= Solution
The <binomial series> gives, for $|z|<14/5$,
$$
(14-5z)^{-1/2}
=\frac1{\sqrt{14}}\sum_{n=0}^\infty
\frac{\binom{2n}n}{4^n}\left(\frac5{14}\right)^n z^n.
$$
All coefficients are nonnegative, and $g(1)=2/3+1/3=1$. Thus define
$$
P(T=0)=\frac23+\frac1{\sqrt{14}},\qquad
P(T=n)=\frac1{\sqrt{14}}\binom{2n}n\left(\frac5{56}\right)^n\quad(n\geq1).
$$
These numbers sum to $1$, so they define a <probability distribution> on the nonnegative integers with the desired <probability generating function>.
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