Q69 / 100 · BITSAT 2024 (Memory Based Paper 1) · Mathematics

For two events A and B , if $\mathrm{P}(\mathrm{A})=\mathrm{P}\left(\frac{\mathrm{A}}{\mathrm{B}}\right)=\frac{1}{4}$ and $P\left(\frac{B}{A}\right)=\frac{1}{2}$, then which of the following is not true?

  • A
    A and B are independent
  • B
    $\mathrm{P}\left(\frac{\mathrm{A}^{\prime}}{\mathrm{B}}\right)=\frac{3}{4}$
  • C
    $\mathrm{P}\left(\frac{\mathrm{B}^{\prime}}{\mathrm{A}^{\prime}}\right)=\frac{1}{2}$
  • Correct answer
    D
    None of these

Solution

$\mathrm{P}\left(\frac{\mathrm{B}}{\mathrm{A}}\right)=\frac{1}{2} \Rightarrow \frac{\mathrm{P}(\mathrm{B} \cap \mathrm{A})}{\mathrm{P}(\mathrm{A})}=\frac{1}{2} \Rightarrow \mathrm{P}(\mathrm{B} \cap \mathrm{A})=\frac{1}{8}$ $\mathrm{P}\left(\frac{\mathrm{A}}{\mathrm{B}}\right)=\frac{1}{4} \Rightarrow \frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{B})}=\frac{1}{4} \Rightarrow \mathrm{P}(\mathrm{B})=\frac{1}{2}$ $\mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{1}{8}=\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{B})$ $\therefore$ Events A and B are independent. Now, $\mathrm{P}\left(\frac{\mathrm{A}^{\prime}}{\mathrm{B}}\right)=\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{B})}=\frac{\mathrm{P}(\mathrm{A}) \mathrm{P}(\mathrm{B})}{\mathrm{P}(\mathrm{B})}=\frac{3}{4}$ and $\mathrm{P}\left(\frac{\mathrm{B}^{\prime}}{\mathrm{A}^{\prime}}\right)=\frac{\mathrm{P}\left(\mathrm{B}^{\prime} \cap \mathrm{A}^{\prime}\right)}{\mathrm{P}\left(\mathrm{A}^{\prime}\right)}=\frac{\mathrm{P}\left(\mathrm{B}^{\prime}\right) \mathrm{P}\left(\mathrm{A}^{\prime}\right)}{\mathrm{P}\left(\mathrm{A}^{\prime}\right)}=\frac{1}{2}$

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

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