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

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

Probability - Part 2

Solution:

$P ( E| F )+ P (\bar{ E } | F )$
$=\frac{ P ( E \cap F )+ P (\bar{ E } \cap F )}{ P ( F )}$
$=\frac{ P (( E \cup \bar{ E }) \cap F )}{ P ( F )}$
$=\frac{ P ( F )}{ P ( F )}=1$