How Do You Calculate the Probability of Specific Fits in Multiple Assemblies?

[USN2ENG]

Homework Statement

What is the probability that among a collection of 20 assemblies we have exactly 2 loose-fits and 1 interference fits?

P(loose-fit) = .1335
P(Interference fit) = .083

Homework Equations

The Attempt at a Solution

I am thinking that I just add the two binomial distributions together.

P(Loose = 2) = (20 choose 2)*.1335^2 * (1-.1335)^18
+
P(Interference = 1) = (20 choose 1)*.083^1 * (1-.083)^19

This comes out to: .25677 + .31998 = .57675

But, I feel like I am missing something. Anyone that can confirm or deny will be much appreciated!
Thanks!

 

[USN2ENG]

Just in case anyone comes across this in the future. It is done by using the multinomial distribution. Apparently the binomial is a generalization of the multinomial with k=2. Good to know.

 

[Ray Vickson]

Just in case anyone comes across this in the future. It is done by using the multinomial distribution. Apparently the binomial is a generalization of the multinomial with k=2. Good to know.

It exactly opposite to what you say: the multinomial is a generalization of the binomial, and not the other way round.