A coin is tossed three times. If \( E \) is the event ‘at least two heads’ and \( F \) is the event ‘at most two heads’, then which of the following statements is/are TRUE?
(A) \( P(E)=7/8 \)
(B) \( P(F)=1/2 \)
(C) \( P(E\cap F)=3/8 \)
(D) \( P(E/F)=3/7 \)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (C) and (D) only
The sample space has \( 2^3=8 \) equally likely outcomes.
E = at least two heads = {HHT, HTH, THH, HHH}, 4 outcomes.
F = at most two heads = every outcome except HHH, 7 outcomes.
E ∩ F = exactly two heads = {HHT, HTH, THH}, 3 outcomes.
\( \displaystyle P(E)=\frac{4}{8}=\frac{1}{2} \)
\( \displaystyle P(F)=\frac{7}{8} \)
\( \displaystyle P(E\cap F)=\frac{3}{8} \)
\( \displaystyle P(E/F)=\frac{P(E\cap F)}{P(F)}=\frac{3/8}{7/8} \)
\( \displaystyle =\frac{3}{7} \)
So P(E) = 1/2, P(F) = 7/8, P(E ∩ F) = 3/8 and P(E/F) = 3/7.
Checking the statements
(A) \( P(E)=7/8 \) — Incorrect. P(E) is 1/2; 7/8 is P(F).
(B) \( P(F)=1/2 \) — Incorrect. P(F) is 7/8; 1/2 is P(E).
(C) \( P(E\cap F)=3/8 \) — Correct. Three of the eight outcomes have exactly two heads.
(D) \( P(E/F)=3/7 \) — Correct. \( \frac{3/8}{7/8}=\frac{3}{7} \).