2019 AMC 12B Problems/Problem 19
Problem
There are lily pads in a row numbered 0 to 11, in that order. There are predators on lily pads 3 and 6, and a morsel of food on lily pad 10. Fiona the frog starts on pad 0, and from any given lily pad, has a chance to hop to the next pad, and an equal chance to jump 2 pads. What is the probability that Fiona reaches pad 10 without landing on either pad 3 or pad 6?
Solution
First, notice that Fiona, if she jumps over the predator on pad , \textbf{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 \textit{doesn't} 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 \frac{1}{8}), or she can go (probability of \frac{1}{4}). 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 \textit{must} go (probability of \frac{1}{4}). Any other option results in her death to a predator.
Thus, the final answer is
See Also
2019 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 18 |
Followed by Problem 20 |
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All AMC 12 Problems and Solutions |