Given the probabilities $P(A) = \frac{1}{2}$, $P(B) = \frac{1}{3}$, $P(C) = \frac{1}{5}$, $P(AB) = \frac{1}{10}$, $P(AC) = \frac{1}{15}$, $P(BC) = \frac{1}{20}$, and $P(ABC) = \frac{1}{30}$, we need to find the probabilities of the following events: $A \cup B$, $\overline{AB}$, $A \cup B \cup C$, $\overline{ABC}$, $\overline{AB}C$, and $\overline{AB} \cup C$.

1. Problem Description

Given the probabilities $P(A) = \frac{1}{2}$, $P(B) = \frac{1}{3}$, $P(C) = \frac{1}{5}$, $P(AB) = \frac{1}{10}$, $P(AC) = \frac{1}{15}$, $P(BC) = \frac{1}{20}$, and $P(ABC) = \frac{1}{30}$, we need to find the probabilities of the following events: $A \cup B$, $\overline{AB}$, $A \cup B \cup C$, $\overline{ABC}$, $\overline{AB}C$, and $\overline{AB} \cup C$.

2. Solution Steps

(1) $P(A \cup B)$:

Using the inclusion-exclusion principle:

$P(A \cup B) = P(A) + P(B) - P(AB)$

$P(A \cup B) = \frac{1}{2} + \frac{1}{3} - \frac{1}{10} = \frac{15}{30} + \frac{10}{30} - \frac{3}{30} = \frac{22}{30} = \frac{11}{15}$

(2) $P(\overline{AB})$:

Using De Morgan's Law:

$P(\overline{AB}) = P(\overline{A \cup B}) = 1 - P(A \cup B)$

$P(\overline{AB}) = 1 - \frac{11}{15} = \frac{4}{15}$

(3) $P(A \cup B \cup C)$:

Using the inclusion-exclusion principle:

$P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(AB) - P(AC) - P(BC) + P(ABC)$

$P(A \cup B \cup C) = \frac{1}{2} + \frac{1}{3} + \frac{1}{5} - \frac{1}{10} - \frac{1}{15} - \frac{1}{20} + \frac{1}{30}$

$P(A \cup B \cup C) = \frac{30}{60} + \frac{20}{60} + \frac{12}{60} - \frac{6}{60} - \frac{4}{60} - \frac{3}{60} + \frac{2}{60} = \frac{51}{60} = \frac{17}{20}$

(4) $P(\overline{ABC})$:

$P(\overline{ABC}) = P(\overline{A \cap B \cap C}) = 1 - P(A \cap B \cap C) = 1 - P(ABC)$

$P(\overline{ABC}) = 1 - \frac{1}{30} = \frac{29}{30}$

(5) $P(\overline{AB}C)$:

$P(\overline{AB}C) = P((\overline{A} \cup \overline{B}) \cap C) = P((\overline{A} \cap C) \cup (\overline{B} \cap C))$

$= P(\overline{A} \cap C) + P(\overline{B} \cap C) - P(\overline{A} \cap \overline{B} \cap C) = (P(C) - P(AC)) + (P(C) - P(BC)) - (P(C) - P(AC) - P(BC) + P(ABC))$

$= P(C) - P(AC) + P(C) - P(BC) - P(C) + P(AC) + P(BC) - P(ABC) = P(C) - P(ABC)$

$P(\overline{AB}C) = \frac{1}{5} - \frac{1}{30} = \frac{6}{30} - \frac{1}{30} = \frac{5}{30} = \frac{1}{6}$

Alternatively, $P(\overline{AB}C) = P((\overline{A} \cup \overline{B})C) = P(C) - P(ABC) = \frac{1}{5} - \frac{1}{30} = \frac{1}{6}$

(6) $P(\overline{AB} \cup C)$:

$P(\overline{AB} \cup C) = P(\overline{AB}) + P(C) - P(\overline{AB} \cap C) = \frac{4}{15} + \frac{1}{5} - \frac{1}{6}$

$P(\overline{AB} \cup C) = \frac{4}{15} + \frac{3}{15} - \frac{1}{6} = \frac{7}{15} - \frac{1}{6} = \frac{14}{30} - \frac{5}{30} = \frac{9}{30} = \frac{3}{10}$

$P(\overline{AB} \cup C) = P((\overline{A} \cup \overline{B}) \cup C)$

Using $P(\overline{AB}) = \frac{4}{15}$ and $P(\overline{AB} C) = \frac{1}{6}$,

$P(\overline{AB} \cup C) = P(\overline{AB}) + P(C) - P(\overline{AB} C)$

$= \frac{4}{15} + \frac{1}{5} - \frac{1}{6} = \frac{4}{15} + \frac{3}{15} - \frac{1}{6} = \frac{7}{15} - \frac{1}{6} = \frac{14}{30} - \frac{5}{30} = \frac{9}{30} = \frac{3}{10}$

3. Final Answer

$P(A \cup B) = \frac{11}{15}$

$P(\overline{AB}) = \frac{4}{15}$

$P(A \cup B \cup C) = \frac{17}{20}$

$P(\overline{ABC}) = \frac{29}{30}$

$P(\overline{AB}C) = \frac{1}{6}$

$P(\overline{AB} \cup C) = \frac{3}{10}$