Patrick L. answered 8d
There are 9 components with 3 of them defective (not working).
You are randomly selecting 2 components.
A = first component is defective
B = second component is defective
a.) P(A) = 3/9 = 1/3 or 0.3333 [the probability of drawing a first component is defective)
b.) P(B | A) = 2/8 = 1/4 or 0.25 [the probability of drawing a second component is defective, given that the first component is defective]
c.) P(A and B) = P(A) × P(B) = (1/3)*(1/4) = 1/12 or 0.08333
d.) Events A and B are independent because picking the first component will affect on picking the second component when there is no replacement. In this case, you don't put back the component in the original sample every time you draw the next component.