Q264 / 270 · NDA 2024 (Phase 1) · Mathematics

$\mathrm{A}, \mathrm{B}$, and C are three events such that $\mathrm{P}(\mathrm{A})=0.6$, $\mathrm{P}(\mathrm{B})=0.4, \mathrm{P}(\mathrm{C})=0.5, \mathrm{P}(\mathrm{A} \cup \mathrm{B})=0.8, \mathrm{P}(\mathrm{A} \cap \mathrm{C})$ $=0.3$ and $\mathrm{P}(\mathrm{A} \cap \mathrm{B} \cap \mathrm{C})=0.2$ and $\mathrm{P}(\mathrm{A} \cup \mathrm{B} \cup \mathrm{C})$ $\geq 0.85$. What is the maximum value of $\mathrm{P}(\mathrm{B} \cap \mathrm{C})$ ?

  • A
    0.1
  • B
    0.2
  • Correct answer
    C
    0.35
  • D
    0.45

Solution

$\begin{aligned} \mathrm{P}(\mathrm{~A})=0.6, \mathrm{P}(\mathrm{~B}) & =0.4, \mathrm{P}(\mathrm{C})=0.5 \\ \mathrm{P}(\mathrm{~A} \cap \mathrm{~B}) & =\mathrm{P}(\mathrm{~A})+\mathrm{P}(\mathrm{~B})-\mathrm{P}(\mathrm{~A} \cup \mathrm{~B}) \\ & =0.6+0.4-0.8=0.2 \end{aligned}$ Now, $\begin{aligned} \mathrm{P}(\mathrm{~A} \cup \mathrm{~B} \cup \mathrm{C})= & \mathrm{P}(\mathrm{~A})+\mathrm{P}(\mathrm{~B})+\mathrm{P}(\mathrm{C}) \\ & -\mathrm{P}(\mathrm{~A} \cup \mathrm{~B})-\mathrm{P}(\mathrm{~A} \cap \mathrm{C}) \\ & -\mathrm{P}(\mathrm{~B} \cap \mathrm{C})+\mathrm{P}(\mathrm{~A} \cap \mathrm{~B} \cap \mathrm{C}) \\ = & 1.5-0.2-0.3-\mathrm{P}(\mathrm{~B} \cap \mathrm{C})+0.2 \end{aligned}$ $\begin{array}{rlrl}\Rightarrow & \mathrm{P}(\mathrm{A} \cup \mathrm{B} \cup \mathrm{C}) & =1.2-\mathrm{P}(\mathrm{B} \cap \mathrm{C}) \\ \because & 0.85 & \leq \mathrm{P}(\mathrm{A} \cup \mathrm{B} \cup \mathrm{C}) \leq 1 \\ \Rightarrow & 0.85 & \leq 1.2-\mathrm{P}(\mathrm{B} \cap \mathrm{C}) \leq 1 \\ \Rightarrow & -0.35 & \leq-\mathrm{P}(\mathrm{B} \cap \mathrm{C}) \leq-0.2 \\ \Rightarrow & 0.2 & \leq \mathrm{P}(\mathrm{B} \cap \mathrm{C}) \leq 0.35\end{array}$ Maximum value of $\mathrm{P}(\mathrm{B} \cap \mathrm{C})=0.35$

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

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