In a simultaneous throw of two coins, the probability of getting at least one head is-

Answer: Option D

Here S = {HH, HT, TH, TT}
Let E = event of getting at least one head = {HT, TH, HH}
$$\therefore P\left(E\right)=\frac{n(E)}{n(S)}=\frac{3}{4}$$