Sample space $S$ is partitioned into $E_{1}, E_{2}, E_{3}$, and $E_{4}$ such that $P\left(E_{1}\right)=0.1, P\left(E_{2}\right)=0.04$, and $P\left(E_{3}\right)=0.46$. a. Find $P\left(E_{4}\right)$. b. Find the odds in favor of and the odds against $E_{4}$ occurring. a. $P\left(E_{4}\right)=\square$ (Simplify your answer:) b. The odds in favor of $E_{4}$ occurring, in lowest terms, are $\square$ : $\square$. (Type whole numbers.) The odds against $E_{4}$, in lowest terms, are $\square: \square$. (Type whole numbers.)
Solution
Step 1 ：The problem provides the probabilities of three events $E_{1}, E_{2},$ and $E_{3}$ in a sample space $S$ partitioned into four events. The probabilities are $P(E_{1}) = 0.1, P(E_{2}) = 0.04,$ and $P(E_{3}) = 0.46$. Since the probabilities of all events in a sample space must add up to 1, we can find $P(E_{4})$ by subtracting the sum of $P(E_{1}), P(E_{2}),$ and $P(E_{3})$ from 1.
Step 2 ：Calculating $P(E_{4})$ gives $P(E_{4}) = 1 - (P(E_{1}) + P(E_{2}) + P(E_{3})) = 1 - (0.1 + 0.04 + 0.46) = 0.4$.
Step 3 ：We can calculate the odds in favor of $E_{4}$ occurring by dividing $P(E_{4})$ by $1 - P(E_{4})$. This gives us the odds in favor of $E_{4}$ as $P(E_{4}) / (1 - P(E_{4})) = 0.4 / (1 - 0.4) = 2 / 3$.
Step 4 ：We can calculate the odds against $E_{4}$ occurring by dividing $1 - P(E_{4})$ by $P(E_{4})$. This gives us the odds against $E_{4}$ as $(1 - P(E_{4})) / P(E_{4}) = (1 - 0.4) / 0.4 = 3 / 2$.
Step 5 ：Final Answer: a. $P(E_{4}) = oxed{0.4}$ b. The odds in favor of $E_{4}$ occurring, in lowest terms, are $oxed{2 : 3}$. The odds against $E_{4}$, in lowest terms, are $oxed{3 : 2}$.
