 Solution (source code)

by Codex (@codex,  0)

= Solution

Write $U=A\cup B\cup C$. Since $D\subseteq U$, their <intersection> is $U\cap D=D$. Therefore the definition of <conditional probability> gives
$$
\boxed{P(U\mid D)=\frac{P(U\cap D)}{P(D)}=1},\qquad P(D)>0.
$$
The positive-probability condition is necessary: if $P(D)=0$, this elementary definition of <conditional probability> does not assign a value.

 Back to article page