MHT CET · Maths · Probability
If \(\mathrm{A}\) and \(\mathrm{B}\) are two events such that \(\mathrm{P}(\mathrm{A})=\frac{1}{3}\), \(\mathrm{P}(\mathrm{B})=\frac{1}{5}, \mathrm{P}(\mathrm{A} \cup \mathrm{B})=\frac{1}{3}\), then the value of \(\mathrm{P}\left(\mathrm{A}^{\prime} / \mathrm{B}^{\prime}\right)+\mathrm{P}\left(\mathrm{B}^{\prime} / \mathrm{A}^{\prime}\right)\) is
- A \(\frac{5}{6}\)
- B \(1\)
- C \(\frac{1}{6}\)
- D \(\frac{11}{6}\)
Answer & Solution
Correct Answer
(D) \(\frac{11}{6}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} \mathrm{P}\left(\mathrm{A}^{\prime} / \mathrm{B}^{\prime}\right) & =\frac{\mathrm{P}\left(\mathrm{A}^{\prime} \cap \mathrm{B}^{\prime}\right)}{\mathrm{P}\left(\mathrm{B}^{\prime}\right)} \\ & =\frac{1-\mathrm{P}(\mathrm{A} \cup \mathrm{B})}{1-\mathrm{P}(\mathrm{B})} \\ & =\frac{1-\frac{1}{3}}{1-\frac{1}{5}}=\frac{\frac{2}{3}}{\frac{3}{5}}=\frac{5}{6} \\ \mathrm{P}\left(\mathrm{B}^{\prime} / \mathrm{A}^{\prime}\right) & =\frac{\mathrm{P}\left(\mathrm{A}^{\prime} \cap \mathrm{B}^{\prime}\right)}{\mathrm{P}\left(\mathrm{A}^{\prime}\right)} \\ & =\frac{1-\mathrm{P}(\mathrm{A} \cup \mathrm{B})}{1-\mathrm{P}(\mathrm{A})} \\ & =\frac{\frac{2}{3}}{1-\frac{1}{3}}=\frac{\frac{2}{3}}{\frac{2}{3}}=1 \\ \mathrm{P}\left(\mathrm{A}^{\prime} / \mathrm{B}^{\prime}\right) & +\mathrm{P}\left(\mathrm{B}^{\prime} / \mathrm{A}^{\prime}\right)=\frac{5}{6}+1=\frac{11}{6}\end{aligned}\)
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