Question 870692
You can feel comfortable using n = 10
Might recommend stattrek.com(binomial P) for a good reference and check to Your work.
Note:  Using n = 10
Yes, n = 10 (taking from a closet/drawer with 10 batteries in them)
AND the probability referred to those 10 alone. (Not in general, so to speak)
1) closet: p(not charged) = .5,  n =10,  
P(x = 4) = 10C4(.5^4)(.5^6) = 210(.5^4)(.5^6) =  .20
2) drawer: p(not charged) = .5
i) n = 10, P(x = 4)= 10C4(.5^4)(.5^6) = 210(.5^4)(.5^6) =  .20
ii)n = 10, P(x =2) = 10C2(.5^2)(.5^8) = .0439
    .0439 + .0439 = .0878
iii)p = .5, n = 20, P(x = 4) =20C4(.5^4)(.5^16) = .0046
3) Compare the options of the previous question (2) towards the mean number of charged batteries that you took. 
As p(charged) = .5, 
the computations as to the various P(charged batteries) would have same results