Let A and B be independent events with \(P(A) = \displaystyle\frac{2}{5}\) and \(P(B) = \displaystyle\frac{1}{4}\). Consider the following statements:
(A) \(P(A\cap B') = \displaystyle\frac{3}{10}\)
(B) \(P(A\cup B) = \displaystyle\frac{11}{20}\)
(C) \(P(A'|B) = \displaystyle\frac{3}{4}\)
(D) \(P(B|A') = \displaystyle\frac{1}{4}\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A), (B) and (D) only
If A and B are independent, so are A and B’, A’ and B, and A’ and B’. Here
\( \displaystyle P(A') = \frac{3}{5},\quad P(B') = \frac{3}{4} \)
\( \displaystyle P(A\cap B') = \frac{2}{5}\times\frac{3}{4} = \frac{3}{10} \)
\( \displaystyle P(A\cap B) = \frac{2}{5}\times\frac{1}{4} = \frac{1}{10} \)
By the addition theorem:
\( \displaystyle P(A\cup B) = \frac{2}{5}+\frac{1}{4}-\frac{1}{10} \)
\( \displaystyle = \frac{8+5-2}{20} = \frac{11}{20} \)
For independent events a condition does not change the probability, so \(P(A'|B) = P(A') = \frac{3}{5}\) and \(P(B|A') = P(B) = \frac{1}{4}\).
So \(P(A\cap B') = \frac{3}{10}\), \(P(A\cup B) = \frac{11}{20}\), \(P(A'|B) = \frac{3}{5}\) and \(P(B|A') = \frac{1}{4}\).
Checking the statements
(A) \(P(A\cap B') = \frac{3}{10}\) — Correct. \(\frac{2}{5}\times\frac{3}{4} = \frac{3}{10}\).
(B) \(P(A\cup B) = \frac{11}{20}\) — Correct. \(\frac{8}{20}+\frac{5}{20}-\frac{2}{20} = \frac{11}{20}\).
(C) \(P(A'|B) = \frac{3}{4}\) — Incorrect. \(P(A'|B) = P(A') = \frac{3}{5}\); \(\frac{3}{4}\) is P(B’).
(D) \(P(B|A') = \frac{1}{4}\) — Correct. \(P(B|A') = P(B) = \frac{1}{4}\).