If a is an integer lying in [- 5, 30], then the probability that the graph of y = x² + 2 (a + 4)x - 5a + 64 is strictly above the x-axis is

$\frac{3}{5}$${\mathrm{x}}^2+2(\mathrm{a}+4)\mathrm{x}-5\mathrm{a}+64\ge 0$

If D < 0, then

$(\mathrm{a}+4{)}^2-(-5\mathrm{a}+64)<0$

$\begin{array}{l}\mathrm{or} {\mathrm{a}}^2+13\mathrm{a}-48<0\\ \mathrm{or} (\mathrm{a}+16)(\mathrm{a}-3)<0\\ \Rightarrow -16<\mathrm{a}<3\iff -5\le \mathrm{a}\le 2\end{array}$

Then, the favorable cases is equal to the number of integers in the interval [-5, 2], i.e., 8.

Total number of cases is equal to the number of integers in the interval [ -5, 30], i.e., 36.

Hence, the required probability is 8/36 = 2/9.