Computing Expectation and Variance by Conditioning [on hold]

Six nickels are tossed, and the total number N of heads is observed. Then
N dimes are tossed, and the total number Z of tails among the dimes is observed.
Find the following values (either in fraction numbers or in numbers up to 4 decimal
points), and make your steps clear. (Do NOT use the Conditional Variance Formula
in any problem.)

(a) E[Z]

(b) Var(Z)

(c) P(Z = 2)

(d) E[Var(Z|N)]

(e) Var[E(Z|N)]

I got answers to these:

a) 162/128

because E[Z] = ∑6k=0\sum_{k=0}^6 E[Z|N=k]P{N=k}

= (61)(1/2)1(1−1/2)5\binom{6}{1} (1/2)^1(1-1/2)^5 E[Z|N=1] + \binom{6}{2} (1/2)^2(1-1/2)^4\binom{6}{2} (1/2)^2(1-1/2)^4 E[Z|N=2] + …

Note: E[Z|N=1]= 1(1/2) and E[Z|N=2]= 2(1/2) … because it is binomial


because E[Z^2]E[Z^2] = \sum_{k=0}^6 E[Z^2|N=k]P(N=k)\sum_{k=0}^6 E[Z^2|N=k]P(N=k)

and the second moment for a binomial is np(1-p) +n^2 p^2np(1-p) +n^2 p^2



var(E[Z|N])= E[(E[Z|N])^2(E[Z|N])^2] – (E[E[Z|N]])^2(E[E[Z|N]])^2 = E[(E[Z|N])^2(E[Z|N])^2] – (E[Z])^2(E[Z])^2

and E[(E[Z|N])^2]E[(E[Z|N])^2] = \sum_{k=0}^6 (E[Z|N=k])^2 P(N=k)\sum_{k=0}^6 (E[Z|N=k])^2 P(N=k)


However, I know they can’t all be correct (I’m not even sure any of them are correct) because Var(Z) = E[Var(Z|N)] – Var(E[Z|N]) doesn’t work. Can someone tell me where I am going wrong? I just showed my thought process because I don’t have the patients to rewrite all my work.