Probability of getting head = $\frac{1}{2}$
Probability of not getting head = probability of getting tail = $\frac{1}{2}$
Coin is tossed odd number of times untill it’s get first head,
i.e., required outcomes of these trials = {$H,TTH,TTTTH,...$}
So, required probability of getting these outcomes $= \frac{1}{2}+(\frac{1}{2})^3+(\frac{1}{2})^5+….$
$$=\frac{\frac{1}{2}}{1-\frac{1}{4}}=\frac{2}{3}$$
Correct Answer: $C$