i:x̄i∈Cj (1− yi) ≥ zj for all j 0 ≤ yi ≤ 1 for all i 0 ≤ zj ≤ 1 for all j. We were investigating how well the algorithm does, that sets xi to true with probability yi, independently of the other xi’s, where y is an optimal solution to the lp-relaxation. Without loss of generality, consider the clause Cj = (x1 ∨ x2 ∨ · · · ∨ xk): P (Cj is not satisfied) = P ( x1, x2, . . . , xk are not satisfied)...