= Solution
The all-red event $B$ is a <subset> of the same-colour event $A$. Its probability is $(1/2)^3=1/8$, while the three disjoint monochromatic cases give
$$
\mathbb P(A)=\left(\frac12\right)^3+\left(\frac13\right)^3+\left(\frac16\right)^3=\frac16.
$$
Thus the <conditional probability> is
$$
\boxed{\mathbb P(B\mid A)=\frac{\mathbb P(B)}{\mathbb P(A)}=\frac34.}
$$
The monochromatic outcomes are not equally likely, so counting colours alone would give the wrong conditional answer.
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