Question

For three non-impossible events $A,B$ and $C$ and $P(A\cap B\cap C)=0$, $P(A\cup B\cup C)=\frac{3}{4}$, $P(A\cap B)=\frac{1}{3}$ and $P\left(C\right)=\frac{1}{6}$. The probability, exactly one of $A$ or $B$ occurs but $C$ doesn't occur is -

• A. $\frac{1}{12}$
• B. $\frac{5}{6}$
• C. $\frac{1}{4}$
• D. $\frac{2}{3}$

Answer 1

The correct option is C $\frac{1}{4}$


$P\left(\right(A\cap \overline{B})\cup (\overline{A}\cap B)\cap \overline{C})=a+bP(A\cap B\cap C)=0=gP(A\cap B)=d+g=\frac{1}{3}\Rightarrow d=\frac{1}{3}P\left(C\right)=c+f+g+e=\frac{1}{6}\Rightarrow c+e+f=\frac{1}{6}P(A\cup B\cup C)=a+b+c+d+e+f+g=\frac{3}{4}\Rightarrow a+b=\frac{1}{4}$