Difference between revisions of "2019 AMC 12B Problems/Problem 16"
m (Reverted (rev.)) |
|||
(36 intermediate revisions by 12 users not shown) | |||
Line 1: | Line 1: | ||
==Problem== | ==Problem== | ||
− | + | There are lily pads in a row numbered <math>0</math> to <math>11</math>, in that order. There are predators on lily pads <math>3</math> and <math>6</math>, and a morsel of food on lily pad <math>10</math>. Fiona the frog starts on pad <math>0</math>, and from any given lily pad, has a <math>\frac{1}{2}</math> chance to hop to the next pad, and an equal chance to jump <math>2</math> pads. What is the probability that Fiona reaches pad <math>10</math> without landing on either pad <math>3</math> or pad <math>6</math>? | |
− | <math>\textbf{(A) }\frac{15}{256}\qquad\textbf{(B) }\frac{1}{16}\qquad\textbf{(C) }\frac{15}{128}\qquad\textbf{(D) }\frac{1}{8}\qquad\textbf{(E) }\ | + | <math>\textbf{(A) } \frac{15}{256} \qquad \textbf{(B) } \frac{1}{16} \qquad \textbf{(C) } \frac{15}{128}\qquad \textbf{(D) } \frac{1}{8} \qquad \textbf{(E) } \frac14</math> |
==Solution 1== | ==Solution 1== | ||
− | + | Firstly, notice that if Fiona jumps over the predator on pad <math>3</math>, she must land on pad <math>4</math>. Similarly, she must land on <math>7</math> if she makes it past <math>6</math>. Thus, we can split the problem into <math>3</math> smaller sub-problems, separately finding the probability. Fiona skips <math>3</math>, the probability she skips <math>6</math> (starting at <math>4</math>) and the probability she ''doesn't'' skip <math>10</math> (starting at <math>7</math>). Notice that by symmetry, the last of these three sub-problems is the complement of the first sub-problem, so the probability will be <math>1 - \text{the probability obtained in the first sub-problem}</math>. | |
− | + | In the analysis below, we call the larger jump a <math>2</math>-jump, and the smaller a <math>1</math>-jump. | |
− | For the first | + | For the first sub-problem, consider Fiona's options. She can either go <math>1</math>-jump, <math>1</math>-jump, <math>2</math>-jump, with probability <math>\frac{1}{8}</math>, or she can go <math>2</math>-jump, <math>2</math>-jump, with probability <math>\frac{1}{4}</math>. These are the only two options, so they together make the answer <math>\frac{1}{8}+\frac{1}{4}=\frac{3}{8}</math>. We now also know the answer to the last sub-problem is <math>1-\frac{3}{8}=\frac{5}{8}</math>. |
− | For the second | + | For the second sub-problem, Fiona ''must'' go <math>1</math>-jump, <math>2</math>-jump, with probability <math>\frac{1}{4}</math>, since any other option would result in her death to a predator. |
− | Thus, the final answer is <math>\frac{3}{8} \cdot \frac{1}{4} \cdot \frac{5}{8} = \frac{15}{256} | + | Thus, since the three sub-problems are independent, the final answer is <math>\frac{3}{8} \cdot \frac{1}{4} \cdot \frac{5}{8} = \boxed{\textbf{(A) }\frac{15}{256}}</math>. |
==Solution 2== | ==Solution 2== | ||
+ | Observe that since Fiona can only jump at most <math>2</math> places per move, and still wishes to avoid pads <math>3</math> and <math>6</math>, she must also land on numbers <math>2</math>, <math>4</math>, <math>5</math>, and <math>7</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 \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>. | ||
+ | |||
+ | ==Solution 3 (recursion)== | ||
+ | |||
+ | Let <math>p_n</math> be the probability of landing on lily pad <math>n</math>. Observe that if there are no restrictions, we would have | ||
+ | <cmath>p_n = \frac{1}{2} \cdot p_{n-1} + \frac{1}{2} \cdot p_{n-2}</cmath> | ||
+ | |||
+ | This is because, given that Fiona is at lily pad <math>n-2</math>, there is a <math>\frac{1}{2}</math> probability that she will make a <math>2</math>-jump to reach lily pad <math>n</math>, and the same applies for a <math>1</math>-jump to reach 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 <math>3</math> or <math>6</math> when considering the probabilities. We obtain the following chart, where an X represents an unused/uncomputed value: | ||
+ | |||
+ | <asy> | ||
+ | |||
+ | unitsize(40); | ||
+ | string[] vals = {"1", "$1/2$", "$3/4$", "X", "$3/8$", "$3/16$", "X", "$3/32$", "$3/64$", "$9/128$", "$15/256$", "X"}; | ||
+ | for(int i =0; i<= 11; ++i) { | ||
+ | draw((i,0)--(i+1,0)--(i+1,1)--(i,1)--cycle); | ||
+ | label((string) i, (i+0.5,0), S); | ||
+ | label(vals[i], (i+0.5, 0.5)); | ||
+ | } | ||
+ | </asy> | ||
+ | |||
+ | Hence the answer is <math>\boxed{\textbf{(A) } \frac{15}{256}}</math>. | ||
+ | |||
+ | Note: If we let <math>p_n</math> be the probability of surviving if the frog is on lily pad <math>n</math>, using <math>p_{10}</math> = 1, we can solve backwards and obtain the following chart: | ||
+ | |||
+ | <asy> | ||
+ | |||
+ | unitsize(40); | ||
+ | string[] vals = {"$15/256$", "$5/128$", "$5/64$", "X", "$5/32$", "$5/16$", "X", "$5/8$", "$3/4$", "$1/2$", "1", "X"}; | ||
+ | for(int i =0; i<= 11; ++i) { | ||
+ | draw((i,0)--(i+1,0)--(i+1,1)--(i,1)--cycle); | ||
+ | label((string) (11-i), (i+0.5,0), S); | ||
+ | label(vals[(11-i)], (i+0.5, 0.5)); | ||
+ | } | ||
+ | </asy> | ||
+ | |||
+ | ==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 | ||
+ | |||
+ | ==Solution 5 (answer choices)== | ||
+ | Note that there is exactly one way to reach lily pad <math>10</math> in <math>8</math> moves, and there is a probability of <math>\frac{1}{256}</math> that this occurs. All other paths take less than <math>8</math> moves, and thus the probability of them occurring is <math>\frac{n}{2^k}</math> for <math>k < 8</math>. Thus, the answer must have a denominator of <math>256</math>, or answer choice <math>\boxed{\textbf{(A)}}</math>. | ||
==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}} |
Latest revision as of 23:33, 13 November 2022
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 land 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
Solution 5 (answer choices)
Note that there is exactly one way to reach lily pad in
moves, and there is a probability of
that this occurs. All other paths take less than
moves, and thus the probability of them occurring is
for
. Thus, the answer must have a denominator of
, or answer choice
.
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.