= Solution
For this <stellar population>, let $m=M/M_\odot$ denote initial <mass> in units of the <solar mass>, and let $\tau=t/\mathrm{Gyr}$ denote present <stellar age>. Constant formation of equal numbers of <stars> per unit time makes the <stellar age> a <uniform distribution> on $[0,10]$ Gyr, so $Y=\tau/10$ is uniform on $[0,1]$. The normalized <initial mass function> has <probability density function> $f(m)=0.1m^{-2}$ for $m>0.1$, since $\int_{0.1}^\infty m^{-2}dm=10$. Consequently
$$
\boxed{Y(t)=\frac{t}{10\,\mathrm{Gyr}},\qquad X(M)=\frac{\int_m^\infty u^{-2}du}{\int_{0.1}^\infty u^{-2}du}=\frac{0.1}{m}.}
$$
These formulae have the stated age and mass domains; outside them the relevant cumulative fractions saturate at zero or one. The <probability integral transform> makes $X$ uniform on $[0,1]$: $m=0.1/X$ gives $f(m)|dm/dX|=1$. The time-independent <initial mass function> and constant number formation rate give <independence> of $X$ and $Y$. All fractions here count objects, including <white dwarfs>, using the stipulated <stellar evolution> law.
A <red giant> has
$$
\frac9m<\tau<\frac{10}m,\qquad 0\leq\tau\leq10.
$$
Thus the <mass> boundaries are $m=9/\tau$ and $m=10/\tau$ for positive $\tau$. Equivalently, for $0.9<m<1$ the <red giant> region runs from $9/m$ to the age cap $10$, and for $m\geq1$ it runs from $9/m$ to $10/m$. There are no <red giants> with $m\leq0.9$. Boundaries have zero <probability> and their endpoint convention does not affect the fractions. In the uniform $(X,Y)$ square the <red giant> region is $9X<Y<10X$, with <triangle> vertices $(0,0)$, $(1/10,1)$ and $(1/9,1)$. The <white dwarf> region is the <triangle> $0<X<Y/10$.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2017/iii/paper-322-population-regions.png]
{title=Mass-age regions and their uniform-coordinate images}
{description=The right panel magnifies the evolved part of the unit square. All of the remaining region at larger X is main sequence.}
At fixed $Y=y$, define the conditional <probabilities> of a <red giant>, <white dwarf> and <main sequence> star by $g(y)$, $w(y)$ and $s(y)$. Their interval widths are
$$
g(y)=\frac{y}{90},\qquad w(y)=\frac{y}{10}=9g(y),\qquad s(y)=1-\frac y9.
$$
<Integration> over the <uniform distribution> of age gives the individual-star fractions
$$
\boxed{p_G=\int_0^1g(y)dy=\frac1{180},\qquad p_W=\int_0^1w(y)dy=\frac1{20}.}
$$
The systems form a <coeval binary population>: the two <binary star> components have the same age. Their masses are independent, so $(X_1,X_2,Y)$ has uniform <probability density function> on $[0,1]^3$ and their states have <conditional independence> given $Y$. Unconditional independence of their states would be incorrect: older systems make both evolved states more likely. The <law of total probability> now gives
$$
\boxed{p_{\geq1G}=\int_0^1\bigl[1-(1-g(y))^2\bigr]dy=\frac1{90}-\frac1{24300}=\frac{269}{24300}.}
$$
The subtraction removes the double counting of systems containing two <red giants>.
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