# probability problem

3 machines (a1,a2,a3) produce parts

a1=45% a2=35% a3=20% 10% of a1 are defective 6% of a2 is defecyive 2% of a3 is defective a)prob of a random part chosen to not be defective
b)part produced by a1

a) i thought of finding the defective first then subtracting it from 1 to get the not defective so...defective= d p(d|a1)=.1 p(d|a2)=.06 p(d|a3)=.02 p(d)= p(dUa1)+p(dUa2)+p(dUa3)=.1x.45+.06x.35+.02x.2 then 1-(above calculation) should give me the not damaged part right?

for b is it just .45? because it asks for the prob that it was produced by a1, so 45% is produced by a1 so 45/100?

-
correct on both counts –  gt6989b Jul 11 '13 at 3:51
i was almost positive about the first part but the second seems to be too easy to be a part of such a problem. –  spiros Jul 11 '13 at 3:55
If you think (b) is too easy, it perhaps meant to ask what is the chance a defective part is produced by $a_1$?
If that's so, it is $$\frac{\mathbb{P}[\text{defective part} \cap \text{produced by} a_1]} {\mathbb{P}[\text{part is defective}} = \frac{0.1 \cdot 0.45}{0.1 \cdot 0.45 + 0.06 \cdot 0.35 + 0.02 \cdot 0.2} = \frac{9}{14} \approx 64.3 \%.$$