I'm trying to find a general formula for the following case :
A production process has one of three possible states: (1) in control, (2) out of control with a type 1 problem, or (3) out of control with a type 2 problem. If the process is in control, an average 6% defectives are produced. If the process is out of control with a type 1 problem 12% defectives are produced. If the process is out of control with a type 2 problem 20% defectives are produced. The production manager, who has been watching the process for a number of hours, assigns probabilities for the three possible conditions: P(in control)=0.7, P(type 1)=0.2 and P(type 2)=0.1.
(i) A sample of one item is taken. It is found to be defective. What are the production manager's revised probabilities for the three possible states of the process?
(ii) Another item is now selected and it also is found to be defective. What are the production manager's new revised probabilities for the three possible states of the process?
(iii) If the production manager samples two items simultaneously and finds them both to be defective, determine the production manager's new revised probabilities for the three possible states of the process.
(iv) Determine the general formula for the above revised probabilities for the three possible states (in control, type 1 and type 2) in the situation where n items are sampled and all n items are found to be defective.
What I've done so far is i. and ii. :
(i)
P( in control | defective ) = 0,042 / (0,042 + 0,024 + 0,02) = 0,488
P( type 1 | defective ) = 0,024 / (0,042 + 0,024 + 0,02) = 0,279
P( type 2 | defective ) = 0,02 / (0,042 + 0,024 + 0,02) = 0,233
And :
(ii)
P( in control | defective ) = 0,029 / (0,029 + 0,033 + 0,047) = 0,266
P( type 1 | defective ) = 0,033 / (0,029 + 0,033 + 0,047) = 0,303
P( type 2 | defective ) = 0,047 / (0,029 + 0,033 + 0,047) = 0,431
And now this where I'm stuck, I don't see how to apply a binomial distribution to find a general formula for n repetition of the event.
In there are $n$ consecutive defectives, I would expect the posterior probabilities to be
$$P(\text{in control}\mid n \text{ defectives}) = \dfrac{0.7 \times 0.06^n}{0.7 \times 0.06^n+0.2 \times 0.12^n+0.1 \times 0.20^n}$$
and similarly for types $1$ and $2$ taking the numerator to be the the relevant part of the denominator.
Eventually I might wonder whether there was a type $3$ out of control state where everything was defective.