If P(A B)=(5)/(6), P(A B)=(1)/(3) and P(not A)=(1)/(2) , then which...

$$\begin{aligned} & P(A \cup B)=\frac{5}{6}, P(A \cap B)=\frac{1}{3}, P\left(A^{\prime}\right)=\frac{1}{2} \\ & P(A)=1-P\left(A^{\prime}\right)=1-\frac{1}{2}=\frac{1}{2} \\ & P(A \cup B)=P(A)+P(B)-P(A \cap B) \end{aligned}$$\frac{5}{6}=\frac{1}{2}+P(B)-\frac{1}{3}$(a)$P(B)=\frac{5}{6}-\frac{1}{2}+\frac{1}{3}=\frac{5-3+2}{6}=\frac{4}{6}=\frac{2}{3}$(b)$P(A \cap B)=\frac{1}{3}=\frac{1}{2} \cdot \frac{2}{3}=P(A) \cdot P(B)\therefore P(A \cap B)=P(A) \cdot P(B)$(c)$P(A \cup B)=\frac{5}{6}$$\begin{aligned} & P(A)+P(B)=\frac{1}{2}+\frac{2}{3}=\frac{7}{6} \\ & \therefore P(A \cup B) < P(A)+P(B) \end{aligned}$$$Hence, option (c) is not correct. (d)$P\left(A^{\prime} \cap B^{\prime}\right)=P\left(A^{\prime}\right) \cdot P\left(B^{\prime}\right)$$\begin{aligned} & 1-P(A \cup B)=1-\frac{5}{6}=\frac{1}{6} \\ & \text{and } P\left(A^{\prime}\right) \cdot P\left(B^{\prime}\right)=(1-P(A)) \cdot(1-P(B)) \\ & =\left(1-\frac{1}{2}\right) \cdot\left(1-\frac{2}{3}\right)=\frac{1}{2} \cdot \frac{1}{3}=\frac{1}{6}\end{aligned}$$$Hence,$P \left(A^{\prime} \cap B^{\prime}\right)=P \left(A^{\prime}\right)\cdotP \left(B^{\prime}\right).