Question

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

• A. $P\left(A/B\right)\ge \left[\left(P\left(A\right)+P\left(B\right)-1\right)/P\left(B\right)\right],P\left(B\right)\ne 0$ is always true
• B. $P\left(A\cap B^{\text{'}}\right)=P\left(A\right)-P\left(A\cap B\right)$ does not hold
• C. $P\left(A\cup B\right)=1-P\left(A^{\text{'}}\right)P\left(B^{\text{'}}\right)$, if $A$ and $B$ are disjoint.
• D. None of the above

Answer 1

Answer: (A), (B)

$P\left(A\cap B^{\text{'}}\right)=P\left(A\right)-P\left(A\cap B\right)$ does not hold

Explanations for the correct option:

Step 1. Write the expression for the $P\left(\frac{A}{B}\right)$:

$\begin{array}{rcl}P\left(\frac{A}{B}\right)& =& \frac{P\left(A\cap B\right)}{P\left(B\right)}\\ & =& \frac{P\left(A\right)+P\left(B\right)-P\left(A\cup B\right)}{P\left(B\right)}\mathbf{[}\mathbf{\because }\mathbf{P}\mathbf{(}\mathbf{A}\mathbf{\cap }\mathbf{B}\mathbf{)}\mathbf{=}\mathbf{P}\mathbf{\left(}\mathbf{A}\mathbf{\right)}\mathbf{+}\mathbf{P}\mathbf{\left(}\mathbf{B}\mathbf{\right)}\mathbf{-}\mathbf{P}\mathbf{(}\mathbf{A}\mathbf{\cup }\mathbf{B}\mathbf{)}\mathbf{]}\\ & =& \frac{P\left(A\right)+P\left(B\right)-P\left(A\cup B\right)-P\left(A^{\text{'}}\cap B^{\text{'}}\right)+P\left(A^{\text{'}}\cap B^{\text{'}}\right)}{P\left(B\right)}\\ & =& \frac{P\left(A\right)+P\left(B\right)-(P\left(A\cup B\right)+P\left(A^{\text{'}}\cap B^{\text{'}}\right))+P\left(A^{\text{'}}\cap B^{\text{'}}\right)}{P\left(B\right)}\\ & =& \frac{P\left(A\right)+P\left(B\right)-1+P\left(A^{\text{'}}\cap B^{\text{'}}\right)}{P\left(B\right)}\mathbf{[}\mathbf{\because }\mathbf{P}\mathbf{(}\mathbf{A}\mathbf{\cup }\mathbf{B}\mathbf{)}\mathbf{+}\mathbf{P}\mathbf{(}{\mathbf{A}}^{\mathbf{\text{'}}}\mathbf{\cap }{\mathbf{B}}^{\mathbf{\text{'}}}\mathbf{)}\mathbf{=}\mathbf{1}\mathbf{]}\\ & =& \frac{P\left(A\right)+P\left(B\right)-1}{P\left(B\right)}+\frac{P\left(A^{\text{'}}\cap B^{\text{'}}\right)}{P\left(B\right)}\\ P\left(\frac{A}{B}\right)& \ge & \frac{P\left(A\right)+P\left(B\right)-1}{P\left(B\right)}\\ & & \end{array}$

Therefore, option (A) is correct.

Explanations for incorrect options:

Option (B): $\mathit{P}\left(A\cap B^{\text{'}}\right)\mathbf{=}\mathit{P}\left(A\right)\mathbf{-}\mathit{P}\left(A\cap B\right)$ does not hold

Step 2. Expression for the $\begin{array}{l}P\left(A\cap B\right)\end{array}$:

$\begin{array}{rcl}P\left(A\cap B\right)& =& P\left(A\right)-P\left(A\cap B\text{'}\right)\\ P\left(A\cap B^{\text{'}}\right)& =& P\left(A\right)-P\left(A\cap B\right)\end{array}$

Here, the expression is correct.

Therefore, the option (B) is correct.

Option (C): $\mathit{P}\left(A\cup B\right)\mathbf{=}\mathbf{1}\mathbf{-}\mathit{P}\left(A^{\text{'}}\right)\mathit{P}\left(B^{\text{'}}\right)$, if A and B are disjoint.

Step 3. Expression for the $P\left(A\cup B\right)$:

$\begin{array}{rcl}P\left(A\cup B\right)& =& 1-P\left(A\text{'}\cap B\text{'}\right)\end{array}$

For disjoint:

$\begin{array}{rcl}P\left(A\text{'}\cap B\text{'}\right)& \ne & P\left(A\text{'}\right)P\left(B\text{'}\right)\end{array}$

Here, the expression is incorrect.

Therefore, the option (C) is incorrect.

Hence, option (A) and (B) is the correct answer.