Question

For any two events $A$ & $B$ in a sample space-

• A. $P\left(\frac{A}{B}\right)\ge \frac{P\left(A\right)+P\left(B\right)-1}{P\left(B\right)},P\left(B\right)\ne 0$ 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 $A$ & $B$ are independent
• D. $P(A\cup B)=1-P\left(\overline{A}\right)P\left(\overline{B}\right)$ , if $A$ & $B$ are disjoint

Answer 1

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

The correct options are
A $P\left(\frac{A}{B}\right)\ge \frac{P\left(A\right)+P\left(B\right)-1}{P\left(B\right)},P\left(B\right)\ne 0$ 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 $A$ & $B$ are independent

Given $A,B$ as two events in a sample place.
1) $P\left(A/B\right)\ge \frac{P\left(A\right)+P\left(B\right)-1}{P\left(B\right)}$
$\Rightarrow \frac{P\left(A\right)+P\left(B\right)-P(A\cap B)}{P\left(B\right)}\ge \frac{P\left(A\right)+P\left(B\right)-1}{P\left(B\right)}P\left(B\right)$(since $P\left(B\right)\ne 0$)
$\Rightarrow P\left(A\right)+P\left(B\right)-P(A\cap B)\ge \left(P\right(A)+P(B)-1)$
$\Rightarrow P(A\cap B)\le 1$which is always true. Hence give condition is true
2) $P(A\cap \overline{B})=P\left(A\right)+P\left(\overline{B}\right)-P(A\cup \overline{B})$
$P(A\cap \overline{B})=P\left(A\right)+P\left(\overline{B}\right)-\left(P\right(\overline{B})+P(A\cap B\left)\right)$ (since $B^c$ means everything except $B$, so union of $B^c$ with $A$, gives additionally the intersection part of $A$ and $B^c$ i.e.,$(A\cap B)$)
$P(A\cap \overline{B})=P\left(A\right)-P(A\cap B)$
3) $P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)=P\left(A\right)+P\left(B\right)-P\left(A\right)P\left(B\right)$
(Since $A,B$ are independent events $P(A\cap B)=P\left(A\right)P\left(B\right)$)