A binomial model would be appropriate. Let $X$ be the number of stolen bicycles that are tracked. Then $X\sim B(1600,0.07)$. You can then compute $$P(X>100)=1-P(X\leq 100)=1-0.1290=\fbox{0.8710} \text{ (to 4d.p.)}$$
A binomial model would be appropriate. Let $X$ be the number of stolen bicycles that are tracked. Then $X\sim B(1600,0.07)$. You can then compute $$P(X>100)=1-P(X\leq 100)=1-0.1290=\fbox{0.8710} \text{ (to 4d.p.)}$$