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

The correct option is B ≤ 1 / 4 We have =P{( E _ 1 ∪ E _ 2 ) ∩( E _ 1 nbsp;) ∩( E _ 2 nbsp;) nbsp;} =P{( E _ 1 ∪ E _ 2 ) ∩( ( E _ ...

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Intersection of Sets

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

\leq \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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Similar questions

E_{1}

E_{2}, P {(E_{1} \cup E_{2}) \cap (_{1}) \cap (_{2})}

A

B

P (¯ ¯¯¯¯¯¯¯¯¯¯¯¯ ¯ A \cup B) = \frac{1}{6},

P (A \cap B) = \frac{1}{4}

P (¯ ¯¯ ¯ A) = \frac{1}{4}

¯ ¯¯ ¯ A

A

A

B

P (B) = \frac{3}{4}, P (A \cap B \cap ¯ ¯¯ ¯ C) = \frac{1}{3}

P (¯ ¯¯ ¯ A \cap B \cap ¯ ¯¯ ¯ C) = \frac{1}{3}

P (B \cap C)

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},

P (E_{1}) = \frac{1}{6}

P (E_{2}) = \frac{1}{3}

P (E_{3}) = \frac{1}{6}

E_{1}

E_{2}

E_{3}

E_{4}

E_{4}

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