If \(P(\mathrm{A}) = \displaystyle\frac{3}{10}\), \(P(\mathrm{B}) = \displaystyle\frac{2}{5}\) and \(P(\mathrm{A} \cup \mathrm{B}) = \displaystyle\frac{3}{5}\), then \(P\left(\displaystyle\frac{\mathrm{B}}{\mathrm{A}}\right) + P\left(\displaystyle\frac{\mathrm{A}}{\mathrm{B}}\right)\) is equal to:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle\frac{7}{12}\)
By the addition rule, \(P(\mathrm{A} \cup \mathrm{B}) = P(\mathrm{A}) + P(\mathrm{B}) - P(\mathrm{A} \cap \mathrm{B})\), so
\( \displaystyle P(\mathrm{A} \cap \mathrm{B}) = \frac{3}{10} + \frac{4}{10} - \frac{6}{10} = \frac{1}{10} \)
By the definition of conditional probability:
\( \displaystyle P\left(\frac{\mathrm{B}}{\mathrm{A}}\right) = \frac{P(\mathrm{A} \cap \mathrm{B})}{P(\mathrm{A})} = \frac{1/10}{3/10} = \frac{1}{3} \)
\( \displaystyle P\left(\frac{\mathrm{A}}{\mathrm{B}}\right) = \frac{P(\mathrm{A} \cap \mathrm{B})}{P(\mathrm{B})} = \frac{1/10}{4/10} = \frac{1}{4} \)
Adding:
\( \displaystyle \frac{1}{3} + \frac{1}{4} = \frac{7}{12} \)