# Cross section for 2 particles that decay independently

This maybe is a naive basic question but I want to be sure. If I want to calculate the cross section of the process $$p\bar{p} \rightarrow W^+HX \rightarrow e^+\nu_e b\bar{b} X$$ ($$H$$ is a Higgs particle, $$p$$ a proton), I'm going to need the branching ratio of $$W^+H \rightarrow e^+\nu_e b\bar{b}$$ defined as:

$$BR(W^+H \rightarrow e^+\nu_e b\bar{b})= \frac{\sigma(W^+H \rightarrow e^+\nu_e b\bar{b})}{\sigma_{total}} \tag1$$

where $$\sigma$$ is the cross section of $$W^+$$'s decay into $$e^+ \nu_e$$ and $$H$$ into $$b\bar{b}$$; and $$\sigma_{total}$$ is the sum of the cross sections of all possible decays of $$W^+$$ and $$H$$.

Then, for the reaction $$W^+H \rightarrow e^+\nu_e b\bar{b}$$ I have two Feynman diagrams.

My question is:

$$\sigma(W^+H \rightarrow e^+\nu_e b\bar{b}) ) = \sigma(W^+ \rightarrow e^+ \nu_e) + \sigma(H \rightarrow b\bar{b})\tag2$$

and use it in Eq. (1) since each cross section is an area or should I use $$BR(W^+H \rightarrow e^+\nu_e b\bar{b}) = BR(W^+ \rightarrow e^+ \nu_e)·BR(H \rightarrow b\bar{b})\tag3$$

since each BR is the probability of the corresponding process in brackets and these two decays are independent?