SSC CGL 2023 PYQ — Question 10

The correct answer is option 3 i.e. $\frac{29}{25}$.

Given that

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

Formula used

$P(\bar{A}\cup \bar{B})$ = 1 - $P({A}∩{B})$

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

Substituting the values we get

$P({A}∩{B})$ = 1 - $\frac{13}{27}$ = $\frac{14}{27}$ 

Hence, P(B) = $\frac{5}{9} - \frac{11}{18} + \frac{14}{27}$ = $\frac{25}{54}$

Hence, $P(\bar{B})$ = 1 - $\frac{25}{54}$ =$\frac{29}{54}$

Hence, the odds against event B = $P(\bar{B})$ / P(B) = $\frac{29}{25}$.