One ticket is drawn from a purse containing four tickets with numbers 223, 133, 212, 322. Let \( A_{i} \) be the event that \( i^{th} \) digit of the number on drawn ticket is 2. Then which of the following events are independent ?
A. \( A_{1} \) and \( A_{2} \)
B. \( A_{2} \) and \( A_{3} \)
C. \( A_{1} \) and \( A_{3} \)
D. \( A_{1} \), \( A_{2} \) and \( A_{3} \)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → A, B and C only
Each ticket has probability \( \frac{1}{4} \). The tickets favourable to each event are
\( \displaystyle A_{1}=\{223,212\} \)
\( \displaystyle A_{2}=\{223,322\} \)
\( \displaystyle A_{3}=\{212,322\} \)
So \( P(A_{1})=P(A_{2})=P(A_{3})=\frac{1}{2} \); two of them are independent when their intersection has probability \( \frac{1}{4} \), and all three need \( P(A_{1}\cap A_{2}\cap A_{3})=\frac{1}{8} \) as well.
Checking the statements
(A) \( A_{1} \) and \( A_{2} \) — Correct. Only 223 is common, so \( P(A_{1}\cap A_{2})=\frac{1}{4} \).
(B) \( A_{2} \) and \( A_{3} \) — Correct. Only 322 is common, so \( P(A_{2}\cap A_{3})=\frac{1}{4} \).
(C) \( A_{1} \) and \( A_{3} \) — Correct. Only 212 is common, so \( P(A_{1}\cap A_{3})=\frac{1}{4} \).
(D) \( A_{1} \), \( A_{2} \) and \( A_{3} \) — Incorrect. No ticket has 2 in all three places, so \( P(A_{1}\cap A_{2}\cap A_{3})=0 \), not \( \frac{1}{8} \).
Tip: Pairwise independence does not make three events independent; the triple product condition must hold as well.