Q43 / 120 · VITEEE 2012 · Mathematics

The probability that atleast one of the events A and $\mathrm{B}$ occurs is $0.6$. If $\mathrm{A}$ and $\mathrm{B}$ occur simultaneously with probability $0.2$, then $\mathrm{P}(\overline{\mathrm{A}})+\mathrm{P}(\overline{\mathrm{B}})$ is

  • A
    $0.4$
  • B
    $0.8$
  • Correct answer
    C
    $1.2$
  • D
    $1.4$

Solution

$\mathrm{P}(\mathrm{A} \cup \mathrm{B})=0.6$ and $\mathrm{P}(\mathrm{A} \cap \mathrm{B})=0.2$ we know that $$ \begin{array}{l} \mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B}) \\ \Rightarrow 0.6=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-0.2 \\ \Rightarrow \mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})=0.8 \\ \Rightarrow 1-\mathrm{P}(\overline{\mathrm{A}})+1-\mathrm{P}(\overline{\mathrm{B}})=0.8 \\ \Rightarrow-[\mathrm{P}(\overline{\mathrm{A}})+\mathrm{P}(\overline{\mathrm{B}})]=0.8-2 \\ \Rightarrow \mathrm{P}(\overline{\mathrm{A}})+\mathrm{P}(\overline{\mathrm{B}})=1.2 \end{array} $$

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

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