2019 AMC 12B Problems/Problem 16: Difference between revisions
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For the second mini-problem, Fiona <math>\textit{must}</math> go <math>1, 2</math> (probability of <math>\frac{1}{4}</math>). Any other option results in her death to a predator. | For the second mini-problem, Fiona <math>\textit{must}</math> go <math>1, 2</math> (probability of <math>\frac{1}{4}</math>). Any other option results in her death to a predator. | ||
Thus, the final answer is <math>\frac{3}{8} \cdot \frac{1}{4} \cdot \frac{5}{8} = \frac{15}{256} = \boxed{A} | Thus, the final answer is <math>\frac{3}{8} \cdot \frac{1}{4} \cdot \frac{5}{8} = \frac{15}{256} = the answe, si..</math>\boxed{\textbf{(A) }\frac{15}{256}}$.s | ||
==Solution 2== | ==Solution 2== | ||
- | |||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2019|ab=B|num-b=15|num-a=17}} | {{AMC12 box|year=2019|ab=B|num-b=15|num-a=17}} | ||
{{MAA Notice}} | {{MAA Notice} s} | ||
Revision as of 19:50, 14 February 2019
Problem
Lily pads numbered from
to
lie in a row on a pond. Fiona the frog sits on pad
, a morsel of food sits on pad
, and predators sit on pads
and
. At each unit of time the frog jumps either to the next higher numbered pad or to the pad after that, each with probability
, independently from previous jumps. What is the probability that Fiona skips over pads
and
and lands on pad
?
Solution 1
First, notice that Fiona, if she jumps over the predator on pad
, must land on pad
. Similarly, she must land on
if she makes it past
. Thus, we can split it into
smaller problems counting the probability Fiona skips
, Fiona skips
(starting at
) and
skip
(starting at
). Incidentally, the last one is equivalent to the first one minus
.
Let's call the larger jump a
-jump, and the smaller a
-jump.
For the first mini-problem, let's see our options. Fiona can either go
(probability of
), or she can go
(probability of
). These are the only two options, so they together make the answer
. We now also know the answer to the last mini-problem (
).
For the second mini-problem, Fiona
go
(probability of
). Any other option results in her death to a predator.
Thus, the final answer is
\boxed{\textbf{(A) }\frac{15}{256}}$.s
Solution 2
-
See Also
| 2019 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 15 |
Followed by Problem 17 |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
{{MAA Notice} s}