1 Probability Theory Dr. Deshi Ye

2 Outline:  1. Conditional Probability  2. Bayes’ Theorem

3 3.6 Conditional probability  Probability of an event is meaningful iff it refers to a given sample.  P(A|S): the probability of A given some space S. If respect to more samples.

4 D and I 10 Ex.  500 machines. Improper assemble I= 30;Defective D= 15; Both I and D = 10.  P(D)=?  P(D|I)=?  P(D and I)=? I 20 D and I 10

5 Conditional prob.  Theorem. If A and B are any events is S and P(B) is not empty, the conditional probability of A given B is:

6 Ex.  A coin is flipped twice. If we assume that all four points in the sample space are equally likely, what is the probability that both flips result in heads, given that the first flip does? Solution: Let E = {(H,H)} be the event that both flips land heads, and F={(H,H), (H,T)} denote the event that the first flip lands heads, then the desired probability is given by

7 Ex.  Suppose a family has two children. We assume that the probability of having a baby boy is ½.  Now suppose we wish to find the probability that the family has one boy and one girl, but we also have the information that at least one of the children is a boy.  What is the probability?

8 Ex.  Celine is undecided as to whether to take a French course or a chemistry course. She estimates that her probability of receiving an A grade would be ½ in a French course, and 2/3 in a chemistry course.  If Celine decides to base her decision on the flip of a fair coin, what is the probability that she gets an A in chemistry?

9 Solution  If we let C be the event that Celine takes chemistry and B denote the event that she receive an A in whatever course she takes, then the desired probability is P(CB).

10 Multiplication rule

11 Multiplication rule – Ex.  An ordinary deck of 52 playing cards is randomly divided into 4 piles of 13 cards each. Compute the probability that each pile has exactly 1 Ace?

12 Solution: another approach  Define events E1 = {the ace of spades is in any one of the piles} E2 = {the ace of spades and the ace of hearts are in different piles} E3 = {the aces of spades, hearts, and diamonds are all in different piles} E4 = {all aces are in different piles}  P(E1) = 1, P(E2|E1)=39/51, P(E3|E1E2)=26/50,  P(E4|E1E2E3) = 13/49.  P(E1E2E3E4) = P(E1)P(E2|E1)P(E3|E1E2)P(E4|E1E2E3)  = 0.105

13 Independent  A is independent of B if and only if P(A|B)=P(A) or P(A and B) = P(A) * P(B)

14 EX.  Suppose that we toss 2 dice. Let E denote the event that the sum of the dice is 6 and F denote the event that the first die equals 4.  Q: are the events E and F independent?

15 Solution PP(EF) = P({4, 2}) = 1/36. PP(E) = 5/36 P(F) = 1/6 PP(E)P(F) = 5/216 CConclusion: E and F are not independent!! QQuestion: how about the event E changed to: the sum is 7.

16 Properties IIf E and F are independent, then so are E and IIf E is independent of F and is also independent of G. Is E then necessarily independent of FG? AAns: No!!

17 Generalization  Theorem 3.8 If A and B are any events in S, then if P(A)>0,P(B)>0. If independent

18 Discussion  P(A|B) and P(A ∩ B)  A, B are independent, mutually exclusive, what is the difference?

19 Bayes’ Theorem Event A

20 Bayes’ Theorem  An experiment depends on the outcomes of various intermediate stages.  EX. An assembly plant receives its voltage regulators. 60% from B1, 30% from B2, 10% from B3 Perform according to specifications: B1: 95%, B2: 80%, B3: 65%. Event: a voltage regulator received by the plant performs according to specifications.

21 Tree diagram B1 B2 B3 P(A|B1)=0.95 A A A P(A|B2)=0. 8 P(A|B3)=0.65

22 Solution

23 Rule of total probability  Theorem If are mutually exclusive events of which one must occur, then

24 Special case

25 EX.  A laboratory blood test is 95 percent effective in detecting a certain disease when it is, in fact, present. However, the test also yield a “false positive” result for 1 percent of the health person tested.  Q: If.5 percent of the population actually has the disease, what is probability a person has the disease given that the test result is positive?

26 Solution  Let D be the event that the tested person has the disease.  Let E be the event that the test result is positive.  The desired probability is P(D|E) Only 0.323!

27 Bayes’ Theorem  Theorem. If are mutually exclusive events of which one must occur, then “Effect” A was “caused” by the event B r The prob. Of reaching A via the r-th branch of the tree

28 EX.

29 Expectation  Expectation: If the probability of obtaining the amounts then the mathematical expectation is

30 Motivations  The expected value of x is a weighted average of possible values that X can take on, each value being weighted by the probability that X assumes it.  Frequency interpretation

31 Summary  Conditional probability  Independent of events A, B  Bayes’ rule, B i disjoint