A) \[P(A\cup B)\ge \frac{2}{3}\]
B) \[\frac{1}{6}\le P(A\cap B)\le \frac{1}{2}\]
C) \[\frac{1}{6}\le P(A'\cap B)\le \frac{1}{2}\]
D) All of these
Correct Answer: D
Solution :
We know that \[\operatorname{P}(A\cup B)\ge \max \,\left\{ P(A),P(B) \right\}\,\,=\,\,\frac{2}{3}\] \[\operatorname{P}(A\cup B) \le \,\,min \left\{ P\left( A \right), P\left( B \right) \right\}=\,\,\frac{1}{2}\] \[\operatorname{and}\,\,P(A\cap B)=\,\,P(A)+P(B)-P(A\cup B)\] \[\ge \,\,P\left( A \right)+P\left( B \right)-1=\frac{1}{6}\] \[\Rightarrow \,\,\,\frac{1}{6}\le P(A\cap B)\le \frac{1}{2}\] \[P(A'\,\cap B)=\,P(B)-\,P(A\cap B)\] \[\therefore \,\,\frac{2}{3}-\frac{1}{2}\le P(A'\cap B)\le \frac{2}{3}-\frac{1}{6}\] \[\Rightarrow \,\,\,\,\,\frac{1}{6}\le P(A'\cap B)\le \frac{1}{2}\]
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