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 \( P(\bar{A})=\frac{1}{4}, \) then events \( A \) and \( B \) are
- A. equally likely, but not independent
- B. equally likely and mutually exclusive
- C. mutually exclusive and independent
- D. independent but not equally likely
Step-by-step solution
Given: P(Ā∩B̅) = 1/6 (since overline{A∪B} = Ā∩B̅), P(A∩B) = 1/4, P(Ā)=1/4. Then P(A)=3/4. Using P(A∪B)=1-P(Ā∩B̅)=5/6, and P(A∪B)=P(A)+P(B)-P(A∩B), we get P(B)=1/3. Since P(A)≠P(B), not equally likely. Check independence: P(A)P(B)=(3/4)*(1/3)=1/4 = P(A∩B), so independent. Thus events are independent but not equally likely.
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