Sample space $S$ is partitioned into $E_{1}, E_{2}$, and $E_{3}$ such that $P\left(E_{1}\right)=\frac{1}{9}$ and $P\left(E_{2}\right)=\frac{1}{15}$.
a) Find $P\left(E_{3}\right)$.
b) Find the odds in favor of and the odds against $E_{3}$ occurring.
a) $P\left(E_{3}\right)=\square($ Simplify your answer.)
b) The odds in favor of $E_{3}$ occurring, in lowest terms, are $\square: \square$.
(Simplify your answer. Type whole numbers.)
The odds against $E_{3}$, in lowest terms, are $\square: \square$.
(Simplify your answer. Type whole numbers.)

Step 1 ：Given that the sample space $S$ is partitioned into $E_{1}, E_{2}$, and $E_{3}$ such that $P\left(E_{1}\right)=\frac{1}{9}$ and $P\left(E_{2}\right)=\frac{1}{15}$.

Step 2 ：Since the sum of the probabilities of all events in a sample space is 1, we can find $P\left(E_{3}\right)$ by subtracting the sum of $P\left(E_{1}\right)$ and $P\left(E_{2}\right)$ from 1. So, $P\left(E_{3}\right)=1-P\left(E_{1}\right)-P\left(E_{2}\right)$.

Step 3 ：Substituting the given values, we get $P\left(E_{3}\right)=1-\frac{1}{9}-\frac{1}{15}=\boxed{0.822}$.

Step 4 ：We can find the odds in favor of $E_{3}$ occurring by dividing $P\left(E_{3}\right)$ by $1 - P\left(E_{3}\right)$. So, the odds in favor of $E_{3}$ occurring are $\frac{P\left(E_{3}\right)}{1 - P\left(E_{3}\right)}=\boxed{4.625:1}$.

Step 5 ：We can find the odds against $E_{3}$ occurring by dividing $1 - P\left(E_{3}\right)$ by $P\left(E_{3}\right)$. So, the odds against $E_{3}$ occurring are $\frac{1 - P\left(E_{3}\right)}{P\left(E_{3}\right)}=\boxed{0.216:1}$.