Three coins tossed simultaneously.
S = { HHH, TTT, HHT, TTH, HTH, THT, HTT, THH}
n(S) = 8
Happening of getting exactly two heads be A.
A= {HHT, HTH, THH}
n(A) = 3
P(A) = \(\frac{n(A)}{n(S)} = \frac{3}{8}\)
Event of getting at least one tail be B.
∴ B = {TTT, HHT, TTH, HTH, THT, HTT, THH}
n(B) = 7