geometric distribution given by

$\displaystyle Pr(X=x) = (1/a)(1-(1/a)^{x-1}) $ where a > 1.

Am I right in saying the distribution above is the same as

$\displaystyle Pr(X=x) = p(1-p)^{x-1} $ which is just the normal geometric pdf or does the change from p to 1/a change something else