Correct : 0.81

The batch is accepted only if none of the 4 randomly chosen bulbs is defective. This is a hypergeometric distribution problem — sampling without replacement from a finite population.
Total bulbs N = 100, defective K = 5, non-defective = 95, sample size n = 4.
P(0 defectives) = C(5,0) × C(95,4) / C(100,4)
C(5,0) = 1 (choosing 0 from the 5 defective bulbs)
C(95,4) = (95 × 94 × 93 × 92) / 24 = 3,183,545
C(100,4) = (100 × 99 × 98 × 97) / 24 = 3,921,225
P = 3,183,545 / 3,921,225 = 0.8119 ≈ 0.81
The intuition is clear — with only 5 defectives out of 100, there is a reasonably high chance that a random sample of 4 misses all of them. If we had used the binomial approximation (treating each pick as independent with p = 5/100 = 0.05): P ≈ (0.95)⁴ = 0.8145 — very close to the exact hypergeometric answer of 0.81, confirming the result.
Correct answer: 0.81 ✓