Q257 / 270 · NDA 2015 (Phase 2) · Mathematics

Two events A and B are such that $\mathrm{P}(\operatorname{not} \mathrm{B})=0.8, \mathrm{P}(\mathrm{A} \cup \mathrm{B})=$ 0.5 and $\mathrm{P}(\mathrm{A} \mid \mathrm{B})=0.4$. Then $\mathrm{P}(\mathrm{A})$ is equal to

  • A
    0.28
  • B
    0.32
  • Correct answer
    C
    0.38
  • D
    None of the above

Solution

$\begin{aligned} & \because \mathrm{P}(\overline{\mathrm{B}})=0.8 \Rightarrow \mathrm{P}(\mathrm{B})=0.2 \\ & \mathrm{P}(\mathrm{A} \cup \mathrm{B})=0.5 \& \mathrm{P}(\mathrm{A} \mid \mathrm{B})=0.4 \\ & \because \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\mathrm{P}(\mathrm{B}) \times \mathrm{P}(\mathrm{A} \mid \mathrm{B})=0.2 \times 0.4=0.08 \\ & \& \mathrm{P}(\mathrm{A})=\mathrm{P}(\mathrm{A} \cup \mathrm{B})-\mathrm{P}(\mathrm{B})+\mathrm{P}(\mathrm{A} \cap \mathrm{B}) \\ & \mathrm{P}(\mathrm{A})=0.5-0.2+0.08=0.38\end{aligned}$

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

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