Difference between revisions of "1990 AIME Problems/Problem 9"
m |
(solution) |
||
| Line 1: | Line 1: | ||
== Problem == | == Problem == | ||
| − | 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>. | + | 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>. |
== Solution == | == Solution == | ||
| − | {{ | + | 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. |
| + | |||
| + | Consider the case when <math>5</math> 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 <math>(H)</math>: | ||
| + | |||
| + | :<math>(H)\ T\ (H)\ T\ (H)\ T\ (H)\ T\ (H)\ T\ (H)</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>. | ||
== 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 22:20, 2 March 2007
Problem
A fair coin is to be tossed 
times. Let 
, in lowest terms, be the probability that heads never occur on consecutive tosses. Find 
.
Solution
Clearly, at least 
tails must be flipped; any less, then by the pigeonhole principle there will be heads that appear on consecutive tosses.
Consider the case when 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 
:
There are six slots for the heads to be placed, but only heads remaining. Thus, there are 
possible combinations of 5 heads. Continuing this pattern, we find that there are 
. There are a total of 
possible flips of 
coins, making the probability 
. Thus, our solution is 
.
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 | ||
