Two fair dice are thrown. Let E be the event “the first die shows an even number”, F the event “the sum of the numbers is 7” and G the event “the sum of the numbers is 8”. Consider the following statements:
(A) E and F are independent.
(B) E and G are independent.
(C) F and G are mutually exclusive.
(D) \(P(E|G) = \displaystyle\frac{3}{5}\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A), (C) and (D) only
There are 36 equally likely outcomes. E has 18 of them, so \(P(E) = \frac{1}{2}\).
F = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}, so \(P(F) = \frac{6}{36} = \frac{1}{6}\)
G = {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)}, so \(P(G) = \frac{5}{36}\)
The outcomes with an even first number are
\(E\cap F\) = {(2, 5), (4, 3), (6, 1)}, so \(P(E\cap F) = \frac{3}{36} = \frac{1}{12}\)
\(E\cap G\) = {(2, 6), (4, 4), (6, 2)}, so \(P(E\cap G) = \frac{3}{36} = \frac{1}{12}\)
Two events are independent when the probability of both equals the product of their probabilities; they are mutually exclusive when they have no outcome in common.
So \(P(E\cap F) = \frac{1}{12}\) against \(P(E)P(F) = \frac{1}{12}\), \(P(E\cap G) = \frac{1}{12}\) against \(P(E)P(G) = \frac{5}{72}\), F and G have no common outcome, and \(P(E|G) = \frac{3}{5}\).
Checking the statements
(A) E and F are independent. — Correct. \(P(E\cap F) = \frac{1}{12} = \frac{1}{2}\times\frac{1}{6}\).
(B) E and G are independent. — Incorrect. \(P(E\cap G) = \frac{1}{12}\), but \(\frac{1}{2}\times\frac{5}{36} = \frac{5}{72}\).
(C) F and G are mutually exclusive. — Correct. The sum cannot be 7 and 8 at the same time.
(D) \(P(E|G) = \frac{3}{5}\) — Correct. Of the 5 outcomes with sum 8, three have an even first number: \(\frac{3}{5}\).