Step 1

Let a,b,c denote the lengths of the sides of our triangle. Without loss of generality, since the triangle is isoceles, let $a=c$, so that (a,a,b) describes our triangle, where $a,b\in \mathbb{N}$. Now we know additionally $2a+b=30$, and by the triangle inequality, $b\le 2a$, So $30-2a\le 2a$, i.e. $a\ge 8$, but of course $a\le 15$. Thus the possible triangles are described by $\{(8,8,14),(9,9,12),(10,10,10),(11,11,8),(12,12,6),(13,13,4),(14,14,2),(15,15,0)\}$.

Step 2

Only one of these is equilateral. Now assuming uniform distribution over these possibilities, and depending whether you include the degenerate (15,15,0), the probability is either $\frac{1}{7}$ or $\frac{1}{8}$.

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