A fair coin is tossed till head appears for the first time. The probability that the number of required tosses is odd, is

1. $\frac{1}{3}$
2. $\frac{1}{2}$
3. $\frac{2}{3}$
4. $\frac{3}{4}$
in Others
edited

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: $C$
by (150 points)