2019 AMC 12B Problems/Problem 19

Revision as of 13:25, 14 February 2019 by Robinaldabanx (talk | contribs) (Solution)

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 $\tfrac{1}{2}$ 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 $3$, \textbf{must} land on pad $4$. Similarly, she must land on $7$ if she makes it past $6$. Thus, we can split it into $3$ smaller problems counting the probability Fiona skips $3$, Fiona skips $6$ (starting at $4$) and \textit{doesn't} skip $10$ (starting at $7$). Incidentally, the last one is equivalent to the first one minus $1$.

Let's call the larger jump a $2$-jump, and the smaller a $1$-jump.

For the first mini-problem, let's see our options. Fiona can either go $1, 1, 2$ (probability of \frac{1}{8}), or she can go $2, 2$ (probability of \frac{1}{4}). These are the only two options, so they together make the answer $\frac{3}{8}$. We now also know the answer to the last mini-problem ($\frac{5}{8}$).

For the second mini-problem, Fiona \textit{must} go $1, 2$ (probability of \frac{1}{4}). Any other option results in her death to a predator.

Thus, the final answer is $\frac{3}{8} \cdot \frac{1}{4} \cdot \frac{5}{8} = \frac{15}{256} = \boxed{A}$

See Also

2019 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 18
Followed by
Problem 20
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All AMC 12 Problems and Solutions
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