MHT CET · Maths · Probability
\(\mathrm{A}\) and \(\mathrm{B}\) are independent events with \(\mathrm{P}(\mathrm{A})=\frac{1}{4}\) and \(P(A \cup B)=2 P(B)-P(A)\), then \(P(B)\) is
- A \(\frac{1}{4}\)
- B \(\frac{3}{5}\)
- C \(\frac{2}{3}\)
- D \(\frac{2}{5}\)
Answer & Solution
Correct Answer
(D) \(\frac{2}{5}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{P}(\mathrm{A} \cup \mathrm{B})=2 \mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A}) \)
\( \therefore \mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})=2 \mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A}) \)
\( \therefore \mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{B})=2 \mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A})\)
\(\ldots[\because A\) and \(B\) are independent events \(]\)
\(\therefore \mathrm{P}(\mathrm{B})+\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{B})=2 \mathrm{P}(\mathrm{A}) \)
\( \therefore \mathrm{P}(\mathrm{B})=\frac{2 \mathrm{P}(\mathrm{A})}{(1+\mathrm{P}(\mathrm{A}))}=\frac{2 \times \frac{1}{4}}{\left(1+\frac{1}{4}\right)}=\frac{2}{5}\)
\( \therefore \mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})=2 \mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A}) \)
\( \therefore \mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{B})=2 \mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A})\)
\(\ldots[\because A\) and \(B\) are independent events \(]\)
\(\therefore \mathrm{P}(\mathrm{B})+\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{B})=2 \mathrm{P}(\mathrm{A}) \)
\( \therefore \mathrm{P}(\mathrm{B})=\frac{2 \mathrm{P}(\mathrm{A})}{(1+\mathrm{P}(\mathrm{A}))}=\frac{2 \times \frac{1}{4}}{\left(1+\frac{1}{4}\right)}=\frac{2}{5}\)
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