Difference between revisions of "2019 AMC 12B Problems/Problem 16"
(Tag: Undo) |
m (→Solution 2) |
||
Line 21: | Line 21: | ||
There are two ways to reach lily pad <math>2</math>, namely <math>1</math>-jump, <math>1</math>-jump, with probability <math>\frac{1}{4}</math>, or just a <math>2</math>-jump, with probability <math>\frac{1}{2}</math>. The total is thus <math>\frac{1}{4} + \frac{1}{2} = \frac{3}{4}</math>. Fiona must now make a <math>2</math>-jump to lily pad <math>4</math>, again with probability <math>\frac{1}{2}</math>, giving <math>\frac{3}{4} \cdot \frac{1}{2} = \frac{3}{8}</math>. | There are two ways to reach lily pad <math>2</math>, namely <math>1</math>-jump, <math>1</math>-jump, with probability <math>\frac{1}{4}</math>, or just a <math>2</math>-jump, with probability <math>\frac{1}{2}</math>. The total is thus <math>\frac{1}{4} + \frac{1}{2} = \frac{3}{4}</math>. Fiona must now make a <math>2</math>-jump to lily pad <math>4</math>, again with probability <math>\frac{1}{2}</math>, giving <math>\frac{3}{4} \cdot \frac{1}{2} = \frac{3}{8}</math>. | ||
− | Similarly, Fiona must now make a <math>1</math>-jump to reach lily pad <math>5</math>, again with probability <math>\frac{1}{2}</math>, giving <math>\frac{3}{8} \cdot {1}{2} = \frac{3}{16}</math>. Then she must make a <math>2</math>-jump to reach lily pad <math>7</math>, with probability <math>\frac{1}{2}</math>, yielding <math>\frac{3}{16} \cdot \frac{1}{2} = \frac{3}{32}</math>. | + | Similarly, Fiona must now make a <math>1</math>-jump to reach lily pad <math>5</math>, again with probability <math>\frac{1}{2}</math>, giving <math>\frac{3}{8} \cdot \frac{1}{2} = \frac{3}{16}</math>. Then she must make a <math>2</math>-jump to reach lily pad <math>7</math>, with probability <math>\frac{1}{2}</math>, yielding <math>\frac{3}{16} \cdot \frac{1}{2} = \frac{3}{32}</math>. |
Finally, to reach lily pad <math>10</math>, Fiona has a few options - she can make <math>3</math> consecutive <math>1</math>-jumps, with probability <math>\frac{1}{8}</math>, or <math>1</math>-jump, <math>2</math>-jump, with probability <math>\frac{1}{4}</math>, or <math>2</math>-jump, <math>1</math>-jump, again with probability <math>\frac{1}{4}</math>. The final answer is thus <math>\frac{3}{32} \cdot \left(\frac{1}{8} + \frac{1}{4} + \frac{1}{4}\right) = \frac{3}{32} \cdot \frac{5}{8} = \boxed{\textbf{(A) } \frac{15}{256}}</math>. | Finally, to reach lily pad <math>10</math>, Fiona has a few options - she can make <math>3</math> consecutive <math>1</math>-jumps, with probability <math>\frac{1}{8}</math>, or <math>1</math>-jump, <math>2</math>-jump, with probability <math>\frac{1}{4}</math>, or <math>2</math>-jump, <math>1</math>-jump, again with probability <math>\frac{1}{4}</math>. The final answer is thus <math>\frac{3}{32} \cdot \left(\frac{1}{8} + \frac{1}{4} + \frac{1}{4}\right) = \frac{3}{32} \cdot \frac{5}{8} = \boxed{\textbf{(A) } \frac{15}{256}}</math>. |
Revision as of 19:19, 2 February 2020
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:
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.