Let A and B be two independent events such that P(A) = \( k \), P(B) = \( 2k \) and probability of occurrence of exactly one of them is \( \displaystyle\frac{5}{9} \). Then the value(/s) of \( k \) is/are:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( k=\displaystyle\frac{1}{3},\frac{5}{12} \)
Exactly one of A, B occurs means A and not B, or B and not A. By independence:
\( \displaystyle P(\text{exactly one})=P(A)\,P(\overline{B})+P(\overline{A})\,P(B) \)
\( \displaystyle =k(1-2k)+(1-k)2k \)
\( \displaystyle =3k-4k^2 \)
\( \displaystyle 3k-4k^2=\frac{5}{9} \)
\( \displaystyle 36k^2-27k+5=0 \)
\( \displaystyle (3k-1)(12k-5)=0\ \Rightarrow\ k=\frac{1}{3} \text{ or } k=\frac{5}{12} \)
Both values are admissible: P(B) = 2k is \( \frac{2}{3} \) or \( \frac{5}{6} \), both at most 1.
Why the other options are incorrect
Option (1) \( k=\displaystyle\frac{1}{3} \) — Incorrect. \( k=\frac{1}{3} \) is a solution, but not the only one.
Option (2) \( k=\displaystyle\frac{5}{12} \) — Incorrect. \( k=\frac{5}{12} \) is a solution, but not the only one.
Option (4) \( k=\displaystyle\frac{2}{3},\frac{7}{12} \) — Incorrect. \( k=\frac{2}{3} \) gives P(B) = \( \frac{4}{3} \gt 1 \), which is impossible, and \( k=\frac{7}{12} \) gives P(B) = \( \frac{7}{6} \gt 1 \).