If E and F are independent events such that 0 | vedclass

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(D) Given that $E$ and $F$ are independent events,we have $P(E \cap F) = P(E) \cdot P(F)$.$(a)$ $P(E \cap F^c) = P(E) - P(E \cap F) = P(E) - P(E)P(F)

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All of the above

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(D) Given that $E$ and $F$ are independent events,we have $P(E \cap F) = P(E) \cdot P(F)$.$(a)$ $P(E \cap F^c) = P(E) - P(E \cap F) = P(E) - P(E)P(F) = P(E)(1 - P(F)) = P(E)P(F^c)$. Thus,$E$ and $F^c$ are independent.$(b)$ $P(E^c \cap F^c) = 1 - P(E \cup F) = 1 - [P(E) + P(F) - P(E \cap F)] = 1 - P(E) - P(F) + P(E)P(F) = (1 - P(E))(1 - P(F)) = P(E^c)P(F^c)$. Thus,$E^c$ and $F^c$ are independent.$(c)$ Since $E$ and $F$ are independent,$P(E/F) = P(E)$. Also,since $E^c$ and $F^c$ are independent,$P(E^c/F^c) = P(E^c)$. Therefore,$P(E/F) + P(E^c/F^c) = P(E) + P(E^c) = 1$.Since all statements $(a)$,$(b)$,and $(c)$ are correct,the correct option is $(d)$.

If $E$ and $F$ are independent events such that $0 < P(E) < 1$ and $0 < P(F) < 1,$ then

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