Question
If two events A and B are such that $\text{P}(\overline{\text{A}})=0.3,\text{P(B)}=0.4$ and $\text{P}(\text{A}\cap\overline{\text{B}})=0.5$ find $\text{P}\Big(\frac{\text{B}}{\overline{\text{A}}\cap\overline{\text{B}}}\Big).$
✓
Answer
According to Baye's Theorem
$\text{P}\Big(\frac{\text{B}}{\overline{\text{A}}\cap\overline{\text{B}}}\Big)=\frac{\text{P}(\text{B}\cap(\overline{\text{A}}\cap\overline{\text{B}}))}{\text{P}(\overline{\text{A}}\cap\overline{\text{B}})}$
$=\frac{\text{P}(\text{B}(\overline{\text{A}\cap\text{B}}))}{\text{P}(\overline{\text{A}}\cap\overline{\text{B}})}$
$=\frac{\text{P}(\overline{\text{B}}(\overline{\text{A}\cap\text{B}}))}{\text{P}(\overline{\text{A}}\cap\overline{\text{B}})}$
$=\frac{\text{P}(\overline{\text{B}}\cup(\text{A}\cup\text{B}))}{\text{P}(\overline{\text{A}}\cap\overline{\text{B}})}$
Now $\overline{\text{B}}\cap\text{B}=\cup=\phi$
So, $\text{P}(\overline{\text{B}}\cup(\text{A}\cup\text{B}))=\phi$
$\therefore\ \text{P}\Big(\frac{\text{B}}{\overline{\text{A}}\cap\overline{\text{B}}}\Big)=0$
$\text{P}\Big(\frac{\text{B}}{\overline{\text{A}}\cap\overline{\text{B}}}\Big)=\frac{\text{P}(\text{B}\cap(\overline{\text{A}}\cap\overline{\text{B}}))}{\text{P}(\overline{\text{A}}\cap\overline{\text{B}})}$
$=\frac{\text{P}(\text{B}(\overline{\text{A}\cap\text{B}}))}{\text{P}(\overline{\text{A}}\cap\overline{\text{B}})}$
$=\frac{\text{P}(\overline{\text{B}}(\overline{\text{A}\cap\text{B}}))}{\text{P}(\overline{\text{A}}\cap\overline{\text{B}})}$
$=\frac{\text{P}(\overline{\text{B}}\cup(\text{A}\cup\text{B}))}{\text{P}(\overline{\text{A}}\cap\overline{\text{B}})}$
Now $\overline{\text{B}}\cap\text{B}=\cup=\phi$
So, $\text{P}(\overline{\text{B}}\cup(\text{A}\cup\text{B}))=\phi$
$\therefore\ \text{P}\Big(\frac{\text{B}}{\overline{\text{A}}\cap\overline{\text{B}}}\Big)=0$
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