Difference between revisions of "2019 AMC 12B Problems/Problem 16"
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Consider – independently – every spot that the frog could attain. | Consider – independently – every spot that the frog could attain. | ||
− | + | Given that it can only jump at most <math>2</math> places per move, and still wishes to avoid pads <math>3</math> and <math>6</math>, it must also land on numbers <math>2</math>, <math>4</math>, <math>5</math>, and <math>7</math>. | |
− | There | + | There are two ways to get to that point – one would be <math>(1,2)</math> on the first move, and the other is just <math>(2)</math>. The total sum is then <math>\frac{1}{2} \times \frac{1}{2} + \frac{1}{2} = \frac{3}{4}</math>, which put into our first column and move on. The frog must subsequently go to space <math>4</math>, again with probability <math>\frac{1}{2}</math>. Thus, be sure to multiply by <math>\frac{1}{2}</math> again, yielding the result of <math>\frac{3}{8}. |
− | + | Similarly, multiply your product by </math>\frac{1}{2}<math> once more, to arrive at spot </math>5<math>: </math>\frac{3}{8} \times {1}{2} = \frac{3}{16}<math>. For number </math>7<math>, take another </math>\frac{1}{2}, giving us <math>{3}{16} \times \frac{1}{2} = \frac{3}{32}</math>. | |
+ | Next, we must look at a number of options. For a fuller picture, it would be best to break down the choices. The only possibilities here are <math>(8,9,10), (8,10), and (9,10)</math>, as the path straight to point <math>10</math> is not available. That leaves us with a partial count of <math>\frac{1}{8} + \frac{1}{4} + \frac{1}{4} = frac{5}{8}. Multiply, to find the result of turning output's, answer </math>\frac{3}{32} \times {5}{8} = \boxed{\textbf{(A)} \frac{15}{256}}$. \square | ||
− | + | --anna0kear. | |
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==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}} | ||
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Revision as of 22:33, 14 February 2019
Contents
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 .
Solution 2
Consider – independently – every spot that the frog could attain.
Given that it can only jump at most places per move, and still wishes to avoid pads
and
, it must also land on numbers
,
,
, and
.
There are two ways to get to that point – one would be on the first move, and the other is just
. The total sum is then
, which put into our first column and move on. The frog must subsequently go to space
, again with probability
. Thus, be sure to multiply by
again, yielding the result of $\frac{3}{8}.
Similarly, multiply your product by$ (Error compiling LaTeX. Unknown error_msg)\frac{1}{2}5
\frac{3}{8} \times {1}{2} = \frac{3}{16}
7
\frac{1}{2}, giving us
.
Next, we must look at a number of options. For a fuller picture, it would be best to break down the choices. The only possibilities here are , as the path straight to point
is not available. That leaves us with a partial count of
\frac{3}{32} \times {5}{8} = \boxed{\textbf{(A)} \frac{15}{256}}$. \square
--anna0kear.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
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