If $¯ E$ and $¯ F$ are complementary events of E and F respectively and if $0 < P (F) < 1$ , then

P (E / F) + P (¯ E / F) = 1

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P (E / F) + P (E ¯ F) = 1

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P (¯ E / F) + P (E ¯ F) = 1

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P (E ¯ F) + P (¯ E ¯ F) = 1

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Solution

The correct options are
A $P (E / F) + P (¯ E / F) = 1$
D $P (E ¯ F) + P (¯ E ¯ F) = 1$
$P (E / F) + P (¯ E / F) = \frac{P (E \cap F)}{P (F)} + \frac{P (¯ E \cap F)}{P (F)} = \frac{P (E \cap F) + P (¯ E \cap F)}{P (F)} = \frac{P (F)}{P (F)} = 1$
Also,
$P (E / ¯ F) + P (¯ E / ¯ F) = \frac{P (E \cap ¯ F)}{P (¯ F)} + \frac{P (¯ E \cap ¯ F)}{P (¯ F)} = \frac{P (E \cap ¯ F) + P (¯ E \cap ¯ F)}{P (F)} = \frac{P (¯ F)}{P (¯ F)} = 1$

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If ¯E and ¯F are complementary events of E and F respectively and if 0<P(F)<1, then

The correct options are A P(E/F)+P(¯E/F)=1 D P(E¯F)+P(¯E¯F)=1P(E/F)+P(¯E/F)=P(E∩F)P(F)+P(¯E∩F)P(F)=P(E∩F)+P(¯E∩F)P(F)=P(F)P(F)=1Also,P(E/¯F)+P(¯E/¯F)=P(E∩¯F)P(¯F)+P(¯E∩¯F)P(¯F)=P(E∩¯F)+P(¯E∩¯F)P(F)=P(¯F)P(¯F)=1

If $¯ E$ and $¯ F$ are complementary events of E and F respectively and if $0 < P (F) < 1$ , then