A number is chosen at random from the numbers 1, 2, 3, …, 12. Let A be the event “the number is even”, B the event “the number is a multiple of 3” and C the event “the number is a multiple of 4”. Consider the following statements:
(A) A and B are independent.
(B) B and C are independent.
(C) A and C are independent.
(D) \(P(A|C) = \displaystyle\frac{1}{2}\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A) and (B) only
The events are A = {2, 4, 6, 8, 10, 12}, B = {3, 6, 9, 12} and C = {4, 8, 12}, so
\( \displaystyle P(A) = \frac{1}{2},\quad P(B) = \frac{1}{3},\quad P(C) = \frac{1}{4} \)
The common parts are \(A\cap B = \{6, 12\}\), \(B\cap C = \{12\}\) and \(A\cap C = \{4, 8, 12\} = C\), since every multiple of 4 is even.
Two events are independent when the probability of both equals the product of their probabilities.
So \(P(A\cap B) = \frac{1}{6}\), \(P(B\cap C) = \frac{1}{12}\), \(P(A\cap C) = \frac{1}{4}\) and \(P(A|C) = 1\).
Checking the statements
(A) A and B are independent. — Correct. \(P(A)P(B) = \frac{1}{2}\times\frac{1}{3} = \frac{1}{6} = P(A\cap B)\).
(B) B and C are independent. — Correct. \(P(B)P(C) = \frac{1}{3}\times\frac{1}{4} = \frac{1}{12} = P(B\cap C)\).
(C) A and C are independent. — Incorrect. \(P(A\cap C) = \frac{1}{4}\), but \(P(A)P(C) = \frac{1}{8}\).
(D) \(P(A|C) = \frac{1}{2}\) — Incorrect. Every multiple of 4 is even, so \(P(A|C) = 1\).