Difference between revisions of "2005 AMC 12A Problems/Problem 14"
(solution) |
Willy132423 (talk | contribs) m (→Solution 2 (Alcumus)) |
||
| (5 intermediate revisions by 5 users not shown) | |||
| Line 6: | Line 6: | ||
</math> | </math> | ||
| − | == Solution == | + | == Solutions == |
| + | === Solution 1 === | ||
There are <math>1 + 2 + 3 + 4 + 5 + 6 = 21</math> dots total. [[Casework]]: | There are <math>1 + 2 + 3 + 4 + 5 + 6 = 21</math> dots total. [[Casework]]: | ||
| − | *The dot is removed from an even face. There is a <math>\frac{2+4+6}{21} = \frac{4}{7}</math> chance of this happening. Then there are 4 odd faces, giving us a probability of <math>\frac 47\frac 46 = \frac{8}{21}</math>. | + | *The dot is removed from an even face. There is a <math>\frac{2+4+6}{21} = \frac{4}{7}</math> chance of this happening. Then there are 4 odd faces, giving us a probability of <math>\frac 47 \cdot \frac 46 = \frac{8}{21}</math>. |
| − | *The dot is removed from an odd face. There is a <math>\frac{1+3+5}{21} = \frac{3}{7}</math> chance of this happening. Then there are 2 odd faces, giving us a probability of <math>\frac 37\frac 26 = \frac{1}{7}</math>. | + | *The dot is removed from an odd face. There is a <math>\frac{1+3+5}{21} = \frac{3}{7}</math> chance of this happening. Then there are 2 odd faces, giving us a probability of <math>\frac 37 \cdot \frac 26 = \frac{1}{7}</math>. |
Thus the answer is <math>\frac 8{21} + \frac 17 = \frac{11}{21} \Longrightarrow \mathrm{(D)}</math>. | Thus the answer is <math>\frac 8{21} + \frac 17 = \frac{11}{21} \Longrightarrow \mathrm{(D)}</math>. | ||
| + | |||
| + | === Solution 2 (Alcumus) === | ||
| + | The probability that the top face is odd is <math>1/3</math> if a dot is removed from an odd face, and the probability that the top face is odd is <math>2/3</math> if a dot is removed from an even face. Because each dot has the probability <math>1/21</math> of being removed, the top face is odd with probability<math>\left(\frac{1}{3}\right)\left(\frac{1+3+5}{21}\right) +\left(\frac{2}{3}\right)\left(\frac{2+4+6}{21}\right) = \frac{33}{63} = \boxed{\frac{11}{21}}</math>. | ||
== See also == | == See also == | ||
| Line 18: | Line 22: | ||
[[Category:Introductory Combinatorics Problems]] | [[Category:Introductory Combinatorics Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 16:44, 3 July 2024
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?
Solutions
Solution 1
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 
.
chance of this happening. Then there are 4 odd faces, giving us a probability of 
.
chance of this happening. Then there are 2 odd faces, giving us a probability of 
.Thus the answer is 
.
Solution 2 (Alcumus)
The probability that the top face is odd is 
if a dot is removed from an odd face, and the probability that the top face is odd is 
if a dot is removed from an even face. Because each dot has the probability 
of being removed, the top face is odd with probability
.
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. 
