[Ahmed Abdullah]

In anti-markovnikov's adition;
the per-oxide (HO-OH) split and give rise to HO* free radicals.
HO-OH-------> 2HO*  ( " * " denotes unpaired electron) 
 
These HO* radicals attack HX ;
HO* + HX-----> H-OH + X*
 
Now this X* free radical react with alkene;
 
CH3-CH=CH2 + X*  ------> CH3-C*H-CH2X      -  -  -    -   -   - -  (1)
CH3-C*H-CH2X + HX -----> CH3-CH2-CH2X + X*  - -  -  -  -  -  -  -(2)
 
Ultimately CH3-CH2-CH2X is the dominant product. But there are other possibilities, like CH3-CHX-CH3, CH3CHX-CH2X , CH3CH(OH)CH2(OH), CH3CH2-CH2(OH) etc.
 
But it is predictable that some of these products should be minor.
Like, a least stable carbon free radical is involved in CH3-CHX-CH3 formation. Obviously CH3-CHX-CH3 is not favoured and produced in minute amount.
 
CH3CHX-CH2X can form if the following reaction occur;
 
CH3-C*H-CH2X + X* -----> CH3-CHX-CH2X  ; But HX is more abundant and more available than X* free radical, so this product cannot be major.
 
In the system there are X*, and HO* free radicals. First step involves the formation of HO* , this free radical can attack either the alkene or the HX.
 
1.CH3-CH=CH2 + HO*  ------> CH3-C*H-CH2(OH)
2.                 HX + HO* -----> H-OH + X*
 
After the formation of CH3-C*H-CH2(OH), it is attacked by other reagent. The most potential attacker is the HX since it is in larger proportion in the system.
So following thing happens;
   
 CH3-C*H-CH2(OH) + HX -----> CH3-CH2-CH2(OH) + X*
 
 
CH3-CH2-CH2(OH) forms long before the formation of CH3-C*H-CH2X free radical, which means that  CH3-CH2-CH2(OH) is the dominant product. This argument implies that CH3-CH2-CH2(OH) formation is favoured over CH3-CH2-CH2X.

Any input should be highly appreciated. Thx

[reflux]

I don't know the numbers off the top of my head, but maybe the process leading to one compound has a highly endothermic step which is why it is not observed.  You can use bond dissociation energies to calculate the enthaply of each step.  Those are relatively simple compounds so you can probably find the exact value you want for a particular bond or at least a reasonable estimate.  There should be a table of BDEs in any given introductory o-chem textbook.