When two different coins are tossed randomly, the sample space is given by:
S = {HH, HT, TH, TT}
Therefore, n(S) = 4.
Let E1 = getting no tails, then, E1 = {HH} and, therefore, n(E1) = 1.
Therefore, P(getting no tails) = 1/4.
Let E2 = getting one tail, then, E2 = {HT, TH}, therefore n(E2) = 2.
Therefore P(getting one tail) = 2/4 = 1/2.
Finally getting either no tails or one tail = 1/4 + 1/2 = 3/4
