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 ?
Answer Choices:
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B.
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E.
Solution:
Firstly, notice that if Fiona jumps over the predator on pad , she must land on pad . Similarly, she must land on if she makes it past . Thus, we can split the problem into smaller sub-problems, separately finding the probability. Fiona skips , the probability she skips (starting at ) and the probability she doesn't skip (starting at ). Notice that by symmetry, the last of these three sub-problems is the complement of the first sub-problem, so the probability will be - the probability obtained in the first sub-problem.
In the analysis below, we call the larger jump a -jump, and the smaller a -jump.
For the first sub-problem, consider Fiona's options. She can either go -jump, -jump, -jump, with probability , or she can go -jump, -jump, with probability . These are the only two options, so they together make the answer . We now also know the answer to the last sub-problem is .
For the second sub-problem, Fiona must go -jump, -jump, with probability , since any other option would result in her death to a predator.
Thus, since the three sub-problems are independent, the final answer is .
OR
Observe that since Fiona can only jump at most places per move, and still wishes to avoid pads and , she must also land on numbers , and .
There are two ways to reach lily pad , namely -jump, -jump, with probability , or just a -jump, with probability . The total is thus . Fiona must now make a -jump to lily pad , again with probability , giving .
Similarly, Fiona must now make a -jump to reach lily pad , again with probability , giving . Then she must make a -jump to reach lily pad , with probability , yielding .
Finally, to reach lily pad , Fiona has a few options - she can make consecutive -jumps, with probability , or -jump, -jump, with probability , or -jump, -jump, again with probability . The final answer is thus .
The problems on this page are the property of the MAA's American Mathematics Competitions