Given the voting system {17:14, 5, 2}
The coalitions, their weight, critical player and if they win or loose will be presented below
From the table above, the only coalition cases we have win are {14+5} and {14+5+2}, Player with weight 14 is critical two times, while playerwith weights 5 is critical twice and player with weight 2 is never critical, thus the following extraction can be made
[tex]\begin{gathered} T\Rightarrow\text{The total number of times players are critical = }4 \\ B_{14}\Rightarrow\text{ the number times player 14 is critical= 2} \\ B_5\Rightarrow\text{ the number times player 5 is critical= }2 \\ B_2\Rightarrow\text{the number times player 2 is critical= }0 \end{gathered}[/tex]
Given the formula for the Banzhaf power index for a player to be;
[tex]P_i=\frac{Bi}{T},\Rightarrow\begin{cases}P_i\Rightarrow\text{ player index} \\ B_i\Rightarrow\text{ the number times player is critical} \\ T\Rightarrow\text{ the total number times players are is critical}\end{cases}[/tex]
Thus to calculate the Banzhaf for each player we have
[tex]\begin{gathered} P_{14}=\beta_1=\frac{B_{14}}{T}=\frac{2}{4}=\frac{1}{2} \\ P_5=\beta_2=\frac{B_5}{T}=\frac{2}{4}=\frac{1}{2} \\ P_2=\beta_3=\frac{B_2}{T}=\frac{0}{4}=0 \end{gathered}[/tex]
Hence, the Banzhaf power distributions are
[tex]\beta_1=\frac{1}{2},\beta_2=\frac{1}{2},\beta_3=0[/tex]
b) Given the voting system {19: 16, 9, 1}
The winning coalitions and their respective critical players is given below
[tex]\begin{gathered} \mleft\lbrace16,9\mright\rbrace\Rightarrow critical\text{ player (16,9)} \\ \mleft\lbrace16,9,1\mright\rbrace\Rightarrow\text{ critical players (16,9)} \end{gathered}[/tex]
From the table above, the only coalition cases we have win are {16,9} and {16,9,1}, Player with weight 16 is critical two times, while playerwith weights 9 is critical twice and player with weight 1 is never critical, thus the following extraction can be made
[tex]\begin{gathered} T=2+2+0=4 \\ B_{16}=2 \\ B_9=2 \\ B_1=0 \end{gathered}[/tex]
The Banzhaf distribution fo the voting sysytem is
[tex]\begin{gathered} \beta_1=\frac{B_{16}}{T}=\frac{2}{4}=\frac{1}{2} \\ \beta_2=\frac{B_9}{T}=\frac{2}{4}=\frac{1}{2} \\ \beta_3=\frac{B_1}{T}=\frac{0}{4}=0 \end{gathered}[/tex]
Hence, the Banzhaf distribution power is
[tex]\beta_1=\frac{1}{2},\beta_2=\frac{1}{2},\beta_3=0[/tex]
Conclusion, the Banzhaf power distributions are the same in both (a) and (b),
Option D is right answer
