## Precalculus (6th Edition) Blitzer

Because $P\left( E \right)+P\left( \text{not}\ E \right)=1$, then $P\left( \text{not}\ E \right)=1-P\left( E \right)$ and $P\left( E \right)=1-P\left( \text{not}\ E \right)$.
We know that the sum of the probability of all possible outcomes in any situation is $1$. Therefore, $P\left( E \right)+P\left( \text{not}\ E \right)=1$ And the probability that an event $E$ will not occur is equal to $1$ minus the probability that it will occur. Therefore, $P\left( \text{not}\ E \right)=1-P\left( E \right)$ Thus, the probability that an event $E$ will occur is equal to $1$ minus the probability that it will not occur. So, $P\left( E \right)=1-P\left( \text{not}\ E \right)$