Answer:
The answer is "np(1+p)".
Step-by-step explanation:
[tex]N_1[/tex] = Number of n tossed heads
According to [tex]N_1[/tex], Y = amount of [tex]N_1[/tex] tossing heads
So, [tex]N_1[/tex]~ Bin(n, p) and [tex]N_2[/tex] | [tex]N_1[/tex] ~ Bin([tex]N_1[/tex], p)
[tex]E(N_1+N_2) & = E\Big[N_1+N_2\vert N_1\Big]\\ &[/tex]
[tex]= E(N_1)+E\Big[N_2\vert N_1\Big]\\\\ & = E(N_1) + E(N_1\,p)\\\\ & = np + np.p \\\\ =np+np^2\\\\ =np(1+p)[/tex]