Difference between revisions of "2016 AIME I Problems/Problem 2"
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==Problem 2== | ==Problem 2== | ||
− | Two dice appear to be normal dice with their faces numbered from <math>1</math> to <math>6</math>, but each die is weighted so that the probability of rolling the number <math>k</math> is directly proportional to <math>k</math>. The probability of rolling a <math>7</math> with this pair of dice is <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>. | + | Two dice appear to be normal dice with their faces numbered from <math>1</math> to <math>6</math>, but each die is weighted so that the probability of rolling the number <math>k</math> is directly proportional to <math>k</math>. The probability of rolling a <math>7</math> with this pair of dice is <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m + n</math>. |
+ | |||
==Solution== | ==Solution== | ||
It is easier to think of the dice as <math>21</math> sided dice with <math>6</math> sixes, <math>5</math> fives, etc. Then there are <math>21^2=441</math> possible rolls. There are <math>2\cdot(1\cdot 6+2\cdot 5+3\cdot 4)=56</math> rolls that will result in a seven. The odds are therefore <math>\frac{56}{441}=\frac{8}{63}</math>. The answer is <math>8+63=\boxed{071}</math> | It is easier to think of the dice as <math>21</math> sided dice with <math>6</math> sixes, <math>5</math> fives, etc. Then there are <math>21^2=441</math> possible rolls. There are <math>2\cdot(1\cdot 6+2\cdot 5+3\cdot 4)=56</math> rolls that will result in a seven. The odds are therefore <math>\frac{56}{441}=\frac{8}{63}</math>. The answer is <math>8+63=\boxed{071}</math> | ||
See also [[2006 AMC 12B Problems/Problem 17]] | 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 <math>n</math> is <math>\frac{n}{21}</math> because <math>21=\frac{6 \cdot 7}{2}</math> | ||
+ | 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 <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 | ||
+ | |||
+ | ==Solution 3== | ||
+ | Since the probability of rolling a <math>1</math> is <math>\frac{1}{21}</math>, the probability of rolling a <math>2</math> is <math>\frac{2}{21}</math> the probability of rolling a <math>3</math> is <math>\frac{3}{21}</math> and so on, we can make a chart of probabilities and add them together. Note that we only need the probabilities of <math>1</math> and <math>6</math>, <math>2</math> and <math>5</math>, and <math>3</math> and <math>4</math>, and the rest is symmetry and the others are irrelevant. | ||
+ | |||
+ | We have: <math>2 \cdot (\frac{1}{21} \cdot \frac{2}{7}</math> <math>+</math> <math>\frac{2}{21} \cdot \frac{5}{21}</math> <math>+</math> <math>\frac{1}{7} \cdot \frac{4}{21})</math> <math>=</math> <math>2 \cdot (\frac{2}{147} + \frac{10}{441} + \frac{4}{147})</math> = <math>2 \cdot \frac{4}{63} = \frac{8}{63}</math>. Therefore, the answer is <math>8 + 63</math> = <math>\boxed{071}</math> | ||
+ | |||
+ | ~Arcticturn | ||
== 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}} |
Latest revision as of 14:49, 17 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
Since the probability of rolling a is
, the probability of rolling a
is
the probability of rolling a
is
and so on, we can make a chart of probabilities and add them together. Note that we only need the probabilities of
and
,
and
, and
and
, and the rest is symmetry and the others are irrelevant.
We have:
=
. Therefore, the answer is
=
~Arcticturn
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 |
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