Example 12.Suppose that a machine produces a defective item with probability p(<p<)and produces a nondefective item with probability 1-p.Suppose further tha six items produced by the machine are selected at random and inspected,and that the results (defective or nondefective)for these six items are independent.We shall determine the probability that exactly two of the six items are defective. Solution.It can be assumed that the sample space s contains all possible arrangements of six items,each one of which might be either defective or nondefective Let Dj denote the event that the jth item in the sample is defective. then D is the event that this item is nondefective. Since the outcomes for the six different items are independent.the probability of obtaining any particular sequence of defective and nondefective items will simply be the product of the individual probabilities for the items.For example. P(DD)=P(D)P(D:)P(D )P(D)P(D)P(D) =1-D)p1-D1-D)p1-D) =p(1-D)° It can be seen that the probability of any other particular sequence in S containing two defective items and four nondefective items will also be p(1-p)". Since there are (6 distinct arrangements of two defective items and four nondefective items They of[名】p产l二p∬ 2.3 Bayes'Theorem:inverting conditional probabilities ConsiderP(BA)=P(AnB).Apply multiplication rule to each side: P(BA)P(A)=P(AB)P(B) Then P(BA)= P(A B)P(B) P(A) This is the simplest form of Bayes Theorem.named after Thomas Bayes(1702-61).English clergyman and founder of Bayesian Statistics. Bayes'Theorem allows us to "invert"the conditioning,i.e.toExample 12. Suppose that a machine produces a defective item with probability p (0<p<1) and produces a nondefective item with probability 1-p. Suppose further that six items produced by the machine are selected at random and inspected, and that the results (defective or nondefective) for these six items are independent. We shall determine the probability that exactly two of the six items are defective. Solution. It can be assumed that the sample space S contains all possible arrangements of six items, each one of which might be either defective or nondefective. Let Dj denote the event that the jth item in the sample is defective, then c Dj is the event that this item is nondefective. Since the outcomes for the six different items are independent, the probability of obtaining any particular sequence of defective and nondefective items will simply be the product of the individual probabilities for the items. For example, (1 ) . (1 ) (1 )(1 ) (1 ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 4 1 2 3 4 5 6 1 2 3 4 5 6 p p p p p p p p P D D D D D D P D P D P D P D P D P D c c c c c c c c              It can be seen that the probability of any other particular sequence in S containing two defective items and four nondefective items will also be (1 ) . 2 4 p  p Since there are         2 6 distinct arrangements of two defective items and four nondefective items. The probability of obtaining exactly two defectives is (1 ) . 2 6 2 4 p  p         2.3 Bayes' Theorem: inverting conditional probabilities Then This is the simplest form of Bayes' Theorem, named after Thomas Bayes (1702-61), English clergyman and founder of Bayesian Statistics. Bayes' Theorem allows us to “invert” the conditioning, i.e. to