**Solution:**

Here, the sample space = {H, T}

i.e. n(S) = 2

(i) If A = Event of getting a tail = {T}

Then, n(A) = 1

Hence, the probability of getting a tail = n(A)/ n(S) = 1/2

(ii) Not getting a tail

As we know, P(getting a tail) + P(not getting a tail) = 1

So, P(not getting a tail) = 1 – (1/2) = ½