Answer:
[tex]p2=0.667\\p1=0.167\\p3=0.167\\p=1[/tex]
Explanation:
Let's start writing the sample space for this exercise :
Let be ''M'' an abbreviation for Macrostate
Ω = { M1 , M2 , M3 }
Let be P(M1) the probability of Macrostate 1.
Reading the exercise, we know that ⇒
[tex]P(M1)=P(M3)[/tex]
Let's note this probability as ''p''.
[tex]P(M1)=P(M3)=p[/tex]
Macrostate 2 is four times more likely to occur than either of the other two macrostates ⇒
[tex]P(M2)=4p[/tex]
The sum of all probabilities must be equal to 1 for this sample space.Therefore,
[tex]p+p+4p=1[/tex]
[tex]6p=1[/tex]
[tex]p=\frac{1}{6}[/tex]
Finally :
[tex]P(M1)=\frac{1}{6}[/tex]
[tex]P(M2)=\frac{4}{6}[/tex]
[tex]P(M3)=\frac{1}{6}[/tex]
For Part A :
[tex]p2=\frac{4}{6}=0.667[/tex]
For Part B and C :
[tex]p1=p3=0.167[/tex]
For Part D :
The sum of the probabilities for all macrostates is equal to 1 :
[tex]p=p1+p2+p3=\frac{1}{6}+\frac{4}{6}+\frac{1}{6}=\frac{6}{6}=1[/tex]
