# Conditional probability calculation

This is a homework question about probability that I do not understand, even after discussing it with my teacher. I understand the correct answer, but don't understand why my answer is wrong.

A company buys rejected devices from two manufacturers. 60% of the devices are made by manufacturer A, 40% are made by manufacturer B. 10% of A's devices are defective. 30% of B's devices are defective.

A new shipment arrived. Due to sloppy handling, we don't know which device comes from which manufacturer.

We take a sample of 20 devices. They are tested thorougly. 2 devices are defective.

Question: What is the probability of finding two defects if we don't know who the manufacturer was?

My answer: general probability of finding a defect is $0.6\cdot0.1 + 0.4 \cdot 0.3 = 0.18$. Probability of finding two defects in a sample of 20 is then $\binom{20}{2} \cdot 0.18^2 \cdot 0.82^{18} = 0.1730$

This is not correct. The correct answer is to first find the probability of two defects when we know that the devices are made by A ($0.2852$) or B ($0.0278$). We then do $0.6\cdot 0.2852+0.4\cdot 0.0278=0.18224$.

I would like to understand why my method is wrong, what my error in thinking is.