A fair coin is tossed three times. Let E be the event “the first toss shows a head”, F the event “at least two tosses show a head” and G the event “all three tosses show the same face”. Consider the following statements:
(A) E and G are independent.
(B) F and G are independent.
(C) E and F are independent.
(D) \(P(F|E) = \displaystyle\frac{1}{2}\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A) and (B) only
The 8 outcomes are equally likely. Listing the events:
E = {HHH, HHT, HTH, HTT}
F = {HHH, HHT, HTH, THH}
G = {HHH, TTT}
\( \displaystyle P(E) = \frac{1}{2},\quad P(F) = \frac{1}{2},\quad P(G) = \frac{1}{4} \)
The common outcomes are E ∩ G = {HHH}, F ∩ G = {HHH} and E ∩ F = {HHH, HHT, HTH}:
\( \displaystyle P(E\cap G) = \frac{1}{8},\quad P(F\cap G) = \frac{1}{8},\quad P(E\cap F) = \frac{3}{8} \)
Two events are independent when the chance of both equals the product of their chances:
\( \displaystyle P(E)P(G) = \frac{1}{8},\quad P(F)P(G) = \frac{1}{8},\quad P(E)P(F) = \frac{1}{4} \)
By conditional probability:
\( \displaystyle P(F|E) = \frac{\frac{3}{8}}{\frac{1}{2}} = \frac{3}{4} \)
So E, G are independent, F, G are independent, E, F are not independent, and \(P(F|E) = \frac{3}{4}\).
Checking the statements
(A) E and G are independent. — Correct. \(P(E\cap G) = \frac{1}{8} = \frac{1}{2}\times\frac{1}{4}\).
(B) F and G are independent. — Correct. \(P(F\cap G) = \frac{1}{8} = \frac{1}{2}\times\frac{1}{4}\).
(C) E and F are independent. — Incorrect. \(P(E\cap F) = \frac{3}{8}\) but \(P(E)P(F) = \frac{1}{4}\).
(D) \(P(F|E) = \frac{1}{2}\) — Incorrect. 3 of the 4 outcomes in E have at least two heads, so \(P(F|E) = \frac{3}{4}\).