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Difference between revisions of "1990 AIME Problems/Problem 9"

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== Problem ==
 
== Problem ==
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A [[fair]] coin is to be tossed <math>10_{}^{}</math> times. Let <math>i/j^{}_{}</math>, in lowest terms, be the [[probability]] that heads never occur on consecutive tosses. Find <math>i+j_{}^{}</math>.  
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A [[fair]] coin is to be tossed <math>10_{}^{}</math> times. Let <math>\frac{i}{j}^{}_{}</math>, in lowest terms, be the [[probability]] that heads never occur on consecutive tosses. Find <math>i+j_{}^{}</math>.  
  
 
__TOC__
 
__TOC__
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== Solution ==
 
== Solution ==
 
=== Solution 1 ===
 
=== Solution 1 ===
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:<math>(H)\ T\ (H)\ T\ (H)\ T\ (H)\ T\ (H)\ T\ (H)</math>
 
:<math>(H)\ T\ (H)\ T\ (H)\ T\ (H)\ T\ (H)\ T\ (H)</math>
  
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There are six slots for the heads to be placed, but only <math>6</math> heads remaining. Thus, using [[stars-and-bars]] there are <math>{6\choose5}</math> possible combinations of 5 heads. Continuing this pattern, we find that there are <math>\sum_{i=6}^{11} {i\choose{11-i}} = {6\choose5} + {7\choose4} + {8\choose3} + {9\choose2} + {{10}\choose1} + {{11}\choose0} = 144</math>. There are a total of <math>2^{10}</math> possible flips of <math>10</math> coins, making the probability <math>\frac{144}{1024} = \frac{9}{64}</math>. Thus, our solution is <math>9 + 64 = \boxed{073}</math>.
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There are six slots for the heads to be placed, but only <math>5</math> heads remaining. Thus, using [[stars-and-bars]] there are <math>{6\choose5}</math> possible combinations of <math>5</math> heads. Continuing this pattern, we find that there are <math>\sum_{i=6}^{11} {i\choose{11-i}} = {6\choose5} + {7\choose4} + {8\choose3} + {9\choose2} + {{10}\choose1} + {{11}\choose0} = 144</math>. There are a total of <math>2^{10}</math> possible flips of <math>10</math> coins, making the probability <math>\frac{144}{1024} = \frac{9}{64}</math>. Thus, our solution is <math>9 + 64 = \boxed{073}</math>.
  
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=== Solution 2 ===
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=== Solution 2 (Recursion)===
 
Call the number of ways of flipping <math>n</math> coins and not receiving any consecutive heads <math>S_n</math>. Notice that tails must be received in at least one of the first two flips.
 
Call the number of ways of flipping <math>n</math> coins and not receiving any consecutive heads <math>S_n</math>. Notice that tails must be received in at least one of the first two flips.
  
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<math>\textbf{-RootThreeOverTwo}</math>
 
<math>\textbf{-RootThreeOverTwo}</math>
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== See also ==
 
== See also ==
 
{{AIME box|year=1990|num-b=8|num-a=10}}
 
{{AIME box|year=1990|num-b=8|num-a=10}}
 
{{MAA Notice}}
 
{{MAA Notice}}
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 +
[[Category:Intermediate Combinatorics Problems]]

Latest revision as of 14:03, 24 June 2024

Problem

A fair coin is to be tossed $10_{}^{}$ times. Let $\frac{i}{j}^{}_{}$, in lowest terms, be the probability that heads never occur on consecutive tosses. Find $i+j_{}^{}$.

Solution

Solution 1

Clearly, at least $5$ tails must be flipped; any less, then by the Pigeonhole Principle there will be heads that appear on consecutive tosses.

Consider the case when $5$ tails occur. The heads must fall between the tails such that no two heads fall between the same tails, and must fall in the positions labeled $(H)$:

$(H)\ T\ (H)\ T\ (H)\ T\ (H)\ T\ (H)\ T\ (H)$

There are six slots for the heads to be placed, but only $5$ heads remaining. Thus, using stars-and-bars there are ${6\choose5}$ possible combinations of $5$ heads. Continuing this pattern, we find that there are $\sum_{i=6}^{11} {i\choose{11-i}} = {6\choose5} + {7\choose4} + {8\choose3} + {9\choose2} + {{10}\choose1} + {{11}\choose0} = 144$. There are a total of $2^{10}$ possible flips of $10$ coins, making the probability $\frac{144}{1024} = \frac{9}{64}$. Thus, our solution is $9 + 64 = \boxed{073}$.

Solution 2 (Recursion)

Call the number of ways of flipping $n$ coins and not receiving any consecutive heads $S_n$. Notice that tails must be received in at least one of the first two flips.

If the first coin flipped is a T, then the remaining $n-1$ flips must fall under one of the configurations of $S_{n-1}$.

If the first coin flipped is a H, then the second coin must be a T. There are then $S_{n-2}$ configurations.

Thus, $S_n = S_{n-1} + S_{n-2}$. By counting, we can establish that $S_1 = 2$ and $S_2 = 3$. Therefore, $S_3 = 5,\ S_4 = 8$, forming the Fibonacci sequence. Listing them out, we get $2,3,5,8,13,21,34,55,89,144$, and the 10th number is $144$. Putting this over $2^{10}$ to find the probability, we get $\frac{9}{64}$. Our solution is $9+64=\boxed{073}$.

Solution 3

We can also split the problem into casework.

Case 1: 0 Heads

There is only one possibility.

Case 2: 1 Head

There are 10 possibilities.

Case 3: 2 Heads

There are 36 possibilities.

Case 4: 3 Heads

There are 56 possibilities.

Case 5: 4 Heads

There are 35 possibilities.

Case 6: 5 Heads

There are 6 possibilities.

We have $1+10+36+56+35+6=144$, and there are $1024$ possible outcomes, so the probability is $\frac{144}{1024}=\frac{9}{64}$, and the answer is $\boxed{073}$.

$\textbf{-RootThreeOverTwo}$

See also

1990 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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