It is not quite clear what you mean by probability to obtain an individual from the overlapping area
. This solves for the area of the green zone in your diagram:
Let:
$X_1 \sim N(\mu_1,\sigma_1^2)$ with pdf $f_1(x_1)$ and cdf $ F_1(x_1)$ and
$X_2 \sim N(\mu_2,\sigma_2^2)$ with pdf $f_2(x_2)$ and cdf $ F_2(x_2)$,
where $\mu_1 < \mu_2$. In your example, the 'black variable' corresponds to $X_1$.
Let $c$ denote the point of intersection where the pdf's meet in the green zone of your plot Then, the area of your green intersection zone is simply:
$$P(X_1>c) + P(X_2<c) = 1 - F_1(c) + F_2(c) = 1-\frac{1}{2} \text{erf}\left(\frac{c-\mu _1}{\sqrt{2} \sigma _1}\right)+\frac{1}{2} \text{erf}\left(\frac{c-\mu _2}{\sqrt{2} \sigma _2}\right)$$
where erf(.) is the error function.
Point $c$ is the solution to $f_1(x) = f_2(x)$ within the green zone, which yields:
$$c = \frac{\mu _2 \sigma _1^2-\sigma _2 \left(\mu _1 \sigma _2+\sigma _1 \sqrt{\left(\mu _1-\mu _2\right){}^2+2 \left(\sigma _1^2-\sigma _2^2\right) \log \left(\frac{\sigma _1}{\sigma _2}\right)}\right)}{\sigma _1^2-\sigma _2^2}$$
For your example, with $ {\mu_1 = 5.28, \mu_2 = 8.45,
\sigma_1 = 0.91, \sigma_2 = 1.36}$, this yields: $c = 6.70458...$,
and the area of the green section is: 0.158413 ...