At a plant, $30$% of all the produced parts are subject to a special electronic inspection. It is known that any produced part which was inspected electronically has no defects with probability $0.90.$ For a part that was not inspected electronically this probability is only $0.80.$
Let E be the event that a part in the plant went through an electronic inspection.
Let D be the event that a part produced in the plant is defective.
Find the following
A- $P(E)$
B- $P(D|E) $
C- $P(D^c|E)$
D- $P(D|E^c)$
E- $P(D^c|E^c)$
F- $P(D)$
G- $P(E|D)$
$--------------------------------------$
$$\mathbf{(A)}\\P(E)=.3 \qquad P(E^c)=.7 \\ \mathbf{(B)}\\ P(D|E)= 1-P(D^c|E)=1-.9=.1 \\ \text{Probability that any part was inspected electronically has no defects } = P(D^c|E)=.9 \\ \mathbf{(C)}\\ P(D^c|E) =.9 \\\mathbf{(D)}\\ \text{This is where I am unsure about.}\\\text{ Given that for a part that was not inspected electronically. Probability of no defect is }\\ \Longrightarrow P(D^c|E^c)=.80 \\ P(D|E^c) = 1-P(D^c|E^c)= 1-.8 =.20\\ \color{red}{Or} \text{ is this the correct interpretation?} \\ \mathbf{}\bbox[5px,border:2px solid red] { P(D|E^c) = .8 \qquad P(D^c|E^c) = 1-P(D|E^c) = 1-.8 =.20 } \\ \mathbf{(F)} \\\text{Using Law of Total Probability }\\ P(D) = P(E)\cdot P(D|E)+P(E^c)\cdot P(D|E^c) \\ \mathbf{(G)}\\ \text{ Using Bayes Rule }\\ P(E|D) = \frac{P(D|E)P(E)}{P(D)} \\ $$Ultimately I was trying to understand which interpretation is the correct one. I know that A|B means Event A occurs given that event B occurs, what seems to cause my misunderstanding is the semantics.
