Question

Two coins are thrown at the same time find the probability of getting at least one head

Answer 1

(vi) 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.