Difference between revisions of "2019 AMC 12B Problems/Problem 16"
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− | + | ==Problem== | |
+ | Lily pads numbered from <math>0</math> to <math>11</math> lie in a row on a pond. Fiona the frog sits on pad <math>0</math>, a morsel of food sits on pad <math>10</math>, and predators sit on pads <math>3</math> and <math>6</math>. At each unit of time the frog jumps either to the next higher numbered pad or to the pad after that, each with probability <math>\frac{1}{2}</math>, independently from previous jumps. What is the probability that Fiona skips over pads <math>3</math> and <math>6</math> and lands on pad <math>10</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) }\frac{1}{4}</math> | |
− | + | ==Solution 1== | |
+ | 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>. | ||
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+ | 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 <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>). | ||
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+ | 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> | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | |||
+ | |||
+ | ==See Also== | ||
+ | {{AMC12 box|year=2019|ab=B|num-b=15|num-a=17}} | ||
+ | {{MAA Notice}} |
Revision as of 19:44, 14 February 2019
Contents
[hide]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 |
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