Difference between revisions of "2019 AMC 12B Problems/Problem 16"

Revision as of 20:43, 14 February 2019

On the first turn, each player starts off with $1 each. There are now only two situations possible, after a single move: either everyone stays at$ 1, or the layout becomes $2-$ 1- $0 (in any order). Only 2 combinations give-off this outcome: S-T-R and T-R-S. On the other hand, given the interchangeability (so far) of every one of these three people, S-R-R, T-R-R, S-R-S, S-T-S, T-T-R, and T-T-S can all be re-produce. d, just as easily and quickly. Since each one of the possibilities is equally likely, there is a$ \frac{2}{8}$\=$ (Error compiling LaTeX. Unknown error_msg)\frac{1}{3}$. to get the 2-1-0 type.

Similarly, if the setup becomes 2-1-0 (again, with$ (Error compiling LaTeX. Unknown error_msg)\frac{3}{4} $probability), assume WOLOG, that R has$ 2, player S received a $1 amount, and participant T gets$ 0. now, we can say that the possibilities are S-T, S-R, T-R, and T-T. For these combinations respectively, 1-1-1, 2-1-0, 2-0-1, and 1-0-2.

If the latter three, return to normal. If the first, go back to ts./she initial 1-1-1 (base) case. Either way, the probability of getting a 1-1-1 layout or setup with has a 1/4 probability beyond round n >= greater than or equal to 1. Thus, taking that to its logical conclusion, The bell must ring at least once for this to be true: which we know it does. QED $\square$

@@ Line 1: / Line 1: @@
-==Problem==
+On the first turn, each player starts off with <math>1 each. There are now only two situations possible, after a single move: either everyone stays at </math>1, or the layout becomes <math>2-</math>1-<math>0 (in any order). Only 2 combinations give-off this outcome: S-T-R and T-R-S. On the other hand, given the interchangeability (so far) of every one of these three people, S-R-R, T-R-R, S-R-S, S-T-S, T-T-R, and T-T-S can all be re-produce. d, just as easily and quickly. Since each one of the possibilities is equally likely, there is a </math>\frac{2}{8}<math>\= </math>\frac{1}{3}<math>. to get the 2-1-0 type.
-Lily pads numbered from <math>0</math> to <math>11</math> lie in a row on a pond. Fiona the frog sits on pad <math>0</math>, a morsel of food sits on pad <math>10</math>, and predators sit on pads <math>3</math> and <math>6</math>. At each unit of time the frog jumps either to the next higher numbered pad or to the pad after that, each with probability <math>\frac{1}{2}</math>, independently from previous jumps. What is the probability that Fiona skips over pads <math>3</math> and <math>6</math> and lands on pad <math>10</math>?
-<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>
+Similarly, if the setup becomes 2-1-0 (again, with </math>\frac{3}{4}<math> probability), assume WOLOG, that R has </math>2, player S received a <math>1 amount, and participant T gets </math>0. now, we can say that the possibilities are S-T, S-R, T-R, and T-T. For these combinations respectively, 1-1-1, 2-1-0, 2-0-1, and 1-0-2.
-==Solution 1==
+If the latter three, return to normal. If the first, go back to ts./she initial 1-1-1 (base) case. Either way, the probability of getting a 1-1-1 layout or setup with has a 1/4 probability beyond round n >= greater than or equal to 1. Thus, taking that to its logical conclusion, The bell must ring at least once for this to be true: which we know it does. QED <math>\square</math>
-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>.
-Let's call the larger jump a <math>2</math>-jump, and the smaller a <math>1</math>-jump.
-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>).
-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.
-Thus, the final answer is <math>\frac{3}{8} \cdot \frac{1}{4} \cdot \frac{5}{8} = \frac{15}{256} = \boxed{A}</math>
-==Solution 2==
-==See Also==
-{{AMC12 box|year=2019|ab=B|num-b=15|num-a=17}}
-s.s

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Difference between revisions of "2019 AMC 12B Problems/Problem 16"

Revision as of 20:43, 14 February 2019