= Solution
Condition on the common age $Y=y$ of the <coeval binary population>. Each component is a <red giant> with conditional <probability> $g(y)=y/100$, independently of the other component at that age. The conditional <probability> of at least one giant is $1-(1-g)^2=2g-g^2$. Integrating over the uniform age distribution gives
$$
\Pr(\text{at least one giant})=\int_0^1\left(\frac{2y}{100}-\frac{y^2}{10000}\right)dy
=\frac1{100}-\frac1{30000}.
$$
Thus \b[the fraction is $299/30000=0.996\overline6\%$, just below $1\%$]. The subtracted term removes the systems counted twice because both components are <red giants>. Using the square of the marginal giant fraction would incorrectly ignore their shared age.
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