• # question_answer Direction: For the following questions, chose the correct answer from the codes [a], [b] [c] and [d] defined as follows. Let us define a two events A and B such that $0<P(A),\,P(B)<1$. Statement I The conditional probability relation between A and B is $P\left( \frac{A}{B} \right)+P\left( \frac{\overline{A}}{\overline{B}} \right)=\frac{3}{2}$ Statement II If the event B is already occurred, then $P\left( \frac{A}{B} \right)=\frac{P(A\cap B)}{P(B)}$and$P(\overline{B})=P(A\cap B)+P(\overline{A}\cap \overline{B})$. A)  Statement I is true, Statement II is also true and Statement II is the correct explanation of Statement I. B)  Statement I is true, Statement II is true and Statement II is not the correct explanation of Statement I. C)  Statement I is true, Statement II is false. D)  Statement I is false, Statement II is true.

I.P $\left( \frac{A}{\overline{B}} \right)+P\left( \frac{\overline{P}}{\overline{B}} \right)=\frac{P(A\cap \overline{B})}{P(\overline{B})}+\frac{P(\overline{A}\cap \overline{B})}{P(\overline{B})}$ $=\frac{P(A\cap \overline{B})+P(\overline{A}\cap \overline{B})}{P(\overline{B})}$$=\frac{P(\overline{B})}{P(\overline{B})}=1$ II. By definition,$P\left( \frac{A}{B} \right)=\,\frac{P(A\cap B)}{P(B)}$ $P(\overline{B})=P[(A\cup \overline{A})\cap \overline{B}]$ $=P[(A\cap \overline{B})\cup \,(\overline{A}\cap \overline{B})]$ $=P(A\cap \overline{B})+P(\overline{A}\cap \overline{B})$