Explanation:
the basic probabilty of being born with that certain trait is 5/9.
that means the probabilty of being born without that certain trait is 1 - 5/9 = 4/9.
the probability of 2 children being born with that trait and 3 being born without that trait is
5/9 × 5/9 × 4/9 × 4/9 × 4/9
but as you can see, this only covers a specific case, e.g. when the first 2 children have the trait, and the younger siblings don't.
for the general probability of that questioned result we need to multiply this with the number of different combinations : in how many ways can I pick 2 children out of 5, when the sequence of the 2 picked children is irrelevant ?
this is C (5, 2) = 5!/((5-2)!×2) = 5!/(3!×2) = 5×4/2 = 5×2 = 10
so, the probability is
10×(5/9)²×(4/9)³ = 10×5²×4³/9⁵ = 0.270961405...
≈ 0.271