Question

A pair of coins tossed at once, find the probability of getting atleast one
head.
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Answer 1

getting at least 1 head:

Let E6 = event of getting at least 1 head. Then,

E6 = {HT, TH, HH} and, therefore, n(E6) = 3.

Therefore, P(getting at least 1 head) = P(E6) = n(E6)/n(S) = ¾.

Answer 2

Let E4 = event of getting no head. Then,

E4 = {TT} and, therefore, n(E4) = 1.

Therefore, P(getting no head) = P(E4) = n(E4)/n(S) = ¼.