It is shown that every family of mutually singular measures in a conditional probability distribution is countable or else there is a perfect set of measures which form a strongly orthogonal family. Theorem: Let $X$ and $Y$ be complete separable metric spaces and $\mu$ a conditional probability distribution on $X \times \mathscr{B}(Y)$. Then either (1) there is a nonempty compact perfect subset $P$ of $X$ and a Borel subset $D$ of $X \times Y$ so that if $x$ and $y$ are distinct elements of $P$, then $\mu(x, D_x) = 1, \mu(y, D_x) = 0$, and $D_x \cap D_y = \phi$ or else (2) if $K$ is a subset of $X$ so that $\{\mu(x, \cdot):x \in K\}$ is a pairwise orthogonal family, then $K$ is countable.