Question

For any two events $A$ and $B$ in a sample space

• A. $P\left(A/B\right)\ge \frac{P\left(A\right)+P\left(B\right)-1}{P\left(B\right)},P\left(B\right)\ne 0$, is always true.
• B. $P(A\cap B)=P\left(A\right)-P(\stackrel{\_}{A}\cap \stackrel{\_}{B})$ does not hold.
• C. $P(A\cup B)=1-P\left(\stackrel{\_}{A}\right)P\left(\stackrel{\_}{B}\right)$, if $A$ and $B$ are independent
• D. $P(A\cup B)=1-P\left(\stackrel{\_}{A}\right)P\left(\stackrel{\_}{B}\right)$, if $A$ and $B$ are disjoint.

Answer 1

Answer: (A), (B), (C)

$P\left(A/B\right)\ge \frac{P\left(A\right)+P\left(B\right)-1}{P\left(B\right)},P\left(B\right)\ne 0$, is always true.
$P(A\cap B)=P\left(A\right)-P(\stackrel{\_}{A}\cap \stackrel{\_}{B})$ does not hold.
$P(A\cup B)=1-P\left(\stackrel{\_}{A}\right)P\left(\stackrel{\_}{B}\right)$, if $A$ and $B$ are independent

Going with the options:
(a) $P\left(\frac{A}{B}\right)=\frac{P(A\cap B)}{P\left(B\right)}$
Now, $P(A\cap B)=P\left(A\right)+P\left(B\right)-P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cup B)-P(\overline{A}\cap \overline{B})+P(\overline{A}\cap \overline{B})$
Thus, $P(A\cap B)=P\left(A\right)+P\left(B\right)-1+P(\overline{A}\cap \overline{B})=>P(A\cap B)\ge P\left(A\right)+P\left(B\right)-1$
(Using the relation $P(A\cup B)+P(\overline{A}\cap \overline{B})=1$)
Hence, $P\left(\frac{A}{B}\right)\ge \frac{P\left(A\right)+P\left(B\right)-1}{P\left(B\right)}$. So, (a) is true.
(b) $P(A\cap B)=P\left(A\right)-P(A\cap \overline{B})$. So, (b) is true.
(c) $P(A\cup B)=1-P(\overline{A}\cap \overline{B})=1-P\left(\overline{A}\right)P\left(\overline{B}\right)$ if A and B are independent. So, (c) is true.
(d) Unless A and B are mentioned as independent, $P(\overline{A}\cap \overline{B})$ cannot be written as $P\left(\overline{A}\right)P\left(\overline{B}\right)$. So, (d) is not true.