A tyre company records distances before replacement in 1000 cases.

Distance (km)	Number of cases
Less than 4000	20
4001 to 9000	210
9001 to 14000	325
More than 14000	445

Find the probability that a randomly chosen tyre lasts:

(A) Less than 4000 km.

(B) Between 4000 and 14000 km.

(C) More than 14000 km.

Total cases = 20 + 210 + 325 + 445 = 1000

(A) $P(\text{less than } 4000 \text{ km}) = \frac{20}{1000} = \frac{1}{50}$

(B) P(between 4000 and 14000 km)

$$= \frac{210 + 325}{1000}$$

$$= \frac{535}{1000} = \frac{107}{200}$$

(C) $P(\text{more than } 14000 \text{ km}) = \frac{445}{1000} = \frac{89}{200}$