If $$P(A \cup B) = \frac{5}{9}, \quad P(\overline{A} \cup \overline{B}) = \frac{13}{27}, \quad P(A) = \frac{11}{18}$$ then the odds against the event of B are:

Correct option is D

Given:

$$P(A \cup B) = \frac{5}{9}, \quad P(\overline{A} \cup \overline{B}) = \frac{13}{27}, \quad P(A) = \frac{11}{18}$$​

Concept and Formula Used:

Concept of Probability;

​$$P(E) = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \\\text{Where, } P(E) = \text{Probability of events,} \\P(\overline{A} \cup \overline{B}) = P(\overline{A}) + P(\overline{B}) - P(\overline{A} \cap \overline{B}) \\$$​​

Solution:

As per the question,

$$P(B) = 1 - P(\overline{B}) \\\text{And } P(\overline{A} \cap \overline{B}) = 1 - P(A \cup B) = 1 - \frac{5}{9} = \frac{4}{9} \\\text{So,} \\P(\overline{A} \cup \overline{B}) = P(\overline{A}) + P(\overline{B}) - P(\overline{A} \cap \overline{B}) \\\frac{13}{27} = \frac{7}{18} + P(\overline{B}) - \frac{4}{9} \\P(\overline{B}) = \frac{29}{54} \\\text{And} \\P(B) = 1 - \frac{29}{54} = \frac{25}{54} \\\text{So, } \frac{P(\overline{B})}{P(B)} = \frac{29}{25}$$​