Alkane free radical halogenation I do not understand how we obtain the product in option one. 
According to Free radical stability order tertiary carbon radical is more stable than secondary carbon radical which is more stable than primary carbon radical. So hydrogen selectivity should also follow the same order. Then why is it that we get option(1) instead of option (2). 
I’m only in high school so please try to keep it simple.
This question is based on relative rates of reaction of different $\ce{H}$ with halogens. In the free radical mechanism of chlorination of alkane, relative rate of reaction of $\ce{3^\circ : 2^\circ : 1^\circ}$ $\ce{H}$ is in order $5:3.8: 1$.

Using this, we get an idea about product amount. To get a rough estimate of relative amounts of different products, we multiply number of identical hydrogen and its rate of reaction with chlorine.

So there is only $1$ $3^\circ$ $\ce{H}$ and its rel. rate is $5$. So relative amount is $5$. For the $2^\circ$ $\ce{H}$, there are $2$ of them and their relative rate is $3.8$. So its relative amount is $2 \cdot 3.8 = 7.6$. Similar calculation for rest of the hydrogens will show you why given answer is correct.

Refer: Wikipedia