  Let $a>1$ be an integer. Denote by $l_a(p)$ the multiplicative order of $a$
modulo primes $p$. We prove that if $\frac{x}{\log x\log\log x}=o(y)$, then
$$\frac 1 y \sum_{a\leq y}\sum_{p\leq x}\frac{1}{l_a(p)}=\log x + C\log\log
x+O\left(\frac x {y \log\log x}\right) $$ which is an improvement over a
theorem by Felix ~\cite{Fe}. Additionally, we also prove two other average
results If $\log^2 x = o(\psi(x))$ and $x^{1-\delta}\log^3 x = o(y)$, then
$$\frac1y \sum_{a<y} \sum_{\substack{{p<x} \\ {l_a(p)>\frac{x}{\psi(x)}}}} 1 =
\pi(x) + O\left(\frac{x\log x}{\psi(x)}\right) + O\left(\frac{x^{2 -
\delta}\log^2 x}y\right).$$ Furthermore, if $x^{1-\delta}\log^3 x = o(y)$, then
$$\frac1y\sum_{a<y} \sum_{\substack{{p<x} \\ {p\nmid a}}}l_a(p) =
c\textrm{Li}(x^2) + O\left( \frac{x^2}{\log^A x} \right) + O\left(\frac{x^{3
-\delta}\log^2 x}y\right)$$ where $$c = \prod_p \left(1-\frac
p{p^3-1}\right).$$
