Difference between revisions of "2005 AMC 12A Problems/Problem 14"
m (→Solution: cdot) |
(→See also) |
||
Line 18: | Line 18: | ||
[[Category:Introductory Combinatorics Problems]] | [[Category:Introductory Combinatorics Problems]] | ||
+ | {{MAA Notice}} |
Revision as of 20:20, 3 July 2013
Problem
On a standard die one of the dots is removed at random with each dot equally likely to be chosen. The die is then rolled. What is the probability that the top face has an odd number of dots?
Solution
There are dots total. Casework:
- The dot is removed from an even face. There is a chance of this happening. Then there are 4 odd faces, giving us a probability of .
- The dot is removed from an odd face. There is a chance of this happening. Then there are 2 odd faces, giving us a probability of .
Thus the answer is .
See also
2005 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 13 |
Followed by Problem 15 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.