A and B are mutually exclusive events of a random experiment and $\mathrm{P}(\mathrm{B}) \neq 1$, then…

A and B are mutually exclusive events of a random experiment and $\mathrm{P}(\mathrm{B}) \neq 1$, then $\mathrm{P}\left(\mathrm{A} \mid \mathrm{B}^{\mathrm{C}}\right)=$
  1. $\frac{\mathrm{P}(\mathrm{A})}{1-\mathrm{P}(\mathrm{B})}$
  2. $\frac{\mathrm{P}(\mathrm{B})}{1-\mathrm{P}(\mathrm{A})}$
  3. $\frac{\mathrm{P}(\mathrm{A})}{1+\mathrm{P}(\mathrm{B})}$
  4. $\frac{\mathrm{P}(\mathrm{A})}{\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})}$

Solution

$\because \mathrm{A}$ and $\mathrm{B}$ are mutually exclusive events. $\begin{aligned} & \therefore P(A \cap B)=0 \\ & \text { Now, } P\left(A \mid B^c\right)=\frac{P\left(A \cap B^c\right)}{P\left(B^c\right)}=\frac{P(A)-P(A \cap B)}{1-P(B)} \\ & =\frac{P(A)}{1-P(B)}\end{aligned}$

Asked in: AP EAMCET 2023 (17 May Shift 2)

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