Q262 / 270 · NDA 2024 (Phase 1) · Mathematics

Let A and B be two events such that $\mathrm{P}(\mathrm{A} \cup \mathrm{B}) \geq$ 0.75 and $0.125 \leq \mathrm{P}(\mathrm{A} \cap \mathrm{B}) \leq 0.375$. What is the maximum value of $\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})$ ?

  • A
    0.75
  • Correct answer
    B
    1.125
  • C
    1.375
  • D
    1.625

Solution

Given : $\quad \mathrm{P}(\mathrm{A} \cup \mathrm{B}) \geq 0.75$ and $0.125 \leq \mathrm{P}(\mathrm{A} \cap \mathrm{B}) \leq 0.375$ $\mathrm{P}(\mathrm{A} \cap \mathrm{B}) \leq 0.375$ $\begin{aligned} & \Rightarrow \quad \mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cup \mathrm{B}) \leq 0.375 \\ & \Rightarrow \quad \mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B}) \leq 0.375+\mathrm{P}(\mathrm{A} \cup \mathrm{B}) \\ & \Rightarrow \quad \mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B}) \leq 0.375+0.75 \\ & \Rightarrow \quad \mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B}) \leq 1.125\end{aligned}$ Maximum value of $\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})=1.125$.

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

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