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| − | ==Problem==
| + | #REDIRECT[[2019_AMC_10B_Problems/Problem_22]] |
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| − | 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 <math>\tfrac{1}{2}</math> 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?
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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>
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| − | ==Solution==
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| − | First, notice that Fiona, if she jumps over the predator on pad <math>3</math>, \textbf{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 \textit{doesn't} 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.
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| − | 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>).
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| − | 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.
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| − | 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==
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| − | {{AMC12 box|year=2019|ab=B|num-b=18|num-a=20}}
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