| 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.
Correct Answer: A
Solution :
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,\[P\left( \frac{A}{B} \right)=\,\frac{P(A\cap B)}{P(B)}\] \[P(\overline{B})=P[(A\cup \overline{A})\cap \overline{B}]\] \[=P[(A\cap \overline{B})\cup \,(\overline{A}\cap \overline{B})]\] \[=P(A\cap \overline{B})+P(\overline{A}\cap \overline{B})\]
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