Q. If $A, B$ and $C$ are exhaustive events satisfying $P(A \cup B \cap \bar{C})=\frac{1}{5}, P(B \cap C)-P(A \cap B \cap C)=\frac{1}{15}$ and $P ( A \cap C )=\frac{1}{10}$, then $P \left( C \cap( A \cup B )^{\prime}\right)$ is equal to
Probability - Part 2
Solution:
From above figure,
$P\left(C \cap(A \cup B)'\right)=1-\left(\frac{1}{10}+\frac{1}{15}+\frac{1}{5}\right)$
$=\frac{30-3-2-6}{30}=\frac{19}{30} $
