A fair die is thrown once. Let E be the event “the number is odd”, F the event “the number is less than 3” and G the event “the number is a prime”. Consider the following statements:
(A) E and F are independent.
(B) E and G are independent.
(C) F and G are independent.
(D) \(P(G|E) = \displaystyle\frac{2}{3}\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A), (C) and (D) only
The events are E = {1, 3, 5}, F = {1, 2} and G = {2, 3, 5}, so
\( \displaystyle P(E) = \frac{1}{2},\quad P(F) = \frac{1}{3},\quad P(G) = \frac{1}{2} \)
The common parts are \(E\cap F = \{1\}\), \(E\cap G = \{3, 5\}\) and \(F\cap G = \{2\}\):
\( \displaystyle P(E\cap F) = \frac{1}{6},\quad P(E\cap G) = \frac{1}{3},\quad P(F\cap G) = \frac{1}{6} \)
Two events are independent when the probability of both equals the product of their probabilities.
So \(P(E\cap F) = \frac{1}{6}\) against \(P(E)P(F) = \frac{1}{6}\), \(P(E\cap G) = \frac{1}{3}\) against \(P(E)P(G) = \frac{1}{4}\), \(P(F\cap G) = \frac{1}{6}\) against \(P(F)P(G) = \frac{1}{6}\), and \(P(G|E) = \frac{2}{3}\).
Checking the statements
(A) E and F are independent. — Correct. \(P(E\cap F) = \frac{1}{6} = \frac{1}{2}\times\frac{1}{3}\).
(B) E and G are independent. — Incorrect. \(P(E\cap G) = \frac{1}{3}\), but \(\frac{1}{2}\times\frac{1}{2} = \frac{1}{4}\).
(C) F and G are independent. — Correct. \(P(F\cap G) = \frac{1}{6} = \frac{1}{3}\times\frac{1}{2}\).
(D) \(P(G|E) = \frac{2}{3}\) — Correct. Of the odd numbers 1, 3, 5, the primes are 3 and 5, giving \(\frac{2}{3}\).