Online Course Discussion Forum
2019 amc12b #16
there are lily pads in a row numbered 
to 
, in that order. There are predators on lily pads 
and 
, and a morsel of food on lily pad 
. Fiona the frog starts on pad , and from any given lily pad, has a 
chance to hop to the next pad, and an equal chance to jump 
pads. What is the probability that Fiona reaches pad without landing on either pad or pad ?
for this one, i separated this into 3 parts: before pad 3, between 3 and 6, after 6. And there are 2 ways for the frog to jump: 1 pad or 2 pads.
I figured out that for before pad 3 there are 2 ways: 1-1-2 or 2-2 so it's 1/8+1/4=3/8.
For between 3 and 6 there is only 1 way:1-2, so it's 1/4
For the after 6 part , the answer says it should be 5/8 but I don't understand how this works. Is it supposed to be solved by stars and bars or something else?
Good job on the earlier parts!
In fact, you want to use the same method for the last part. After jumping over pad 6, the frog is at pad 7 and must get to pad 10. Remember, however, that since there is a pad 11, the frog can still jump over the food on pad 10, which we need to avoid.
Hence, the possible jumps are 1-1-1, 1-2, or 2-1, with probability 1/8+1/4+1/4 = 5/8 for the last part. Then we just need to multiply all the parts together for the final answer.
Note: The method above works great, as the different cases are not too bad. I also, however, would recommend trying a recursive approach to the problem. If $P_N$ is the probability the frog gets to pad $N$, we know $P_0 = 1$ and $P_3=P_6=0$. How can you write $P_N$ in terms of the previous pads?
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