16. Let Q_n denote the probability that in n tosses of a fair coin no run of 3 consecutive heads appears. Show that

$$

Q_n = \frac{1}{2} Q_{n-1} + \frac{1}{2}Q_{n-2} + \frac{1}{8}Q_{n-3}

$$

$$

Q_0 = Q_1 = Q_2 = 1

$$

Find Q₈.

HINT: Condition on the first tail.