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

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== Solution ==
 
== Solution ==
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=== Solution 1 ===
 
Clearly, at least <math>5</math> tails must be flipped; any less, then by the [[pigeonhole principle]] there will be heads that appear on consecutive tosses.  
 
Clearly, at least <math>5</math> tails must be flipped; any less, then by the [[pigeonhole principle]] there will be heads that appear on consecutive tosses.  
  
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There are six slots for the heads to be placed, but only <math>5</math> heads remaining. Thus, there are <math>{6\choose5}</math> possible combinations of 5 heads. Continuing this pattern, we find that there are <math>\displaystyle\sum_{i=6}^{11} {i\choose{11-i}} = {6\choose5} + {7\choose4} + {8\choose3} + {9\choose2} + {{10}\choose1} + {{11}\choose0} = 6 + 35 + 56 + 36 + 10 + 1 = 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 = 73</math>.  
 
There are six slots for the heads to be placed, but only <math>5</math> heads remaining. Thus, there are <math>{6\choose5}</math> possible combinations of 5 heads. Continuing this pattern, we find that there are <math>\displaystyle\sum_{i=6}^{11} {i\choose{11-i}} = {6\choose5} + {7\choose4} + {8\choose3} + {9\choose2} + {{10}\choose1} + {{11}\choose0} = 6 + 35 + 56 + 36 + 10 + 1 = 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 = 73</math>.  
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 +
=== Solution 2 ===
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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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 +
If the first coin flipped is a T, then the remaining <math>n-1</math> flips must fall under one of the configurations of <math>S_{n-1}</math>.
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If the first coin flipped is a H, then the second coin must be a T. There are then <math>S_{n-2}</math> configurations.
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Thus, <math>S_n = S_{n-1} + S_{n-2}</math>. By counting, we can establish that <math>S_1 = 2</math> and <math>S_2 = 3</math>. Therefore, <math>S_3 = 5,\ S_4 = 8</math>, forming the [[Fibonacci sequence]]. Listing them out, we get <math>2,3,5,8,13,21,34,55,89,144</math>, and the 10th number is <math>144</math>. Putting this over <math>2^{10}</math> to find the probability, we get <math>\frac{9}{64}</math>.
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=1990|num-b=8|num-a=10}}
 
{{AIME box|year=1990|num-b=8|num-a=10}}

Revision as of 09:33, 3 March 2007

Problem

A fair coin is to be tossed $10_{}^{}$ times. Let $i/j^{}_{}$, 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, there are ${6\choose5}$ possible combinations of 5 heads. Continuing this pattern, we find that there are $\displaystyle\sum_{i=6}^{11} {i\choose{11-i}} = {6\choose5} + {7\choose4} + {8\choose3} + {9\choose2} + {{10}\choose1} + {{11}\choose0} = 6 + 35 + 56 + 36 + 10 + 1 = 144$. 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 $9 + 64 = 73$.

Solution 2

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 $\frac{9}{64}$.

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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