$A$ tyre manufacturing company kept a record of the distance covered before a tyre needed to be replaced. The table shows the results of $1000$ cases.

Distance (in $km$)	less than $4000$	$4000$ to $9000$	$9001$ to $14000$	more than $14000$
Frequency	$20$	$210$	$325$	$445$

If you buy a tyre of this company,what is the probability that:
$(i)$ it will need to be replaced before it has covered $4000 \, km$?
$(ii)$ it will last more than $9000 \, km$?
$(iii)$ it will need to be replaced after it has covered somewhere between $4000 \, km$ and $14000 \, km$?

(N/A) $(i)$ The total number of trials $= 1000$.
The frequency of a tyre that needs to be replaced before it covers $4000 \, km$ is $20$.
So,$P(\text{tyre to be replaced before it covers } 4000 \, km) = \frac{20}{1000} = 0.02$.
$(ii)$ The frequency of a tyre that will last more than $9000 \, km$ is the sum of frequencies for the ranges $9001$ to $14000$ and more than $14000$,which is $325 + 445 = 770$.
So,$P(\text{tyre will last more than } 9000 \, km) = \frac{770}{1000} = 0.77$.
$(iii)$ The frequency of a tyre that requires replacement between $4000 \, km$ and $14000 \, km$ is the sum of frequencies for the ranges $4000$ to $9000$ and $9001$ to $14000$,which is $210 + 325 = 535$.
So,$P(\text{tyre requiring replacement between } 4000 \, km \text{ and } 14000 \, km) = \frac{535}{1000} = 0.535$.