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. | ||
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+ | Case 1: Two <math>1 \cdot 1</math> tiles fill the space between lilypads <math>0</math> and <math>2</math>. | ||
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+ | 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> | ||
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+ | Case 2: A single <math>1 \cdot 2</math> tile fills the space between lilypads <math>0</math> and <math>2</math>. | ||
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+ | 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> | ||
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+ | 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.