Solution
= Solution
With replacement, successive draws are <independent>, with red <probability> $2/7$ and green <probability> $5/7$. The first red on draw $n$ requires $n-1$ greens followed by one red. Hence
$$
\boxed{\mathbb P(N=n)=\left(\frac57\right)^{n-1}\frac27,\qquad n=1,2,\ldots.}
$$
This is a <geometric distribution>. Its total <probability> is one because the <geometric series> sums to $(2/7)/(1-5/7)=1$.