All probability situations can be reduced to two possible outcomes: success or failure.

When we express the outcomes in this way we say that they are complementary.

The sum of the probability of an event and its complement is always $1$.

If $\mathrm{P}(\mathrm{E})$ is the probability of an event $\mathrm{E}$ happening and $\mathrm{P}(\bar{\mathrm{E}})$ is the probability of that event not happening then

$$\mathrm{P}(\mathrm{E}) +\mathrm{P}(\bar{\mathrm{E}}) = 1.$$

Rearranging this equation gives:

$$\mathrm{P}(\bar{\mathrm{E}}) = 1 - \mathrm{P}(\mathrm{E})$$

When we throw the ball we can say that there are two possible outcomes: either the ball goes through the hoop or the ball does not go through the hoop (the ball misses the hoop).

Let $\mathrm{H}$ be the event that the ball goes through the hoop. Then

$$\mathrm{P}(\mathrm{H}) + \mathrm{P}(\bar{\mathrm{H}}) = 1.$$

But we are given that $\mathrm{P}(\bar{\mathrm{H}}) = \displaystyle\frac{1}{\var{Hoop}}$.

Rearranging the above equation to obtain $\mathrm{P}(\mathrm{H})$.

$$\begin{align} \mathrm{P}(\mathrm{H}) &= 1 - \mathrm{P}(\bar{\mathrm{H}}) \\[0.5em] &= 1 - \displaystyle\frac{1}{\var{Hoop}}\\[0.5em] &= \simplify[fractionNumbers]{{1-1/{Hoop}}}. \end{align}$$