Question

If $A,B$ and $C$ are exhaustive events satisfying $P((A\cup B)\cap C)=\frac{1}{5}$, $P(B\cap C)-P(A\cap B\cap C)=\frac{1}{15}$ and $P(A\cap C)=\frac{1}{10}$ then $P(C\cap (A\bar{\cup }B))$ is equal to

• A. $\frac{17}{30}$
• B. $\frac{18}{30}$
• C. $\frac{19}{30}$
• D. $\frac{20}{30}$

Answer 1

Answer: (C)

We have,
$P\left(\left(A\cup B\right)\cap \bar{C}\right)=\frac{1}{5}$
$P\left(B\cap C\right)-P\left(A\cap B\cap C\right)=\frac{1}{15}$
$P\left(A\cap C\right)=\frac{1}{10}$
$P(C\cap (A\cup \bar{U}))=?$
Now,

Then,
$P\left(C\right)+P\left(\left(A\cup B\right)\cap \bar{C}\right)=1$
$\Rightarrow P\left(C\right)=1-\frac{1}{5}=\frac{4}{5}$
And, $\begin{array}{l}P(C\cap (A\cup \bar{B}))=P(C)-[P(B\cap C)-P(A\cap B\cap C)+P(A\cap C)]\\ \Rightarrow P(C\cap (A\bar{\cup }B))=\frac{4}{5}-\left(\frac{1}{15}+\frac{1}{10}\right)=\frac{19}{30}\end{array}$