Two coins are tossed together. Let E be the event “head on the first coin”, F be the event “head on the second coin” and G be the event “both coins show the same face”.
(A) E and F are independent events
(B) E and F are mutually exclusive events
(C) E and G are independent events
(D) \(P(G|E) = \displaystyle\frac{1}{2}\)
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 sample space is HH, HT, TH, TT, each with probability \(\frac{1}{4}\).
E = {HH, HT}, F = {HH, TH}, G = {HH, TT}
\( \displaystyle P(E) = P(F) = P(G) = \frac{1}{2} \)
Each pair has only HH in common:
\( \displaystyle P(E \cap F) = P(E \cap G) = \frac{1}{4} \)
Two events are independent when the probability of both equals the product of their probabilities, and mutually exclusive when they have no outcome in common.
So \(P(E \cap F) = P(E \cap G) = \frac{1}{4} = \frac{1}{2} \times \frac{1}{2}\), E and F share the outcome HH, and \(P(G|E) = \frac{1/4}{1/2} = \frac{1}{2}\).
Checking the statements
(A) E and F are independent events — Correct. \(P(E \cap F) = \frac{1}{4} = P(E)P(F)\).
(B) E and F are mutually exclusive events — Incorrect. Both happen in the outcome HH.
(C) E and G are independent events — Correct. \(P(E \cap G) = \frac{1}{4} = P(E)P(G)\).
(D) \(P(G|E) = \frac{1}{2}\) — Correct. Given E = {HH, HT}, only HH gives the same face, so it is 1 out of 2.