Q63 / 125 · BITSAT 2011 · Mathematics

If \(A\) and \(\mathrm{B}\) are mutually exclusive events and if \(\mathrm{P}(\mathrm{B})=\frac{1}{3}, \mathrm{P}(\mathrm{A} \cup \mathrm{B})=\frac{13}{21}\), then \(\mathrm{P}(\mathrm{A})\) is equal to

  • A
    \(1 / 7\)
  • B
    \(4 / 7\)
  • Correct answer
    C
    \(2 / 7\)
  • D
    \(5 / 7\)

Solution

For mutually exclusive events \(\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B}) \Rightarrow \mathrm{P}(\mathrm{A})=\frac{2}{7}\)

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

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