For any two independent events E_1 and E_2, P | vedclass

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Question

(A) By De Morgan's Law,we know that $\bar{E}_1 \cap \bar{E}_2 = \overline{E_1 \cup E_2}$.Substituting this into the expression,we get $(E_1 \cup E_2)

Vedclass · Accepted Answer

$< \frac{1}{4}$

Vedclass · Answer

(A) By De Morgan's Law,we know that $\bar{E}_1 \cap \bar{E}_2 = \overline{E_1 \cup E_2}$.Substituting this into the expression,we get $(E_1 \cup E_2) \cap (\overline{E_1 \cup E_2})$.Since the intersection of any set $A$ and its complement $\bar{A}$ is the empty set $\phi$,we have $(E_1 \cup E_2) \cap (\overline{E_1 \cup E_2}) = \phi$.The probability of an empty set is $P(\phi) = 0$.Since $0 < \frac{1}{4}$,the correct option is $A$.

For any two independent events $E_1$ and $E_2,$ $P\{(E_1 \cup E_2) \cap (\bar{E}_1 \cap \bar{E}_2)\}$ is

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