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(D) We know that for any event $A$ and a given event $B$ with $P(B) > 0$,the conditional probability satisfies $P(A/B) + P(\overline{A}/B) = 1$. For o

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Both $(a)$ and $(c)$

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(D) We know that for any event $A$ and a given event $B$ with $P(B) > 0$,the conditional probability satisfies $P(A/B) + P(\overline{A}/B) = 1$. For option $(a)$: $P(E/F) + P(\overline{E}/F) = \frac{P(E \cap F)}{P(F)} + \frac{P(\overline{E} \cap F)}{P(F)} = \frac{P(E \cap F) + P(\overline{E} \cap F)}{P(F)} = \frac{P((E \cup \overline{E}) \cap F)}{P(F)} = \frac{P(S \cap F)}{P(F)} = \frac{P(F)}{P(F)} = 1$. Thus,$(a)$ is true. For option $(c)$: Similarly,replacing $F$ with $\overline{F}$ (given $P(\overline{F}) = 1 - P(F) > 0$),we have $P(E/\overline{F}) + P(\overline{E}/\overline{F}) = 1$. Thus,$(c)$ is true. Therefore,both $(a)$ and $(c)$ are correct.

If $\overline{E}$ and $\overline{F}$ are the complementary events of events $E$ and $F$ respectively and if $0 < P(F) < 1$,then

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