Question

If $A$ and $B$ are independent events such that $P\left(A\right)>0,P\left(B\right)>0$, then

• A. $A$ and $B$ are mutually exclusive
• B. $A$ and $\overline{B}$ are dependent
• C. $\overline{A}$ and $B$ are dependent
• D. $P\left(\frac{A}{B}\right)+P\left(\frac{\overline{A}}{B}\right)=1$

Answer 1

The correct option is D $P\left(\frac{A}{B}\right)+P\left(\frac{\overline{A}}{B}\right)=1$

Since $A$ and $B$ are independent events, therefore
$P(A\cap B)=P\left(A\right)P\left(B\right),P\left(\frac{A}{B}\right)=P\left(A\right)$ and $P\left(\frac{\overline{A}}{B}\right)=P\left(\overline{A}\right).$
Now,$P(A\cap B)=P\left(A\right)P\left(B\right)\ne 0$
$\Rightarrow A$ and $B$ are not mutually exclusive
Since $A\cap B$ and $A\cap \overline{B}$ are mutually exclusive events such that
$(A\cap B)\cup (A\cap \overline{B})=A.$ Therefore,
$P(A\cap B)+P(A\cap \overline{B})=P\left(A\right)$
$\Rightarrow P\left(A\right)P\left(B\right)+P(A\cap \overline{B})=P\left(A\right)$
$\Rightarrow P(A\cap \overline{B})=P\left(A\right)-P\left(A\right)P\left(B\right)=P\left(A\right)(1-P\left(B\right))=P\left(A\right)P\left(\overline{B}\right)$
So, $A$ and $\overline{B}$ are independent events.
Similarly, $\overline{A}$ and $B$ are also independent events.
Now, $P\left(\frac{A}{B}\right)=P\left(A\right)$ and $P\left(\frac{\overline{A}}{B}\right)=P\left(\overline{A}\right)$
$\Rightarrow P\left(\frac{A}{B}\right)+P\left(\frac{\overline{A}}{B}\right)=P\left(A\right)+P\left(\overline{A}\right)=1.$