Difference between revisions of "1995 AIME Problems/Problem 15"
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== Problem == | == Problem == | ||
| − | Let <math> | + | Let <math>p_{}</math> be the [[probability]] that, in the process of repeatedly flipping a fair coin, one will encounter a run of <math>5</math> heads before one encounters a run of <math>2</math> tails. Given that <math>p_{}</math> can be written in the form <math>m/n</math> where <math>m_{}</math> and <math>n_{}</math> are relatively prime positive integers, find <math>m+n</math>. |
| + | __TOC__ | ||
== Solution == | == Solution == | ||
| − | Think of the problem as a sequence of H's and T's. No two T's can occur in a row, so the sequence is blocks of 1 to 4 H's separated by T's and ending in 5 H's. Since the first letter could be T or the sequence could start with a block of H's, the total probability is 3/2 that of it had to start with and H. The answer to the problem is then the sum of all numbers of the form <math> | + | === Solution 1 === |
| + | Think of the problem as a sequence of <tt>H</tt>'s and <tt>T</tt>'s. No two <tt>T</tt>'s can occur in a row, so the sequence is blocks of <math>1</math> to <math>4</math> <tt>H</tt>'s separated by <tt>T</tt>'s and ending in <math>5</math> <tt>H</tt>'s. Since the first letter could be <tt>T</tt> or the sequence could start with a block of <tt>H</tt>'s, the total probability is <math>3/2</math> that of it had to start with and <tt>H</tt>. | ||
| + | |||
| + | The answer to the problem is then the sum of all numbers of the form <math>\frac 32 \left( \frac 1{2^a} \cdot \frac 12 \cdot \frac 1{2^b} \cdot \frac 12 \cdot \frac 1{2^c} \cdots \right) \cdot \left(\frac 12\right)^5</math>, where <math>a,b,c \ldots </math> are all numbers <math>1-4</math>, since the blocks of <tt>H</tt>'s can range from <math>1-4</math> in length. The sum of all numbers of the form <math>(1/2)^a</math> is <math>1/2+1/4+1/8+1/16=15/16</math>, so if there are n blocks of <tt>H</tt>'s before the final five <tt>H</tt>'s, the answer can be rewritten as the sum of all numbers of the form <math>\frac 32\left( \left(\frac {15}{16}\right)^n \cdot \left(\frac 12\right)^n \right) \cdot \left(\frac 1{32}\right)=\frac 3{64}\left(\frac{15}{32}\right)^n</math>, where <math>n</math> ranges from <math>0</math> to <math>\infty</math>, since that's how many blocks of <tt>H</tt>'s there can be before the final five. This is an infinite geometric series whose sum is <math>\frac{3/64}{1-(15/32)}=\frac{3}{34}</math>, so the answer is <math>\boxed{037}</math>. | ||
| + | |||
| + | === Solution 2 === | ||
| + | Let <math>p_H, p_T</math> respectively denote the probabilities that a string of <tt>H</tt>'s and <tt>T</tt>'s are successful. A successful string can either start with <tt>H</tt>, or it can start with <tt>T</tt> and then continue with a string starting with <tt>H</tt> (as there cannot be <math>2</math> <tt>T</tt>'s in a row). Thus | ||
| + | |||
| + | <center><math>p_T = \frac 12p_H.</math></center> | ||
| + | |||
| + | A successful string starting with <tt>H</tt> must start with a block of <math>1,2,3,4</math> <tt>H</tt>'s, then a <tt>T</tt>, then a successful string starting with <tt>H</tt>, or reach a block of <math>5</math> <tt>H</tt>'s. Thus, | ||
| + | |||
| + | <center><math>p_H = \left(\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16}\right) \cdot \left(\frac 12\right) p_H + \frac{1}{32} \Longrightarrow p_H = \frac{1}{17}.</math></center> | ||
| + | |||
| + | The answer is <math>p_H + p_T = \frac{3}{2}p_H = \frac{3}{34}</math>, and <math>m+n = \boxed{037}</math>. | ||
== See also == | == See also == | ||
| − | + | {{AIME box|year=1995|num-b=14|after=Last question}} | |
| − | + | ||
| + | [[Category:Intermediate Combinatorics Problems]] | ||
Revision as of 21:25, 11 August 2008
Problem
Let 
be the probability that, in the process of repeatedly flipping a fair coin, one will encounter a run of 
heads before one encounters a run of 
tails. Given that can be written in the form 
where 
and 
are relatively prime positive integers, find 
.
Solution
Solution 1
Think of the problem as a sequence of H's and T's. No two T's can occur in a row, so the sequence is blocks of 
to 
H's separated by T's and ending in H's. Since the first letter could be T or the sequence could start with a block of H's, the total probability is 
that of it had to start with and H.
The answer to the problem is then the sum of all numbers of the form 
, where 
are all numbers 
, since the blocks of H's can range from in length. The sum of all numbers of the form 
is 
, so if there are n blocks of H's before the final five H's, the answer can be rewritten as the sum of all numbers of the form 
, where 
ranges from 
to 
, since that's how many blocks of H's there can be before the final five. This is an infinite geometric series whose sum is 
, so the answer is 
.
Solution 2
Let 
respectively denote the probabilities that a string of H's and T's are successful. A successful string can either start with H, or it can start with T and then continue with a string starting with H (as there cannot be T's in a row). Thus
A successful string starting with H must start with a block of 
H's, then a T, then a successful string starting with H, or reach a block of H's. Thus,
The answer is 
, and 
.
See also
| 1995 AIME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 14 |
Followed by Last question | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
