability
, or she can go
-jump,
-jump, with probability
. These are the only two options, so they together make the answer
. We now also know the answer to the last sub-problem is
.
For the second sub-problem, Fiona must go -jump,
-jump, with probability
, since any other option would result in her death to a predator.
Thus, since the three sub-problems are independent, the final answer is .
Solution 2
Observe that since Fiona can only jump at most places per move, and still wishes to avoid pads
and
, she must also land on numbers
,
,
, and
.
There are two ways to reach lily pad , namely
-jump,
-jump, with probability
, or just a
-jump, with probability
. The total is thus
. Fiona must now make a
-jump to lily pad
, again with probability
, giving
.
Similarly, Fiona must now make a -jump to reach lily pad
, again with probability
, giving
. Then she must make a
-jump to reach lily pad
, with probability
, yielding
.
Finally, to reach lily pad , Fiona has a few options - she can make
consecutive
-jumps, with probability
, or
-jump,
-jump, with probability
, or
-jump,
-jump, again with probability
. The final answer is thus
.
Solution 3 (recursion)
Let be the probability of landing on lily pad
. Observe that if there are no restrictions, we would have
This is because, given that Fiona is at lily pad , there is a
probability that she will make a
-jump to reach lily pad
, and the same applies for a
-jump to reach 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
or
when considering the probabilities. We obtain the following chart, where an X represents an unused/uncomputed value:
Hence the answer is .
Note: If we let be the probability of surviving if the frog is on lily pad
, using
= 1, we can solve backwards and obtain the following chart:
Solution 4 (simple casework bash)
This is equivalent to finding the probability for each of the valid ways of tiling a -by-
rectangular grid (with one end being lilypad
and the other being lilypad
) with tiles of size
and
that satisfies the conditions mentioned in the problem. Obviously, for Fiona to skip pads
and
, a
tile must be placed with one end at lilypad
and the other at lilypad
, and another
must be placed with one end at lilypad
and the other at lilypad
. Thus, since only a
tile can fit between the two aforementioned
tiles, we will place it there. Now, we can solve this problem with simple casework.
Case 1: Two tiles fill the space between lilypads
and
.
There are two ways to permute a placement of a tile and a
tile between lilypads
and
, so our probability for this sub-case is
. In the other subcase where the space between lilypads
and
is completely filled with
tiles, there is trivially only one tiling, thus the probability for this sub-case is
The total probability for this case is
Case 2: A single tile fills the space between lilypads
and
.
Note that the combined probability for this case will be double that of Case , since a single
tile takes up one less tile than two
tiles. Thus, the probability for this case is
Summing our cases up, we obtain .
-fidgetboss_4000
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.