16. Two bullets are placed in two consecutive chambers of a 6-chamber pistol. The cylinder is then spun. The pistol is fired but the first shot is a blank. Let $p$ denote the probability that the second shot is also a blank if the cylinder is spun after the first shot and let $\mathrm{q}$ denote the probability that the second shot is also a blank if the cylinder is not spun after the first shot. Find the smallest integer $N$ such that
$$
N \geq \frac{100 p}{q} .
$$
16. Answer: 89
$p=\operatorname{Prob}\left(2^{\text {nd }}\right.$ shot blank $\mid$ cylinder spun again $)=4 / 6$, and
$q=\operatorname{Prob}\left(2^{\text {nd }}\right.$ shot blank $\mid$ cylinder not spun again $)=3 / 4$.
Therefore $\frac{100 p}{q}=\frac{800}{9} \approx 88.8$, and hence the smallest $N .=89$.