A fair die is tossed repeatedly until a six is obtained. Let $$X$$ denote the number of tosses required and let $$a = P(X = 3)$$, $$b = P(X \geq 3)$$ and $$c = P(X \geq 6 \mid X > 3)$$. Then $$\frac{b + c}{a}$$ is equal to _______.

A fair die: $$P(\text{six}) = 1/6$$, $$P(\text{not six}) = 5/6$$.

$$a = P(X = 3) = (5/6)^2 \cdot (1/6) = 25/216$$

$$b = P(X \geq 3) = (5/6)^2 = 25/36$$ (first two tosses are not six)

$$c = P(X \geq 6 \mid X > 3)$$: Given that first 3 tosses are not six, the probability that we need at least 6 tosses means tosses 4 and 5 are also not six.

$$c = (5/6)^2 = 25/36$$

$$\frac{b + c}{a} = \frac{25/36 + 25/36}{25/216} = \frac{50/36}{25/216} = \frac{50}{36} \times \frac{216}{25} = \frac{50 \times 6}{25} = 12$$

The answer is $$\boxed{12}$$.