Difference between revisions of "2019 AMC 12B Problems/Problem 16"
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==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 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-prolems 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 sub-problem, | + | For the first sub-problem, consider Fiona's options. She can either go <math>1, 1, 2</math>, with probability <math>\frac{1}{8}</math>, or she can go <math>2, 2</math>, 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 sub-problem, Fiona <math>\textit{must}</math> go <math>1, 2</math> <math> | + | For the second sub-problem, Fiona <math>\textit{must}</math> go <math>1, 2</math>, 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 | + | 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== | ||
− | Consider – independently – every | + | Consider – independently – every lily pad that Fiona could reach. |
− | Given that | + | Given that she 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 | + | There are two ways to achieve this - one would be <math>(1,2)</math> on her first move, and the other is just <math>(2)</math>. The total sum is then <math>\frac{1}{2} \times \frac{1}{2} + \frac{1}{2} = \frac{3}{4}</math>, which put into our first column and move on. The frog must subsequently go to space <math>4</math>, again with probability <math>\frac{1}{2}</math>. Thus, be sure to multiply by <math>\frac{1}{2}</math> again, yielding the result of <math>\frac{3}{8}</math>. |
Similarly, multiply your product by <math>\frac{1}{2}</math> once more, to arrive at spot <math>5</math>: <math>\frac{3}{8} \times {1}{2} = \frac{3}{16}</math>. For number <math>7</math>, take another <math>\frac{1}{2}</math>, giving us <math>\frac{3}{16} \times \frac{1}{2} = \frac{3}{32}</math>. | Similarly, multiply your product by <math>\frac{1}{2}</math> once more, to arrive at spot <math>5</math>: <math>\frac{3}{8} \times {1}{2} = \frac{3}{16}</math>. For number <math>7</math>, take another <math>\frac{1}{2}</math>, giving us <math>\frac{3}{16} \times \frac{1}{2} = \frac{3}{32}</math>. |
Revision as of 23:01, 18 February 2019
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-prolems 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 , with probability
, or she can go
, 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 go
, 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
Consider – independently – every lily pad that Fiona could reach.
Given that she 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 achieve this - one would be on her 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 lily pad
, 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, where an X represents an unused/uncomputed value:
As we can see, the answer is
(Solution by vedadehhc)
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 |
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