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. There's no other way to tell the two types apart from one another.
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?

Let's call F the event that 16 of the 30 tortillas are failures, A the event that you choose a type 1 part and B the event that you choose a type 2 part.

So, the probability that you picked a Type 1 part given that 16 of the 30 tortillas are failures is calculated as:

P(A/F)=P(A∩F)/P(F)

Where P(F) = P(A∩F) + P(B∩F)

Then, the probability that a type 1 part created 16 failures can be calculated using the binomial distribution as:

[tex]P(x)=\frac{n!}{x!(n-x)!}*p^{x}*(1-p)^{n-x}[/tex]

Where x is the number of failures, n is the total number of tortillas and p is the failure rate, so:

[tex]P(16)=\frac{30!}{16!(30-16)!}*0.4^{16}*(1-0.4)^{30-16}=0.0489[/tex]

Therefore, The probability P(A∩F) that you choose a type 1 part and this part created 16 square tortillas is:

(0.3)(0.0489) = 0.0147

Because 0.3 is the probability to choose a type 1 part and 0.0489 is the probability that a type 1 part created 16 square tortillas.

At the same way, the probability that a type 2 part created 16 failures is:

[tex]P(16)=\frac{30!}{16!(30-16)!}*0.75^{16}*(1-0.75)^{30-16}=0.0054[/tex]

Therefore, P(B∩F) is: (0.7)(0.0054) = 0.0038

Finally, P(F) and P(A/F) are equal to:

P(F) = 0.0147 + 0.0038 = 0.0185

P(A/F) = 0.0147/0.0185 = 0.7946