Difference between revisions of "2006 AIME A Problems/Problem 5"

Revision as of 15:31, 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 $F$ 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 $F$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers, find $m+n.$

Solution

For now, assume that face $F$ has a 6 on it so that the face opposite $F$ has a 1 on it. Let $A(n)$ be the probability of rolling a number $n$ on one die and let $B(n)$ be the probability of rolling a number $n$ on the other die. One way of getting a 7 is to get a 2 on die $A$ and a 5 on die $B$ . The probability of this happening is $A(2)\cdot B(5)=\frac{1}{6} \cdot \frac{1}{6}=\frac{1}{36}=\frac{8}{288}$ . Conversely, one can get a 7 by getting a 2 on die $B$ and a 5 on die $A$ , the probability of which is also $\frac{8}{288}$ . Getting 7 with a 3 on die $A$ and a 4 on die $B$ also has a probability of $\frac{8}{288}$ , as does getting a 7 with a 4 on die $A$ and a 3 on die $B$ . Subtracting all these probabilities from $\frac{47}{288}$ leaves a $\frac{15}{288}=\frac{5}{96}$ chance of getting a 1 on die $A$ and a 6 on die $B$ or a 6 on die $A$ and a 1 on die $B$ :

$A(6)\cdot B(1)+B(6)\cdot A(1)=\frac{5}{96}$

Since the two dice are identical, $B(1)=A(1)$ and $B(6)=A(6)$ so

$A(1)\cdot A(6)+A(1)\cdot A(6)=\frac{5}{96}$

$2\cdot A(1)\cdot A(6)=\frac{5}{96}$

$A(1)\cdot A(6)=\frac{5}{192}$

But we know that $A(2)=A(3)=A(4)=A(5)=\frac{1}{6}$ and that $\sum_{n=1}^6 A(n)=1$ , so:

$A(1)+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+A(6)=\frac{6}{6}$

$A(1)+A(6)+\frac{4}{6}=\frac{6}{6}$

$A(1)+A(6)=\frac{2}{6}$

$A(1)+A(6)=\frac{1}{3}$

Now, combine the two equations:

$A(1)=\frac{1}{3}-A(6)$

$A(6)\cdot(\frac{1}{3}-A(6))=\frac{5}{192}$

$\frac{A(6)}{3}-A(6)^2=\frac{5}{192}$

$A(6)^2-\frac{A(6)}{3}+\frac{5}{192}=0$

$A(6)=\frac{-\frac{-1}{3}\pm\sqrt{\frac{-1}{3}^2-4\cdot1\cdot\frac{5}{192}}}{2\cdot1}$

$A(6)=\frac{\frac{1}{3}\pm\sqrt{\frac{1}{9}-\frac{5}{48}}}{2}$

$A(6)=\frac{\frac{1}{3}\pm\sqrt{\frac{16}{144}-\frac{15}{144}}}{2}$

$A(6)=\frac{\frac{1}{3}\pm\sqrt{\frac{1}{144}}}{2}$

$A(6)=\frac{\frac{1}{3}\pm\frac{1}{12}}{2}$

$A(6)=\frac{\frac{4}{12}\pm\frac{1}{12}}{2}$

$A(6)=\frac{\frac{5}{12}}{2}, \frac{\frac{3}{12}}{2}$

$A(6)=\frac{5}{24}, \frac{3}{24}$

$A(6)=\frac{5}{24}, \frac{1}{8}$

We know that $A(6)>\frac{1}{6}$ , so it can't be $\frac{1}{8}$ . Therefore, it has to be $\frac{5}{24}$ and the answer is $5+24=29$ .

Note also that the initial assumption that face $F$ 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.

@@ Line 55: / Line 55: @@
 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>.
+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.
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Difference between revisions of "2006 AIME A Problems/Problem 5"

Revision as of 15:31, 2 August 2006

Problem

Solution

See also