The probability that a randomly chosen factor of $$10^{19}$$ is a multiple of $$10^{15}$$ is 

Given number is $$10^{19}$$ , this can be presented as $$2^{19}\times\ \ 5^{19}$$ .

so, total no of factors = (19+1)(19+1) = $$20^2$$.

To calculate no of factors which are multiples of $$10^{15}$$ , lets write $$10^{19}$$ = $$10^{15}\times\ 2^4\times\ 5^4$$ . This implies no of factors which are multiple of $$10^{15}$$ are equal to no of factors of $$2^4\times\ 5^4$$ . 

Therefore, total no of factors which are multiple of $$10^{15}$$ are :

$$2^4=\left\{2^0,2^1,...2^4\right\}$$

$$5^4=\left\{5^0,5^1,...5^4\right\}$$

Therefore , it equal to = (4+1)x(4+1) = 25

The probability = $$\dfrac{\ 25}{20^2}\ =\ \ \dfrac{\ 1}{16}$$