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

m (→‎Solution 2)
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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>. Our solution is <math>9+64=\boxed{073}</math>.
 
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>. Our solution is <math>9+64=\boxed{073}</math>.
  
 +
=== 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 <math>1+10+36+56+35+6=144</math>, and there are <math>1024</math> possible outcomes, so the probability is <math>\frac{144}{1024}=\frac{9}{64}</math>, so the answer is <math>\boxed{073}</math>.
 +
 +
\textbf{-RootThreeOverTwo}
 
== 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}}

Revision as of 00:37, 21 November 2018

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, 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

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 $\frac{144}{1024}=\frac{9}{64}$, so 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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