1 2 Outlines Probability Basic definitions: Randomization - PowerPoint PPT Presentation

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2 Outlines • Probability • Basic definitions:  Randomization experiment  Sample spaces  Elementary outcomes • Basic operations—conditional probability • Bayes Theorem

Randomness 3  Things may happen randomly, for examples o Comparison of treatment effects in clinical trials o Calculation of the risk of breast cancer  Probability o Study of randomness o Language of uncertainty

4 Study of randomness 4  Random Experiment : is the process of observing the outcome of a chance event.

5 Basic Definitions 5  A random experiment for which the outcome cannot be predicted with certainty  But all possible outcomes can be identified prior to its performance  And it may be repeated under the same conditions

6 Basic Definitions 6  Elementary outcomes: are all possible results of the random experiment.  Sample space: is the set or collection of all the elementary outcomes : {Head , Tail}

7 Another ex.: Throw a single dice 7

8 Another ex.: Throw a single die 8 Sample space?

9 Another ex.: Throw a pair of dice 9 Sample space?

10 Another ex.: Throw a pair of dice 10 Sample space?

11 Another ex.: Throw a pair of dice 11 Sample space?

12 Probability theory 12  For a random experiment with n elementary outcomes O 1 , O 2 , ... O n , we assign a numerical weight or probability to each outcome.  P(O i )= the likelihood of the occurrence of an event  Toss a fair coin: P(H)=P(T)=0.5  Throw one die: P(die=5)=1/6  Throw two dice: P(Black 5, white 2)=1/36

13 Properties of probability 13 Let Ω be the sample space for a probability measure P o 0 ≤ P(Oi) ≤ 1, probability is number between 0 and 1; it is non-negative o P(O 1 )+ P(O 2 )+...+P(On)= P( Ω ) = 1 A null set, denoted as ∅, has no elements and P( ∅ ) o = 0

14 Basic Operations

15 Event An Event is a set of elementary outcomes. The 15 probability of an event is the sum of the probabilities of the elementary outcomes in the set. For the dice example Event description Event’s elementary outcomes probability A: dice add to 3 B:dice add to 6 C:White die shows 1 D: Black die shows 1

16 Event An Event is a set of elementary outcomes. The 16 probability of an event is the sum of the probabilities of the elementary outcomes in the set. For the dice example: Event description Event’s elementary outcomes probability 2 A: dice add to 3 {(1,2), (2,1)} P A = ( ) 36 B:dice add to 6 {(1,5),(2,4),(3,3),(4,2),(5,1)} 5 P B = ( ) 36 6 C:White die shows 1 {(1,1),(1,2),(1,3),(1,4),(1,5),(1,6)} P C = ( ) 36 D: Black die shows 1 {(1,1),(2,1),(3,1),(4,1),(5,1),(6,1)} 6 P D = ( ) 36

17 Basic definition 17  Let Ω denote the set comprised of the totality of all elements in our space of interest o If A ⊂ Ω , Ā (complement of A) is the set of all elements of which do not belong to A Ω Ā A

18 Combining events using logical operations 18  For two events A and B, o A or B( A ∪ B) : The event A or the event B occurs (or both do) o A and B (A ∩ B) : The event A and the event B both occur o Not A ( Ā ) : The event A does not occur

Example 19  Let A = {1, 2, 3} and B = {3, 4, 5} o A ∩ B = o A ∪ B =  Let Ω = {1, 2, 3, 4, 5, 6, 7, 8, ...}: the positive integers, and let A = {2, 4, 6, 8, . . .} o Ā =  A = {1, 2, 3} and B = {1, 2, 3, 4} o Does A ⊂ B?

20 Example 20  Let C is the event White die is1 and D is the event Black die is 1 P(C)= P(D)= P(C ∩ D)= P(C ∪ D)=

21 Example 21  Let C is the event White die is1 and D is the event Black die is 1 P(C)= P(D)= P(C ∩ D)= P(C ∪ D)=

22 Addition Rule 22  P(C ∪ D)= P(C) + P(D)- P(C ∩ D) P(C)=1/6 P(D)=1/6 P(C ∩ D)= 1/36 P(C ∪ D)=11/36

23 Mutually exclusive 23  If the overlap of event A and B is empty, i.e. P(A ∩ B)= 0 or A ∩ B = Ø , in this case, we say event A and B are mutually exclusive. A: the dice add to 3 B: the dice add to 6

24 Special Addition Rule 24  If A and B are mutually exclusive, then  P(A ∪ B)= P(A) + P(B)

25 Subtraction Rule 25  P( Ā ) = 1 – P(A)  A: a double 1 is not thrown  P(A)= ?

26 Subtraction Rule 26  P( Ā ) = 1 – P(A)  A: a double-1 is not thrown  P( Ā )=1/36 So P(A)=1- 1/36 = 35/36

27 Conditional Probability 27 Questions: 1. Throw a pair of dice. What’s the probability that the faces sum to 3 (event A)? P(A)=?

