Let X = x ,   y ∈ ℤ × ℤ   :   x 2 8 + y 2 20 < 1   and  …

Given,

Equation of ellipse $\frac{x^2}{8}+\frac{y^2}{20}=1$

And equation of parabola $y^2=5\mathrm{x}$

Now intersection point of ellipse and parabola will be $\left(2,\sqrt{10}\right) \& \left(2,-\sqrt{10}\right)$

Now plotting the region $\frac{x^2}{8}+\frac{y^2}{20}<1 \mathrm{and} y^2<5\mathrm{x}$ we get,

Now randomly chosen points will be between origin and the intersecting point,

So, the integer points inside region are $$\begin{gathered} \left\{\left(2, 1\right), \left(2, -1\right), \left(2, 2\right), \left(2, -2\right),\left(2, 3\right), \left(2, -3\right), \\ \left(2, 0\right), \left(1, 1\right),\left(1, -1\right), \left(1, 2\right), \left(1, -2\right), \left(1, 0\right)\right\} \end{gathered}$$

Total number of ways to select three points $={}^{12}C_3=220$

Required number of triangle $=4\times {}^{7}C_1+9\times {}^{5}C_1=73$

{Note: Points are taken such a way that distance between two points are multiple of $2$ for example if we take points on line $x=1$ we can take following points $\left\{\left(1,2\right) \& \left(1,0\right), \left(1,0\right) \& \left(1,-2\right), \left(1,1\right) \& \left(1,-1\right), \left(1,2\right) \& \left(1,-2\right)\right\}$

so there will be $4$ pairs from line $x=1$ and will chose another point from $x=2$ line so triangle will have a positive area. Also note that height of the triangle will always be $1$}

Hence, probability will be $=\frac{73}{220}$