alleryAt least one '6' in 4 throws of a fair die:
A4/6
B2/6
C1 - (5/6)⁴ ≈ 0.518
D2/3
Answer & Solution
Correct answer: C. 1 - (5/6)⁴ ≈ 0.518
Complementary: P(no 6 in 4) = (5/6)⁴ = 625/1296. P(at least one 6) = 1 - 625/1296 = 671/1296 ≈ 0.518.
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