Difference between revisions of "2012 UNCO Math Contest II Problems/Problem 8"
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== Solution == | == Solution == | ||
| − | + | The probability that a six is rolled on any given roll is <math>1/6</math>, so on average it will take <math>6</math> rolls to roll a six, meaning there will be an average of five rolls before the first <math>6</math>. Since these <math>5</math> rolls could be any of <math>1,2,3,4,</math> or <math>5</math>, their average value is <math>3</math>, so the answer is <math>3 \cdot 5=15</math> | |
== See Also == | == See Also == | ||
| − | {{ | + | {{UNCO Math Contest box|n=II|year=2012|num-b=7|num-a=9}} |
[[Category:Intermediate Combinatorics Problems]] | [[Category:Intermediate Combinatorics Problems]] | ||
Latest revision as of 14:09, 1 August 2020
Problem
An ordinary fair die is tossed repeatedly until the face with six dots appears on top. On
average, what is the sum of the numbers that appear on top before the six? For example, if
the numbers 
are the numbers that appear, then the sum of the numbers before the
six appears is 
. Do not include the 
in the sum.
Solution
The probability that a six is rolled on any given roll is 
, so on average it will take rolls to roll a six, meaning there will be an average of five rolls before the first . Since these 
rolls could be any of 
or , their average value is 
, so the answer is 
See Also
| 2012 UNCO Math Contest II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 7 |
Followed by Problem 9 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 | ||
| All UNCO Math Contest Problems and Solutions | ||
