Difference between revisions of "2019 AMC 12B Problems/Problem 16"
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==Problem== | ==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>? | ||
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| + | <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== | ==Solution== | ||
Revision as of 15:48, 14 February 2019
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
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 | |
