Q77 / 160 · AP EAMCET 2023 (16 May Shift 1) · Mathematics

$A$ and $B$ are independent events of a random experiment if and only if

  • A
    $\mathrm{P}(\mathrm{A} \mid \mathrm{B}) \neq \mathrm{P}(\mathrm{A} \cap \mathrm{B})$
  • B
    $\mathrm{P}(\mathrm{A} \mid \mathrm{B})=\mathrm{P}(\mathrm{B} \mid \mathrm{A})$
  • C
    $\mathrm{P}(\mathrm{A} \mid \mathrm{B}) \neq \mathrm{P}\left(\mathrm{A} \mid \mathrm{B}^{\mathrm{C}}\right)$
  • Correct answer
    D
    $P(A \mid B)=P\left(A \mid B^C\right)$

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) $P(A \mid B)=P\left(A \mid B^C\right)$

Hence, the correct answer is option: 'D'.

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