Difference between revisions of "2019 AMC 12B Problems/Problem 16"
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==Solution 1== | ==Solution 1== | ||
− | First, notice that Fiona, if she jumps over the predator on pad <math>3</math>, | + | First, notice that Fiona, if she jumps over the predator on pad <math>3</math>, 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 it into <math>3</math> smaller problems counting the probability Fiona skips <math>3</math>, Fiona skips <math>6</math> (starting at <math>4</math>) and <math>\textit{doesn't}</math> skip <math>10</math> (starting at <math>7</math>). Incidentally, the last one is equivalent to the first one minus <math>1</math>. |
Let's call the larger jump a <math>2</math>-jump, and the smaller a <math>1</math>-jump. | Let's call the larger jump a <math>2</math>-jump, and the smaller a <math>1</math>-jump. | ||
− | For the first mini-problem, let's see our options. Fiona can either go <math>1, 1, 2</math> (probability of \frac{1}{8}), or she can go <math>2, 2</math> (probability of \frac{1}{4}). These are the only two options, so they together make the answer <math>\frac{3}{8}</math>. We now also know the answer to the last mini-problem (<math>\frac{5}{8}</math>). | + | For the first mini-problem, let's see our options. Fiona can either go <math>1, 1, 2</math> (probability of <math>\frac{1}{8}</math>), or she can go <math>2, 2</math> (probability of <math>\frac{1}{4}</math>). These are the only two options, so they together make the answer <math>\frac{3}{8}</math>. We now also know the answer to the last mini-problem (<math>\frac{5}{8}</math>). |
− | For the second mini-problem, Fiona \textit{must} go <math>1, 2</math> (probability of \frac{1}{4}). Any other option results in her death to a predator. | + | For the second mini-problem, Fiona <math>\textit{must}</math> go <math>1, 2</math> (probability of <math>\frac{1}{4}</math>). Any other option results 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} = \boxed{A}</math> | Thus, the final answer is <math>\frac{3}{8} \cdot \frac{1}{4} \cdot \frac{5}{8} = \frac{15}{256} = \boxed{A}</math> | ||
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==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}} |
Revision as of 20:42, 14 February 2019
Contents
Problem
Lily pads numbered from to
lie in a row on a pond. Fiona the frog sits on pad
, a morsel of food sits on pad
, and predators sit on pads
and
. At each unit of time the frog jumps either to the next higher numbered pad or to the pad after that, each with probability
, independently from previous jumps. What is the probability that Fiona skips over pads
and
and lands on pad
?
Solution 1
First, notice that Fiona, if she jumps over the predator on pad , must land on pad
. Similarly, she must land on
if she makes it past
. Thus, we can split it into
smaller problems counting the probability Fiona skips
, Fiona skips
(starting at
) and
skip
(starting at
). Incidentally, the last one is equivalent to the first one minus
.
Let's call the larger jump a -jump, and the smaller a
-jump.
For the first mini-problem, let's see our options. Fiona can either go (probability of
), or she can go
(probability of
). These are the only two options, so they together make the answer
. We now also know the answer to the last mini-problem (
).
For the second mini-problem, Fiona go
(probability of
). Any other option results in her death to a predator.
Thus, the final answer is
Solution 2
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.