AP EAMCET · Maths · Probability
If \(\mathrm{A}, \mathrm{B}\) and \(\mathrm{C}\) are three events of a random experiment with \(\mathrm{P}(\mathrm{A})=0.4, \mathrm{P}(\mathrm{B})=0.3\) and \(\mathrm{P}(\mathrm{A} \cap \mathrm{B})=0.2\), then the probability that neither A nor B occurs is
- A \(0.5\)
- B \(0.15\)
- C \(0.13\)
- D \(0.12\)
Answer & Solution
Correct Answer
(A) \(0.5\)
Step-by-step Solution
Detailed explanation
Given, \(\mathrm{P}(\mathrm{A})=0.4, \mathrm{P}(\mathrm{B})=0.3\) \(\begin{aligned} & P(A \cap B)=0.2 \\ & \text { Now, } P(A \cup B)=P(A)+P(B)-P(A \cap B) \\ & \Rightarrow P(A \cup B)=0.4+0.3-0.2=0.5 \end{aligned}\) Since, probability of neither A nor B occurs…
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