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junior college 2 | H2 Maths
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Date Posted: 3 years ago
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It is known that :
P(A∪B∪C) = P(A) + P(B) + P(C) - P(A∩B) - P(A∩C) - P(B∩C) + P(A∩B∩C)
(You can try to eliminate the overlaps to see why this is the case)
Since they are pairwise independent,
P(A∪B∪C) = P(A) + P(B) + P(C) - P(A)×P(B) - P(A)×P(C) - P(B)×P(C) + P(A∩B∩C)
0.71 = 0.5 + 0.3 + 0.2 - 0.5 × 0.3 - 0.5 × 0.2 - 0.3 × 0.2 + P(A∩B∩C)
0.71 = 1 - 0.15 - 0.1 - 0.06 + P(A∩B∩C)
0.71 = 0.69 + P(A∩B∩C)
P(A∩B∩C) = 0.02
But P(A) × P(B) × P(C) = 0.5 × 0.3 × 0.2 = 0.03
Since P(A∩B∩C) ≠ P(A) × P(B) × P(C), the three events are not independent
P(A∪B∪C) = P(A) + P(B) + P(C) - P(A∩B) - P(A∩C) - P(B∩C) + P(A∩B∩C)
(You can try to eliminate the overlaps to see why this is the case)
Since they are pairwise independent,
P(A∪B∪C) = P(A) + P(B) + P(C) - P(A)×P(B) - P(A)×P(C) - P(B)×P(C) + P(A∩B∩C)
0.71 = 0.5 + 0.3 + 0.2 - 0.5 × 0.3 - 0.5 × 0.2 - 0.3 × 0.2 + P(A∩B∩C)
0.71 = 1 - 0.15 - 0.1 - 0.06 + P(A∩B∩C)
0.71 = 0.69 + P(A∩B∩C)
P(A∩B∩C) = 0.02
But P(A) × P(B) × P(C) = 0.5 × 0.3 × 0.2 = 0.03
Since P(A∩B∩C) ≠ P(A) × P(B) × P(C), the three events are not independent
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