CA Foundation · Quantitative Aptitude · Probability
A, B and C are three events with P(A) = 0.5, P(B) = 0.4, P(C) = 0.3, P(A∩B) = 0.2, P(B∩C) = 0.1, P(A∩C) = 0.15 and P(A∩B∩C) = 0.05. What is P(A ∪ B ∪ C)?
The probability is 0.80. Sum the single probabilities to get 1.2, subtract the three pairwise intersections totalling 0.45, then add back the triple intersection 0.05, giving 0.80.
- A0.75Correct
- B0.80
- C0.70
- D1.20
Explanation
P(A∪B∪C) = 0.5+0.4+0.3 - (0.2+0.1+0.15) + 0.05 = 1.2 - 0.45 + 0.05 = 0.80. Option 0.75 omits adding back the triple intersection (0.75 would result from 1.2-0.45).
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