Why? For each clause,there are 8 possible assignments for these three variables,and only 1 fails. ▣ E.g.x1V x2 Vx3:only (x1,x2,x3)=(0,0,0)fails. E.g.V x2Vx3:only (x1,x2,x3)=(1,0,1)fails. Thus if you assign randomly,then with each clause fails with probability only 1/8. Thus the expected number of satisfied clauses is 7m/8. ▣ m:number of clauses 8Why? ◼ For each clause, there are 8 possible assignments for these three variables, and only 1 fails. ❑ E.g. 𝑥1 ∨ 𝑥2 ∨ 𝑥3: only 𝑥1, 𝑥2, 𝑥3 = (0,0,0) fails. ❑ E.g. 𝑥1 ∨ 𝑥2 ∨ 𝑥3 : only 𝑥1, 𝑥2, 𝑥3 = (1,0,1) fails. ◼ Thus if you assign randomly, then with each clause fails with probability only 1/8. ◼ Thus the expected number of satisfied clauses is 7𝑚/8. ❑ 𝑚: number of clauses 8