Difference between revisions of "2016 AIME I Problems/Problem 2"
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The probability that 7 is rolled is now <math>2(\frac{1}{21} \cdot \frac{6}{21}+\frac{2}{21} \cdot \frac{5}{21} + \frac{3}{21} \cdot \frac{4}{21})</math> which is equal to <math>\frac{56}{441}=\frac{8}{63}</math>. Therefore the answer is <math>8+63=\boxed{071}</math> | The probability that 7 is rolled is now <math>2(\frac{1}{21} \cdot \frac{6}{21}+\frac{2}{21} \cdot \frac{5}{21} + \frac{3}{21} \cdot \frac{4}{21})</math> which is equal to <math>\frac{56}{441}=\frac{8}{63}</math>. Therefore the answer is <math>8+63=\boxed{071}</math> | ||
~PEKKA | ~PEKKA | ||
| + | |||
| + | ==Solution 3== | ||
== See also == | == See also == | ||
{{AIME box|year=2016|n=I|num-b=1|num-a=3}} | {{AIME box|year=2016|n=I|num-b=1|num-a=3}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 17:24, 15 October 2021
Problem 2
Two dice appear to be normal dice with their faces numbered from 
to 
, but each die is weighted so that the probability of rolling the number 
is directly proportional to . The probability of rolling a 
with this pair of dice is 
, where 
and 
are relatively prime positive integers. Find 
.
Solution
It is easier to think of the dice as 
sided dice with sixes, 
fives, etc. Then there are 
possible rolls. There are 
rolls that will result in a seven. The odds are therefore 
. The answer is 
See also 2006 AMC 12B Problems/Problem 17
Solution 2
Since the probability of rolling any number is 1, and the problem tells us the dice are unfair, we can assign probabilities to the individual faces. The probability of rolling is 
because 
Next, we notice that 7 can be rolled by getting individual results of 1 and 6, 2 and 5, or 3 and 4 on the separate dice.
The probability that 7 is rolled is now 
which is equal to . Therefore the answer is
~PEKKA
Solution 3
See also
| 2016 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
