|
|
| (11 intermediate revisions by 5 users not shown) |
| Line 1: |
Line 1: |
| − | == Problem ==
| + | #REDIRECT [[2006 AIME I Problems/Problem 5]] |
| − | When rolling a certain unfair six-sided die with faces numbered 1, 2, 3, 4, 5, and 6, the probability of obtaining face <math> F </math> 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 <math> F </math> is <math> m/n, </math> where <math> m </math> and <math> n </math> are relatively prime positive integers, find <math> m+n. </math>
| |
| − | | |
| − | | |
| − | == Solution ==
| |
| − | For now, assume that face F has a 6 on it and 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 <math>A(2)*B(5)=\frac{1}{6}*\frac{1}{6}=\frac{1}{36}=\frac{8}{288}</math>. 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 <math>\frac{8}{288}</math>. Getting 7 with a 3 on die A and a 4 on die B also has a probability of <math>\frac{8}{288}</math>, as does getting a 7 with a 4 on die A and a 3 on die B. Subtracting all these probabilities from <math>\frac{47}{288}</math> leaves a <math>\frac{15}{288}=\frac{5}{96}</math> 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:
| |
| − | | |
| − | <math>A(6)*B(1)+B(6)*A(1)=\frac{5}{96}</math>
| |
| − | | |
| − | Since both die are the same, <math>B(1)=A(1)</math> and <math>B(6)=A(6)</math>:
| |
| − | | |
| − | <math>A(6)*A(1)+A(6)*A(1)=\frac{5}{96}</math>
| |
| − | | |
| − | <math>2*A(6)*A(1)=\frac{5}{96}</math>
| |
| − | | |
| − | <math>A(6)*A(1)=\frac{5}{192}</math>
| |
| − | | |
| − | But we know that <math>A(2)=A(3)=A(4)=A(5)=\frac{1}{6}</math> and that <math>\sum_{n=1}^6 A(n)=1</math>, so:
| |
| − | | |
| − | <math>A(1)+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+A(6)=\frac{6}{6}</math>
| |
| − | | |
| − | <math>A(1)+A(6)+\frac{4}{6}=\frac{6}{6}</math>
| |
| − | | |
| − | <math>A(6)+A(1)=\frac{1}{3}</math>
| |
| − | | |
| − | Now, combine the two equations:
| |
| − | | |
| − | <math>A(1)=\frac{1}{3}-A(6)</math>
| |
| − | | |
| − | <math>A(6)*(\frac{1}{3}-A(6))=\frac{5}{192}</math>
| |
| − | | |
| − | <math>\frac{A(6)}{3}-A(6)^2=\frac{5}{192}</math>
| |
| − | | |
| − | <math>A(6)^2-\frac{A(6)}{3}+\frac{5}{192}=0</math>
| |
| − | | |
| − | <math>\begin{eqnarray*}
| |
| − | A(6)=\frac{-\frac{-1}{3}+/-\sqrt{\frac{-1}{3}^2-4*1*\frac{5}{192}}}{2*1}
| |
| − | &=&\frac{5}{24}, /frac{1}{8}
| |
| − | \end{eqnarray*}</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>.
| |
| − | | |
| − | == See also ==
| |
| − | *[[2006 AIME II Problems]]
| |
| − | | |
| − | [[Category:Intermediate Combinatorics Problems]]
| |