28 Conditional Probability 28 Questions: 1. Sample space of A has 36 outcomes. P(A)=

29 Conditional Probability 29 Questions: 1. Sample space of A has 36 outcomes. P(A)=2/36=1/18

30 Conditional Probability 30 Questions: 1. Throw a pair of dice. What’s the probability that the faces sum to 3 (event A)? P(A)= 2/36=1/18 2. Suppose we throw the white die before the black die. The white die comes up 1(event C). What’s the probability of A now? P(A|C)=? : Conditional Probability that event A will occur given event C has already occurred.

31 Conditional Probability 31 Given event C is happened, the sample space is reduced. Reduced sample space of C has 6 outcomes.

32 Conditional Probability 32 Given event C is happened, the sample space is reduced. Reduced sample space of C has 6 outcomes. A: the faces sum to 3 P(A|C)= Event A ∩ C

33 Conditional Probability 33 Given event C is happened, the sample space is reduced. Reduced sample space of C has 6 outcomes. A: the faces sum to 3 P(A|C)=1/6 Event A ∩ C

34 Conditional Probability 34 Event A ∩ C P(A ∩ C)=1/36 A: the faces sum to 3 P(A|C)=1/6 ∩ ( ) P A C Event C, = ( | ) P A C P(C)=1/6 ( ) P C

35 Conditional Probability 35 The Conditional Probability of E given F is  ( ) P E F = ( | ) P E F ( ) P F Fact 1: P E E = ( | ) 1 Fact 2: When E and F are mutually exclusive P E F = ( | ) 0

36 Multiplication Rule 36 = =  ( ) ( | ) ( ) ( | ) ( ) P E F P E F P F P F E P E Independence When P(E)=P(E|F) or equivalently P(F)=P(F|E), E and F are independent When Special Multiplication Rule When E and F are independent =  ( ) ( ) ( ) P E F P E P F

37 37 Independence =  ( ) ( ) ( ) P E F P E P F Event E and F are = ( | ) ( ) P E F P E independent = ( | ) ( ) P F E P F EX. Event C is white die comes up 1; event D is Black 1. die comes up 1. Please show C and D are When independent Event F is the sum of the two dice is 3. Show C 2. and F are not independent

38 Summary  Addition Rule 38 P(C ∪ D)= P(C) + P(D)- P(C ∩D)  Special Addition Rule: when E and F are mutually exclusive P(C ∪ D)= P(C) + P(D)  Subtraction Rule P( Ā) = 1 – P(A)  Multiplication Rule = =  ( ) ( | ) ( ) ( | ) ( ) P E F P E F P F P F E P E  Special Multiplication rule: when E and F are indep. =  ( ) ( ) ( ) P E F P E P F

39 Case of the false positive 39 Suppose a rare disease affects one out of every 1000 people in a population. And suppose that there is a good but not perfect test for this disease If a person has the disease, the test comes back  positive 99% of the time About 2 % of uninfected patients also test positive  Q: If your just tested positive, what are your chances of having the disease?

40 Case of the false positive 40 A: Patient has the disease B: Patient tests positive P(A)= P(B| A)= P(B|Not A)= Q: P(A|B)=

41 Case of the false positive 41 A: Patient has the disease B: Patient tests positive P(A)=0.001 P(B| A)=0.99 P(B|Not A)=0.02 Q: P(A|B)=?

42 Case of the false positive 42 A: Patient has the disease B: Patient tests positive P(A)=0.001 P(B| A)=0.99 P(B|Not A)=0.02 ∩ ( ) P A B Q: by conditional = ( | ) P A B probability ( ) P B

43 Case of the false positive 43 A: Patient has the disease B: Patient tests positive P(A)=0.001 P(B| A)=0.99 P(B|Not A)=0.02 ∩ ( ) ( | ) ( ) P A B P B A P A = = Q: ( | ) P A B ( ) ( ) P B P B By multiplication rule

44 Case of the false positive 44 A: Patient has the disease B: Patient tests positive P(A)=0.001 P(B| A)=0.99 P(B|Not A)=0.02 ∩ ( ) ( | ) ( ) P A B P B A P A = = Q: ( | ) P A B ( ) ( ) P B P B ??

45 Total probability rule 45 A: Disease status B: Test results The sample space can be divided into four mutually exclusive events A not A B ∩A B ∩ not A B not B∩A not B ∩ not A not B = + = +   ( ) ( ) ( ) ( | ) ( ) ( | ) ( ) P B P B A P B A P B A P A P B A P A

46 Bayes Theorem 46 ( ) ( | ) P A P B A = ( | ) P A B + ( | ) ( ) ( | ) ( ) P B A P A P B A P A

47 Case of the false positive A: Patient has the disease 47 B: Patient tests positive P(A)=0.001 P(B| A)=0.99 P(B|Not A)=0.02 ( ) ( | ) P A P B A = ( | ) P A B + Q: ( | ) ( ) ( | ) ( ) P B A P A P B A P A 0.001*0.99 0.00099 = = + 0.99*0.001 0.02*0.999 0.02097 = 0.0472

48 One exercise Ω A2 48 A1  For two events A1 and A2, please shade the area for o A1 ∩ Ā 2