Let A = both good, B = at least one good. \(P(A) = \binom{6}{2}/\binom{10}{2} = 15/45 = 1/3\). \(P(B) = 1 - P(\text{both bad}) = 1 - \binom{4}{2}/\binom{10}{2} = 1 - 6/45 = 39/45\). But the condition is one specific tube tested is good: \(P(\text{other good}|\text{one is good}) = \frac{5}{9}\) using hypergeometric reasoning on remaining 9 tubes (5 good, 4 bad). Answer is \(\dfrac{5}{9}\).