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Answer:
i) Probability that both candidates employed are women = 5/14
ii) Probability that the second candidate is a woman = 5/8
iii) Probability that the first candidate is a woman given that second one is a woman = 4/5
Step-by-step explanation:
Let the probability that a man is employed be P(M) = 3/8
Probability that a woman is employed P(W) = 5/8
a) Probability that both candidates employed are women = (5/8) × (4/7) = 5/14
b) Probability that the second candidate is a woman = (probability that first candidate is a man and second candidate is a woman) + (probability that first candidate is a woman & second candidate is a woman)
= (3/8)(5/7) + (5/8)(4/7) = (15/56) + (20/56) = 35/56 = 5/8
c) Probability that the first candidate is a woman given that second one is a woman
Given that the second candidate was a women, means that the first candidate-women was selected among other four women.
Probability = (4/8)/(5/8) = 4/5
Answer:
See the proof below.
Step-by-step explanation:
Assuming this complete question: "For each given p, let Z have a binomial distribution with parameters p and N. Suppose that N is itself binomially distributed with parameters q and M. Formulate Z as a random sum and show that Z has a binomial distribution with parameters pq and M."
Solution to the problem
For this case we can assume that we have N independent variables 
with the following distribution:
bernoulli on this case with probability of success p, and all the N variables are independent distributed. We can define the random variable Z like this:
From the info given we know that 
We need to proof that 
by the definition of binomial random variable then we need to show that:
The deduction is based on the definition of independent random variables, we can do this:
And for the variance of Z we can do this:
![Var(Z)_ = E(N) Var(X) + Var (N) [E(X)]^2](https://tex.z-dn.net/?f=%20Var%28Z%29_%20%3D%20E%28N%29%20Var%28X%29%20%2B%20Var%20%28N%29%20%5BE%28X%29%5D%5E2%20)
![Var(Z) =Mpq [p(1-p)] + Mq(1-q) p^2](https://tex.z-dn.net/?f=%20Var%28Z%29%20%3DMpq%20%5Bp%281-p%29%5D%20%2B%20Mq%281-q%29%20p%5E2)
And if we take common factor 
we got:
![Var(Z) =Mpq [(1-p) + (1-q)p]= Mpq[1-p +p-pq]= Mpq[1-pq]](https://tex.z-dn.net/?f=%20Var%28Z%29%20%3DMpq%20%5B%281-p%29%20%2B%20%281-q%29p%5D%3D%20Mpq%5B1-p%20%2Bp-pq%5D%3D%20Mpq%5B1-pq%5D)
And as we can see then we can conclude that
