Given the weighted voting system [52: 45, 43, 7, 5], which of the following is a minimal winning coalition?
$I \{ A , B \}$
II $\{ A , C \}$
III $\{ A , B , C \}$
A) I only
B) I and II
C) I, II, and III
D) none

Question

Given the weighted voting system [52: 45, 43, 7, 5], which of the following is a minimal winning coalition? 
$I \{ A , B \}$
II $\{ A , C \}$
III $\{ A , B , C \}$
A) I only
B) I and II
C) I, II, and III
D) none

Quizplus · Accepted Answer

In order for a coalition to be a winning coalition, their total weight must be greater than or equal to the quota, which is calculated as (total weight of all voters / 2) + 1. For this weighted voting system, the quota is (52+45+43+7+5)/2 + 1 = 77.

For coalition I, the total weight is 45 + 52 = 97, which is greater than the quota. Therefore, I is a winning coalition. However, it is not a minimal winning coalition because removing either A or B would make it lose.

For coalition II, the total weight is 45 + 43 = 88, which is greater than the quota. Therefore, II is a winning coalition. Furthermore, it is a minimal winning coalition because if either A or C is removed, the coalition would no longer be able to meet the quota.

For coalition III, the total weight is 45 + 52 + 43 = 140, which is greater than twice the quota (154). Therefore, III is not a minimal winning coalition because removing any one voter would still allow the coalition to meet the quota.

Given the Weighted Voting System [52: 45, 43, 7, 5] I{A,B}I \{ A , B \}I{A,B}

Given the Weighted Voting System [52: 45, 43, 7, 5] $I \{ A , B \}$