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I'm not sure if there is a gap in my stats knowledge or Mathematica knowledge but it's time to ask for help.

I have the following system I am trying to calculate the condition probability of getting to stage 3.

Finding the joint distribution is straightforward:

Block[{A, B, C, p},
 {A, B, C} = {0.2528674027226927, 0.506570580754557, 0.705439330543933};

 p = NProbability[
   c == 1 && b == 1 && 
    a == 1 , {a \[Distributed] BernoulliDistribution[A], 
    b \[Distributed] BernoulliDistribution[B], 
    c \[Distributed] BernoulliDistribution[C]}];

 p
]
(* results 0.0903634 *)    

However, to me this network needs to be modelled conditionally since how many draws make it to stage 3 depends on passing stage 1 and 2. However, I've tried a couple of different ways and my answers don't seem reasonable when compared to the initial values. I was expecting a value under ~0.25. Does anyone notice anything wrong with me code?

Block[{results, prob, A, B, C},
 {A, B, C} = {0.2528674027226927, 0.506570580754557, 0.705439330543933};

 prob = Probability[
   b == 1 \[Conditioned] 
    a == 1, {a \[Distributed] BernoulliDistribution[A], 
    b \[Distributed] BernoulliDistribution[B]}];

 results = 
  Probability[c == 0 \[Conditioned] prob, 
   c \[Distributed] BernoulliDistribution[C]];
 
 results
 ]
 (* returns \[Piecewise]    
             0.294561   0.506571
             0  True.   *)
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    $\begingroup$ I don't really know what your system represents, but purely from the mathematical viewpoint, the probability $0.0904$ seems correct to me. There is only one way to the stage 3: start -> stage 1 -> stage 2 -> stage 3. So the probability for all three steps to occur is the product of probabilities for each step(since they are – I suppose – independent). Therefore: $p = 0.253 \times 0.507 \times 0.705 = 0.0904$. $\endgroup$
    – Domen
    Jul 13, 2021 at 18:38
  • $\begingroup$ Just wondering... Do you want to model a card game like poker or black jack? Whatever you want to model, I suggest editing your question and adding a description of it. Anchoring something in a specific domain will often get you more and better answers. $\endgroup$
    – Jagra
    Jul 13, 2021 at 18:43
  • $\begingroup$ Is what you want the probability of getting to Stage 3? If you want a "conditional" probability, you need to state what you're conditioning on. $\endgroup$
    – JimB
    Jul 13, 2021 at 20:03
  • $\begingroup$ Never use upper-case letters to start names for variables as they can conflict with Mathematica's internal naming. $\endgroup$ Jul 13, 2021 at 21:41

1 Answer 1

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Couldn't you model this as a DiscreteMarkovProcess?

(* states are start(1), stage1(2), stage2(3), stage3(4), missing(5) *)
proc = DiscreteMarkovProcess[{1, 0, 0, 0, 0},
   {{0, 0.253, 0, 0, 0.747},
    {0, 0, 0.507, 0, 0.493},
    {0, 0, 0, 0.705, 0.295},
    {0, 0, 0, 1, 0},
    {0, 0, 0, 0, 1}
    }];

Graph[proc]

PDF[StationaryDistribution[proc], 4]
(* 0.0904311 which also happens to be 0.253 * 0.507 * 0.705 *)

Stage 3 (vertex 4) and Missing (vertex 5) are both absorbing states in the graph. You could also calculate Probability[x[3] == 4, x \[Distributed] proc] and get the same answer. markov chain graph

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