Difference between revisions of "2006 AIME A Problems/Problem 5"
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We know that <math>A(6)>\frac{1}{6}</math>, so it can't be <math>\frac{1}{8}</math>. Therefore, it has to be <math>\frac{5}{24}</math> and the answer is <math>5+24=29</math>. | We know that <math>A(6)>\frac{1}{6}</math>, so it can't be <math>\frac{1}{8}</math>. Therefore, it has to be <math>\frac{5}{24}</math> and the answer is <math>5+24=29</math>. | ||
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+ | Note also that the initial assumption that face <math>F</math> was the face labelled 6 is unnecessary -- we would have carried out exactly the same steps and found exactly the same probability no matter which face it was. | ||
== See also == | == See also == |
Revision as of 16:31, 2 August 2006
Problem
When rolling a certain unfair six-sided die with faces numbered 1, 2, 3, 4, 5, and 6, the probability of obtaining face is greater than 1/6, the probability of obtaining the face opposite is less than 1/6, the probability of obtaining any one of the other four faces is 1/6, and the sum of the numbers on opposite faces is 7. When two such dice are rolled, the probability of obtaining a sum of 7 is 47/288. Given that the probability of obtaining face
is
where
and
are relatively prime positive integers, find
Solution
For now, assume that face has a 6 on it so that the face opposite
has a 1 on it. Let
be the probability of rolling a number
on one die and let
be the probability of rolling a number
on the other die. One way of getting a 7 is to get a 2 on die
and a 5 on die
. The probability of this happening is
. Conversely, one can get a 7 by getting a 2 on die
and a 5 on die
, the probability of which is also
. Getting 7 with a 3 on die
and a 4 on die
also has a probability of
, as does getting a 7 with a 4 on die
and a 3 on die
. Subtracting all these probabilities from
leaves a
chance of getting a 1 on die
and a 6 on die
or a 6 on die
and a 1 on die
:
Since the two dice are identical, and
so
But we know that and that
, so:
Now, combine the two equations:
We know that , so it can't be
. Therefore, it has to be
and the answer is
.
Note also that the initial assumption that face was the face labelled 6 is unnecessary -- we would have carried out exactly the same steps and found exactly the same probability no matter which face it was.