TS EAMCET · Maths · Probability
If \(P(A \cup B)=0.8\) and \(P(A \cap B)=0.3\), then \(P\left(A^C\right)+P\left(B^C\right)\) is equal to
- A 0.3
- B 0.5
- C 0.7
- D 0.9
Answer & Solution
Correct Answer
(D) 0.9
Step-by-step Solution
Detailed explanation
Given, \(P(A \cup B)=0.8\) and \(P(A \cap B)=0.3\) \(\begin{aligned} \because \quad P(A \cup B) & =P(A)+P(B)-P(A \cap B) \\ \Rightarrow P(A)+P(B) & =P(A \cup B)+P(A \cap B) \\ & =0.8+0.3=1.1 \\ \Rightarrow \quad P(A) & =1.1-P(B) \end{aligned}\) Now,…
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