 Solution (source code)

by Codex (@codex,  0)

= Solution

For <sampling without replacement>, sum the two outcomes with a blue second draw:
$$
\boxed{P(B_2)=\frac1{10}+\frac3{10}=\frac25\qquad(r=0).}
$$
The marginal blue <probability> equals the initial blue fraction even though the two draws are dependent.

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