Let A and B be independent events such that \(P(A)=\displaystyle\frac{1}{2}\) and \(P(A\cup B)=\displaystyle\frac{3}{5}\). Which of the following are correct?
(A) \(P(B)=\displaystyle\frac{1}{5}\)
(B) \(P(A\cap B)=\displaystyle\frac{1}{10}\)
(C) \(P(A'\cap B')=\displaystyle\frac{2}{5}\)
(D) \(P(A|B)=\displaystyle\frac{1}{5}\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A), (B) and (C) only
Let P(B) = p. For independent events \(P(A\cap B) = P(A)P(B)\), so the addition theorem gives
\(P(A\cup B) = P(A)+P(B)-P(A)P(B)\)
\( \displaystyle \frac{3}{5} = \frac{1}{2}+p-\frac{p}{2} \)
\( \displaystyle \frac{p}{2} = \frac{1}{10} \)
\( \displaystyle p = \frac{1}{5} \)
Then
\( \displaystyle P(A\cap B) = \frac{1}{2}\times\frac{1}{5} = \frac{1}{10} \)
By De Morgan’s law, \(A'\cap B'\) is the complement of \(A\cup B\):
\( \displaystyle P(A'\cap B') = 1-\frac{3}{5} = \frac{2}{5} \)
For independent events, \(P(A|B) = P(A) = \frac{1}{2}\).
So \(P(B) = \frac{1}{5}\), \(P(A\cap B) = \frac{1}{10}\), \(P(A'\cap B') = \frac{2}{5}\) and \(P(A|B) = \frac{1}{2}\).
Checking the statements
(A) \(P(B)=\frac{1}{5}\) — Correct. \(\frac{p}{2} = \frac{1}{10}\) gives \(p = \frac{1}{5}\).
(B) \(P(A\cap B)=\frac{1}{10}\) — Correct. \(\frac{1}{2}\times\frac{1}{5} = \frac{1}{10}\).
(C) \(P(A'\cap B')=\frac{2}{5}\) — Correct. \(1-\frac{3}{5} = \frac{2}{5}\).
(D) \(P(A|B)=\frac{1}{5}\) — Incorrect. For independent events \(P(A|B) = P(A) = \frac{1}{2}\).