Question

If $\text{A}$ and $\text{B}$ are two events such that $\text{P(AUB)}$$=$$\text{P(A∩B)}$, then the true relation is

• A. $\text{P(A)+P(B)=0}$
• B. $\text{P(A) + P(B) = P(A)P(B/A)}$
• C. $\text{P(A)+P(B) = 2P(A)P(B/A)}$
• D. None of these

Answer 1

The correct option is C

$\text{P(A)+P(B) = 2P(A)P(B/A)}$

Explanation for the correct option :

Step $\mathbf{1}$. Find the value of$\mathbf{\text{P(A)+P(B)}}$

Given, $\text{P(AUB)}$$=$$\text{P(A∩B)}$

We know that,

$\Rightarrow$ $\text{P(A∩B)}$ $=$$\text{P(A)}$$+$$\text{P(B)}$$-$$\text{P(A∩B)}$ $\mathbf{\text{P(AUB) = P(A) + P(B) - P(A∩B)}}$

$\Rightarrow$ $\text{2 P(A∩B)}$$=$$\text{P(A)}$$+$$\text{P(B)}$ $....\left(1\right)$

Step $\mathbf{2}$. Multiply and divide equation $\left(1\right)$ with $\mathbf{\text{P(A)}}$

So, equation $\left(1\right)$ becomes

$\text{2 P}\left(\text{A}\right)\left(\text{P(A∩B)/P}\left(\text{A}\right)\right)$ $=$$\text{P(A)}$$+$$\text{P(B)}$

As, $\mathbf{\text{P(A∩B)/P}}\left(\text{A}\right)$$\mathbf{=}$$\mathbf{\text{P}}\left(\raisebox{1ex}{$\text{B}$}\!\left/ \!\raisebox{-1ex}{$\text{A}$}\right.\right)$

$\Rightarrow$ $\text{2 P}\left(\text{A}\right)$$\text{P}\left(\raisebox{1ex}{$\text{B}$}\!\left/ \!\raisebox{-1ex}{$\text{A}$}\right.\right)$$=$$\text{P(A)}$$+$$\text{P(B)}$

Therefore, $\mathbf{\text{P(A)}}$$\mathbf{+}$$\mathbf{\text{P(B)}}$$\mathbf{=}$$\mathbf{\text{2 P}}\left(\text{A}\right)$$\mathbf{\text{P}}\left(\raisebox{1ex}{$\text{B}$}\!\left/ \!\raisebox{-1ex}{$\text{A}$}\right.\right)$

Hence, option $\mathbf{"}\mathbf{\text{C"}}$ is correct.