Question

f A and B are any two events of a sample space S, then P.A/ D P.A and B/ 􀀀 P.B/.

Answer 1

a)

$$P(A|B)=\dfrac{P(A\cap B)}{P(B)}$$

$$=\dfrac{P(A)+P(B)-P(A\cup B)}{P(B)}$$

$$=\dfrac{P(A)+P(B)-(1-P(\overline{A}\cap \overline{B}))}{P(B)}$$

$$=\dfrac{P(A)+P(B)-1+P(\overline{A}\cap \overline{B})}{P(B)}$$

$$P(\overline{A}\cap \overline{B})\geq0$$

Hence

$$P(A|B)\geq\dfrac{P(A)+P(B)-1}{P(B)}$$

The statement is True.

b)

$$P(A\cap B)=P(A)-P(A\cap\overline{B})$$

Then

$$P(A\cap B)=P(A)-P(\overline{A}\cap\overline{B}), \text{ does not hold}$$

The statement is True.

c)

$$P(A\cup B)=1-P(\overline{A}\cap \overline{B})$$

If $A$ and $B$ are independent, then

$$P(\overline{A}\cap \overline{B})=P(\overline{A})P(\overline{B})$$

Hence

$$P(A\cup B)=1-P(\overline{A}\cap \overline{B})=1-P(\overline{A})P(\overline{B})$$

The statement $P(\overline{A}\cap \overline{B})=P(\overline{A})P(\overline{B}),$ if $A$ and $B$ are independent, is True.

d) Unless $A$ and $B$ are mentioned as independent, $P(\overline{A}\cap \overline{B})$ cannot be written as $P(\overline{A})P(\overline{B}).$

The statement $P(\overline{A}\cap \overline{B})=P(\overline{A})P(\overline{B}),$ if $A$ and $B$ are disjoint, is False.