Total numbers of elementary events are 2 + 2 = 4 

Let E be the event of getting one head 

Favorable outcome are: TH , HT 

Numbers of favorable outcomes are = 2 

P (exactly one head) = P (E) = \(\frac{2}4\) = \(\frac{1}2\)

Two coins are tossed:

Sample space = {HH, HT, TH, TT}

Total number of outcomes = 4

P(getting a tail) = 1/2.

Now,

(i) P(exactly 1 head) = 2/4 = 1/2

(ii) P(at most 1 head) = 3/4

(iii) P(at least 1 head) = 3/4