TS EAMCET · Maths · Probability
If \(P(A \cup B)=0.8\) and \(P(A \cap B)=0.3\), then \(P(\bar{A})+P(\bar{B})\) is equal to:
- A 0.3
- B 0.5
- C 0.8
- D 0.9
Answer & Solution
Correct Answer
(D) 0.9
Step-by-step Solution
Detailed explanation
We have, \(P(A)+P(B)=P(A \cup B)+P(A \cap B)\) \(=0.8+0.3=1.1\) \(\Rightarrow \quad 1-P(\bar{A})+1-P(\bar{B})=1.1\) \(\Rightarrow \quad P(\bar{A})+P(\bar{B})=2-1.1=0.9\)
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