Solution:
We know that,
When two coins are tossed simultaneously, the possible outcomes are {(H, H), (H, T), (T, H), (T, T)}
So, n(S) = 4
(i) The outcomes favourable to the event E, ‘at least one head’ are {(H, H), (H, T), (T, H)}
So, the number of outcomes favourable to E is 3 = n(E)
Hence, P(E) = n(E)/ n(S) = ¾
(ii) The outcomes favourable to the event E, ‘at most one head’ are {(T, H), (H, T), (T, T)}
So, the number of outcomes favourable to E is 3 = n(E)
Hence, P(E) = n(E)/ n(S) = 3/4