$A$ and $B$ are independent events of a random experiment if and only if
$A$ and $B$ are independent events of a random experiment if and only if
- $\mathrm{P}(\mathrm{A} \mid \mathrm{B}) \neq \mathrm{P}(\mathrm{A} \cap \mathrm{B})$
- $\mathrm{P}(\mathrm{A} \mid \mathrm{B})=\mathrm{P}(\mathrm{B} \mid \mathrm{A})$
- $\mathrm{P}(\mathrm{A} \mid \mathrm{B}) \neq \mathrm{P}\left(\mathrm{A} \mid \mathrm{B}^{\mathrm{C}}\right)$
- $P(A \mid B)=P\left(A \mid B^C\right)$
Solution
Given that $A$ and $B$ are independent events.
$\therefore \quad P(A \cap B)=P(A) \cdot P(B)$
$P(A \mid B)=\frac{P(A \cap B)}{P(B)}=\frac{P(A) \cdot P(B)}{P(B)}=P(A)$ ...(i)
And $P(B \mid A)=\frac{P(B \cap A)}{P(A)}=\frac{P(B) \cdot P(A)}{P(A)}=P(B)$
$\begin{aligned} & P\left(A \mid B^C\right)=\frac{P\left(A \cap B^C\right)}{P\left(B^C\right)}=\frac{P(A)-P(A \cap B)}{P\left(B^C\right)} \\ & =\frac{P(A)-P(A) \cdot P(B)}{P\left(B^C\right)}=\frac{P(A)[1-P(B)]}{P\left(B^C\right)}\end{aligned}$
$\Rightarrow P\left(A \mid B^C\right)=\frac{P(A) \cdot P\left(B^C\right)}{P\left(B^C\right)}=P(A)$ ...(ii)
From eqns. (i) and (ii)
$P(A \mid B)=P\left(A \mid B^C\right)$
Asked in: AP EAMCET 2023 (16 May Shift 1)
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