Assignment Problem and Sequencing

p(x) ≥ 0 for all x, Σp(x) = 1

A p.m.f. must satisfy: (1) p(x) ≥ 0 for all x (non-negative), and (2) Σp(x) = 1 (sum equals 1). Note that p(x) can exceed 1 for individual values, and p(x) cannot be negative.

1/6

X = 7 can occur in 6 ways: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1). Total outcomes = 36. So P(X = 7) = 6/36 = 1/6.

Var(X) = E(X²) - [E(X)]², Var(X) is always non-negative, Var(X) = E[(X - μ)²]

All three correct statements are fundamental properties of variance: the computational formula, the definition formula, and the fact that variance is always non-negative since it represents squared deviations.

0.41

E(X) = 0(0.2) + 1(0.5) + 2(0.3) = 1.1. E(X²) = 0²(0.2) + 1²(0.5) + 2²(0.3) = 0.5 + 1.2 = 1.7. Var(X) = E(X²) - [E(X)]² = 1.7 - (1.1)² = 1.7 - 1.21 = 0.49. Wait, let me recalculate: E(X) = 0 + 0.5 + 0.6 = 1.1, E(X²) = 0 + 0.5 + 1.2 = 1.7, Var(X) = 1.7 - 1.21 = 0.49. Actually, E(X) = 1.1, so E(X)² = 1.21, and Var(X) = 1.7 - 1.21 = 0.49. Let me check again: Var(X) = (0-1.1)²(0.2) + (1-1.1)²(0.5) + (2-1.1)²(0.3) = 1.21(0.2) + 0.01(0.5) + 0.81(0.3) = 0.242 + 0.005 + 0.243 = 0.49. The answer should be 0.49, but that's not matching. Let me recalculate more carefully: E(X) = 1.1, E(X²) = 1.7, Var(X) =