Binomial Distribution for successive events

[SSGD]

So I new to probability and need someone to help me out if you could. I wanted to look into the probability of a process being complete if each operation of the process has its own likely hood of success or failure. I know that I should be using a binomial distribution to study the process. Let's say I have 3 operations (A, B, C) that need to be completed in a process. Each operation determines the number of parts that move to the next operations, but the probability of the parts being processed correctly at each operations is independent of the previous operations. So because they are independent (I'm assuming) then...

$P(A \cap B \cap C) = P(A)P(B)P(C)$

This is where I am having issues. The number of parts (n) moving to the next operation is dependent on the last operations but the probability (p) for each operation isn't dependent on the last operation or the next operation.

If my last assumption is true then I want to take the product of all three to find the distribution for the whole process. If I trial n parts through the entire process.

$P(A \cap B \cap C) = BINOMDIST(k,n,p_A)BINOMDIST(k,n,p_B)BINOMDIST(k,n,p_C)$

I need some help with the mode of this product. All I can think of doing is taking the derivative with respect to k and setting it to zero to solve for the mode, but this being discrete I know this is not correct. I have looked for a derivation for the mode of the Binomial Distribution online, but all I find is the solution not the steps to get there. I think if I saw the whole derivation of it I could handle the mode of the product.

Also, do powers of the binomial coefficients simplify. This looks like it could get messy.

Thanks for any help you can offer.

 

[Stephen Tashi]

$P(A \cap B \cap C) = P(A)P(B)P(C)$This is where I am having issues. The number of parts (n) moving to the next operation is dependent on the last operations but the probability (p) for each operation isn't dependent on the last operation or the next operation.

One reason you are having issues is that you haven't defined A,B,C as "events" in a "probability space". You merely used A,B,C as names for processes. So it isn't clear what you mean by notation like $P(A)$.

If I trial n parts through the entire process.

$P(A \cap B \cap C) = BINOMDIST(k,n,p_A)BINOMDIST(k,n,p_B)BINOMDIST(k,n,p_C)$

The probability that a particular part W is finally completed is $p_A \ p_B \ p_C$ so I think you should be using $BINOMDIST(k, n, \ p_A \ p_B \ p_C)$ to find the probability that $k$ parts are totally completed.

 

[SSGD]

P(A) is the probability of k successful parts being produced in operation A if n trials are performed. Same for B and C. Hope this makes sense. As far a probability space I will have to look its definition. Would this be similar to Venn Diagrams or Probability Trees?

 

[haruspex]

P(A) is the probability of k successful parts being produced in operation A if n trials are performed. Same for B and C. Hope this makes sense. As far a probability space I will have to look its definition. Would this be similar to Venn Diagrams or Probability Trees?

If you were to define P(A) as the probability that one part, taken at random, makes it through process A, etc., then you could indeed multiply the three probabilities together to find the probability that a given part makes it through all three. I believe that approach will give you what you are after.
But with the definition you give above, it makes no sense to multiply them together. If only k make it through A, n is no longer interesting in B. What you would end up with is P_A(n,k)P_B(k,l)P_C(l, m) as the probability that k make it through A, l of those make it through B, and m of those make it through C. But I'm guessing you only care about n and m there, not k and l.

 

[SSGD]

Yeah you are right. I am over complicating this problem. The probability of a part making thru A B and C is the product of pA pB and pC. So if I trailed n parts then BINOMDIST(k,n,pApBpC) is the distribution for the entire process. I was mixing up the probability of one part and the probability of k parts in n trials. I drew a probability tree last night and it started to make sense. Hopefully I'm understand this correctly. I just someone to review what I'm doing because I am trying to self teach. Thanks for all the help.

 

[haruspex]

Yeah you are right. I am over complicating this problem. The probability of a part making thru A B and C is the product of pA pB and pC. So if I trailed n parts then BINOMDIST(k,n,pApBpC) is the distribution for the entire process. I was mixing up the probability of one part and the probability of k parts in n trials. I drew a probability tree last night and it started to make sense. Hopefully I'm understand this correctly. I just someone to review what I'm doing because I am trying to self teach. Thanks for all the help.

Yes, that looks right.