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  • State True or False for the statements in the Exercise. If A and B are two events such that P(A) > 0 and P(A) + P(B) >1, then P(B|A)>=1

State True or False for the statements in the Exercise.
If A and B are two events such that P(A) > 0 and P(A) + P(B) >1, then\mathrm{P}(\mathrm{B} \mid \mathrm{A}) \geq 1-\frac{\mathrm{P}\left(\mathrm{B}^{\prime}\right)}{\mathrm{P}(\mathrm{A})}

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Solution

True
\operatorname{As}, \mathrm{P}(\mathrm{B} \mid \mathrm{A})=\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{A})}=\frac{\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{AUB})}{\mathrm{P}(\mathrm{A})} \\\Rightarrow P(B \mid A)=\frac{P(A)+1-P\left(B^{\prime}\right)-P(A \cup B)}{P(A)} \\\Rightarrow \mathrm{P}(\mathrm{B} \mid \mathrm{A})=1-\frac{\mathrm{P}\left(\mathrm{B}^{\prime}\right)}{\mathrm{P}(\mathrm{A})}+\frac{1-\mathrm{P}(\mathrm{AUB})}{\mathrm{P}(\mathrm{A})}
The above equation means that:
\\P(B \mid A) \geq 1-\frac{P\left(B^{\prime}\right)}{P(A)} \\ \frac{1-\mathrm{P}(\mathrm{AUB})}{\mathrm{P}(\mathrm{A})}
we need to add \mathrm{P}(\mathrm{A}) \quad to get the equal term
\Rightarrow LHS is greater.

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