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 the above
Correct Answer: D
Solution :
We have \[P(A\cup B)\ge \max .\]\[\{P(A),\,P(B)=\frac{2}{3}\] \[P(A\cap B)\le \min .\]\[\{P(A),P(B)\}=\frac{1}{2}\] and\[P(A\cap B)=P(A)+P(B)-P(A\cup B)\ge P(A)-P(B)-1=\frac{1}{6}\] \[\Rightarrow \frac{1}{6}\le P(A\cap B)\le \frac{1}{2}\] \[P\,({A}'\cap B)=P(B)-P(A\cap B)\] Hence \[\frac{2}{3}-\frac{1}{2}\le P({A}'\cap B)\le \frac{2}{3}-\frac{1}{6}\] \[\Rightarrow \frac{1}{6}\le P({A}'\cap B)\le \frac{1}{2}\].You need to login to perform this action.
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