Maths · Probability: Probability of an event, addition and multiplication theorems of probability
Let and be two events such that =\frac{1}{6}, P(A \cap B)=\frac{1}{4} \) and =\f
Let \( A \) and \( B \) be two events such that \( P(\overline{A \cup B})=\frac{1}{6}, P(A \cap B)=\frac{1}{4} \) and \( P(\bar{A})=\frac{1}{4}, \) where \( \bar{A} \) stands for the complement of the event A. Then the events \( A \) and \( B \) are?
- A. Independent but not equally likely
- B. Independent and equally likely
- C. Mutually exclusive and independent
- D. Equally likely but not independent
Step-by-step solution
Given P(¬A)=1/4 ⇒ P(A)=3/4. P(¬(A∪B))=1/6 ⇒ P(A∪B)=5/6. Using inclusion-exclusion: P(A∪B)=P(A)+P(B)-P(A∩B) ⇒ 5/6=3/4+P(B)-1/4 ⇒ P(B)=1/3. Check independence: P(A)P(B)=(3/4)*(1/3)=1/4 = P(A∩B). So A and B are independent. But P(A)=3/4 ≠ P(B)=1/3, so not equally likely.
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