A fair coin is tossed three times in succession. If the first toss produces a head, then the probability of getting exactly two heads in three tosses is A. $$\frac{1}{8}$$ B. $$\frac{1}{2}$$ C. $$\frac{3}{8}$$ D. $$\frac{3}{4}$$

The probability of getting exactly two heads in three tosses is $\frac{3}{8}$.

To get exactly two heads, the first toss must be a head, and the second and third tosses must be either both heads or both tails. The probability of the first toss being a head is $\frac{1}{2}$. The probability of the second and third tosses being both heads is $\frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$. The probability of the second and third tosses being both tails is also $\frac{1}{4}$. Therefore, the probability of getting exactly two heads in three tosses is $\frac{1}{2} \cdot \frac{1}{4} + \frac{1}{2} \cdot \frac{1}{4} = \frac{3}{8}$.

Option A is incorrect because it is the probability of getting exactly one head in three tosses. Option B is incorrect because it is the probability of getting at least one head in three tosses. Option C is incorrect because it is the probability of getting two heads in three tosses, regardless of whether the first toss is a head or a tail. Option D is incorrect because it is the probability of getting three heads in three tosses.