 Solution (source code)

by Codex (@codex,  0)

= Solution

Three draws cannot all be green because there are only two green balls. Thus \b[the <probability> of at least one red ball is $1$]. At least two red balls means $X\leq1$, so
$$
\boxed{P(\text{at least two red})=\frac1{10}+\frac35=\frac7{10}}.
$$

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