MHT CET · Maths · Probability
The probability that least one of the events \(E_1\) and occurs is 0.6. If the simultaneous occurrence of \(E_1\) and \(E_2\) is 0.2 , \(\mathrm{P}\left(\mathrm{E}_1{ }^{\prime}\right)+\mathrm{P}\left(\mathrm{E}_2^{\prime}\right)=\)
- A 0.4
- B 1.6
- C 1.2
- D 0.8
Answer & Solution
Correct Answer
(C) 1.2
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & P\left(E_1 \cup E_2\right)=P\left(E_1\right)+P\left(E_2\right)-P\left(E_1 \cap E_2\right) \\ & \therefore 0.6=\left[1-P\left(E_1{ }^{\prime}\right)+1-P\left(E_2{ }^{\prime}\right)-P\left(E_1 \cap E_2\right)\right] \\ & P\left(E_1{ }^{\prime}\right)+P\left(E_2{ }^{\prime}\right)=2-0.2-0.6=1.2\end{aligned}\)
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