Probability (homework) problem involving (I think) binomial distributions and conditional probability.

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Having some trouble with this homework assignment; not looking for a direct answer but some guidance.

I have a tortilla making machine for which I need to replace a part in order to get it working. I evaluate the efficiency of the tortilla maker by the proportion of failures it produces (the number of square tortillas). It can only ever produce round or square tortillas.

I have a box of replacement parts that I need to choose one from and place into my tortilla making machine. These parts come in two types: Type 1 has a failure rate of .4, and Type 2 has a failure rate of .75. I also know that, in that box, 30% of the replacement parts are of type 1.

I choose a replacement part from the box at random, a place it into the machine and I use the machine to make 30 tortillas; of these, I find that 16 of the tortillas it created are square (failures).

Question: What is the probability that I picked a Type 1 part?

Answer so far:

I think this problem involves binomial distributions and conditional probability.

The first thing I did was calculate the probability that a Type 1 and Type 2 part could produce a 16/30 outcome. To do this I used binomial distributions and produced these results:

  • Type 1 : p = .0489 (That is, a Type 1 part is going to produced 16/30 failed tortillas just shy of 5% of the time)
  • Type 2 : p = .0054

My trouble is how exactly to incorporate the information regarding the underlying distribution in the box. Would I use conditional probability here? That is: Calculate:

P(I picked a Type 1 part Given that it produced 16/30 failures)?

Would I need to calculate a general probability of producing 16/30 failures, independent of part Type?

Looking for Guidance! Thanks!