Answer:
There are 10 components with 3 that are defective:
A = event that the first component is defective
B = event that the second component is defective
b) P(A) = this is the probability that the first component is defective:
= number of defective component/total number of component
P(A) = 3/10 = 0.3
b) P(B|A) = the probability that the second component is defective after the first defective component. Since A has occurred:
= number of defective component left/total number of component left
P(B|A) = 2/9 = 0.22
c) P(A ∩ B) = probability that both are defective i.e. A and B occurring
= 3/10 * 2/9
P(A ∩ B) = 6/90 = 0.06
d) P(Ac ∩ B) = probability that A is not defective and B is defective
P(Ac) = 1 - 3/10 = 7/10 = 0.7
∴ P(Ac ∩ B) =7/10 * 2/9 = 14/90 = 0.16
e) P(B) = probability that the second component is defective
= 2/9 = 0.22
f) B is dependent on A because the defective component is not replaced after the first selection. There would be 9 items left after the first selection of which 2 are defective.
Step-by-step explanation:
