Difference between revisions of "2005 AMC 12A Problems/Problem 14"
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− | == 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]]: | ||
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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 == | ||
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[[Category:Introductory Combinatorics Problems]] | [[Category:Introductory Combinatorics Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 17:44, 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
.
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 |
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