CMA Foundation · Fundamentals of Business Mathematics and Statistics · Probability
For three events A, B and C: P(A)=0.5, P(B)=0.4, P(C)=0.3, P(A and B)=0.2, P(B and C)=0.1, P(A and C)=0.15 and P(A and B and C)=0.05. What is the probability that at least one of the events occurs?
The probability is 0.80. Apply the three-event addition theorem: add the single probabilities (1.2), subtract the pairwise intersections (0.45) and add back the triple intersection (0.05), giving 1.2 - 0.45 + 0.05 = 0.80. Omitting the triple term would give 0.75.
- A0.75Correct
- B0.80
- C0.90
- D0.85
Explanation
P(A or B or C) = sum of singles - sum of pairs + triple = (0.5+0.4+0.3) - (0.2+0.1+0.15) + 0.05 = 1.2 - 0.45 + 0.05 = 0.80. Wait, recompute: 1.2 - 0.45 = 0.75; 0.75 + 0.05 = 0.80. Hence the answer is 0.80. Option 0.75 omits adding back the triple intersection.
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