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. $P(A\cap \overline{B})=P\left(A\right)-P(A\cap B)$, does not hold
• C. $P(A\cup B)=1-P\left(\overline{A}\right)P\left(\overline{B}\right)$, if $A$ and $B$ are independent
• D. $P(A\cup B)=1-P\left(\overline{A}\right)P\left(\overline{B}\right)$, if $A$ and $B$ are disjoint

Answer 1

Answer: (A), (C)

The correct options are
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
C $P(A\cup B)=1-P\left(\overline{A}\right)P\left(\overline{B}\right)$, if $A$ and $B$ are independent

We know that
$P\left(\frac{A}{B}\right)=\frac{P(A\cap B)}{P\left(B\right)}$
$=\frac{P\left(A\right)+P\left(B\right)-P(A\cup B)}{P\left(B\right)}$
Since, $P(A\cup B)<1,therefore$
$-P(A\cup B)>-1$
$\Rightarrow P\left(A\right)+P\left(B\right)-+(A\cup B)>P\left(A\right)+P\left(B\right)-1$
$\Rightarrow \frac{P\left(A\right)+P\left(B\right)-P(A\cup B)}{P\left(B\right)}>\frac{P\left(A\right)+P\left(B\right)-1}{P\left(B\right)}$
$\Rightarrow P\left(\frac{A}{B}\right)>\frac{P\left(A\right)+P\left(B\right)-1}{P\left(B\right)}$
Thus option (A) is correct and (C) is also correct.
Since, $P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)$
$P(A\cup B)=P\left(A\right)+P\left(B\right)-P\left(A\right)P\left(B\right)$
If A and B are independent
$=1[1-P(A\left)\right][1-P(B\left)\right]$
$=1-P\left(\overline{A}\right)P\left(\overline{B}\right)$