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

1198 182 Probability - Part 2 Report Error

A

$P (E ∣ F) + P (\overset{ˉ}{E} ∣ F) = 1$

47%

B

$P (E ∣ F) + P (E ∣ \overset{ˉ}{F}) = 1$

34%

C

$P (\overset{ˉ}{E} ∣ F) + P (E ∣ \overset{ˉ}{F}) = 1$

3%

D

$P (E ∣ \overset{ˉ}{F}) + P (\overset{ˉ}{E} ∣ \overset{ˉ}{F}) = 0$

16%

Solution:

$P (E ∣ F) + P (\overset{ˉ}{E} ∣ F)$
$= \frac{P ( E \cap F ) + P ( E ˉ \cap F )}{P ( F )}$
$= \frac{P (( E \cup E ˉ ) \cap F )}{P ( F )}$
$= \frac{P ( F )}{P ( F )} = 1$