A fair coin is tossed till a head appears for the first time. The probability that the number of required tosses is odd, is A. $$\frac{1}{3}$$ B. $$\frac{1}{2}$$ C. $$\frac{2}{3}$$ D. $$\frac{3}{4}$$

The correct answer is $\boxed{\frac{1}{2}}$.

The probability of getting a head on any given toss is $\frac{1}{2}$. The probability of getting an odd number of tosses before the first head is the same as the probability of getting at least one tail before the first head. This is because the probability of getting a tail on any given toss is also $\frac{1}{2}$, and the events “getting a tail” and “getting an odd number of tosses before the first head” are mutually exclusive. Therefore, the probability of getting at least one tail before the first head is $1 – \frac{1}{2} = \boxed{\frac{1}{2}}$.

Option A is incorrect because it is the probability of getting a tail on the first toss. Option B is incorrect because it is the probability of getting a head on the first toss. Option C is incorrect because it is the probability of getting an even number of tosses before the first head. Option D is incorrect because it is the probability of getting a tail on the first toss and then getting a head on the second toss.