For any two independent events E 1 and E 2 -P E 1 - E 2 - E 1 - E 2 is

The correct option is A Since E _ 1 ∩ E _ 2 = E _ 1 ∪ E _ 2 and ( E _ 1 ∪ E _ 2 ) ∩( E _ 1 ∪ E _ 2 nbsp;) =ϕ nbsp;∴P[ ( E ...

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Addition Theorem of Probability for 2 Events

For any two i...

Question

For any two independent events $E_{1}$ and $E_{2}, P {(E_{1} \cup E_{2}) \cap (_{1}) \cap (_{2})}$ is

< \frac{1}{4}

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> \frac{1}{4}

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\geq \frac{1}{2}

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None of these

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Solution

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Similar questions

E_{1}

E_{2}

P (E_{1}) = \frac{1}{4}, P (\frac{E_{2}}{E_{1}}) = \frac{1}{2}

P (\frac{E_{1}}{E_{2}}) = \frac{1}{4},

E_{1} a n d E_{2}

P (E_{1}) = P (E_{2}); P (E_{1} \cup E_{2}) = 1; E_{1} a n d E_{2}

P (E_{1}) =

E_{1}

E_{2}

P (E_{1}) = \frac{1}{4}

P (\frac{E_{2}}{E_{1}}) = \frac{1}{2}

P (\frac{E_{1}}{E_{2}}) = \frac{1}{4}

E_{1}

E_{2}

P (E_{1}) = 1 / 4, P (E_{2} / E_{1}) = 1 / 2

P (E_{1} / E_{2}) = 1 / 4

E^{c}

E

E_{1}, E_{2}

E_{3}

P (E_{1}) > 0

P (E_{1} \cap E_{2} \cap E_{3}) = 0

P (\frac{E_{2}^{c} \cap E_{3}^{c}}{E_{1}})

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