A coin is biased so that the head is 3 times as likely to occur as tail. This coin is tossed until a head or three tails occur. If $$X$$ denotes the number of tosses of the coin, then the mean of $$X$$ is

Given: P(H) = 3/4, P(T) = 1/4. Toss until head or three tails occur. X = number of tosses.

Possible outcomes:

X=1: H with prob 3/4.

X=2: TH with prob (1/4)(3/4) = 3/16.

X=3: TTH or TTT with prob (1/4)²(3/4) + (1/4)³ = 3/64 + 1/64 = 4/64 = 1/16.

$$E[X] = 1 \cdot \frac{3}{4} + 2 \cdot \frac{3}{16} + 3 \cdot \frac{1}{16}$$

$$= \frac{3}{4} + \frac{6}{16} + \frac{3}{16} = \frac{12}{16} + \frac{6}{16} + \frac{3}{16} = \frac{21}{16}$$

The correct answer is Option C: $$\frac{21}{16}$$.