Regarding allylic free radicals and anti-Markovnikov product

I was searching a question "What is the primary product when cyclohexene reacts with bromine in the presence of UV light?" and in a website they said: 



> 
> 3-Bromocyclohexene will be the product because when an alkene reacts
>  with $\ce{Br2}$ in presence of UV light, free radical substitution
>  reaction takes place. In this case, a stable cyclohexene allylic free
>  radical will form which will further react with bromine free radical
>  to form 3-bromocyclohexene.
> 
> 
> 


A reaction that I already studied was propene + $\ce{HBr}$ in presence of $\ce{H2O2}$
and the answer is:[![enter image description here](https://i.stack.imgur.com/egbM7.jpg)](https://i.stack.imgur.com/egbM7.jpg)


The reason is that carbon radical is present on secondary carbon where the neighboring atoms can donate electron density to stabilize the radical.


But based on the first reaction I speculate that the answer should be:[![enter image description here](https://i.stack.imgur.com/UIdaV.jpg)](https://i.stack.imgur.com/UIdaV.jpg)
because here an allylic free radical is formed.


So which is correct?


