Computing Fundamentals 2 Lecture 6 Probability. Lecturer: Patrick Browne

Transcription

1 Computing Fundamentals 2 Lecture 6 Probability Lecturer: Patrick Browne

2 Probability If a die is thrown we consider it certain that it will land, with a random chance that it will show a 6. With s successes out of n experiments f=s/n is called the relative frequency of success. It becomes stable in the long run. It is this long term stability (limit) that forms the basis of probability.

3 Sample Space and Events The sample space S is the set of all possible outcomes of a given experiment. An element or outcome in S is called a sample point (or sample). An event A is a set of outcomes, it is a subset of the sample space. The singleton {a} where a S is called an elementary event. The empty set,, sometimes represents an impossible event.

4 Sample Space and Events An event gives rise to a set hence we can use set operations to combine events. A B is the event that occurs whenever A occurs or B occurs (or both) A B is the event that occurs whenever A and B both occur. A c is the event that occurs whenever A does not occur (called the complement of A) Two events are mutually exclusive if they are disjoint: A B =.

5 Sample Space and Events Toss a die and observe the top number S={1,2,3,4,5,6} A even number event, B odd number event, C prime number event. A={2,4,6} B={1,3,5} C={2,3,5} A C ={2,3,4,5,6} B C = {3,5} C c = {1,4,6} (1 is not prime) A and B are mutually exclusive.

6 Sample Space and Events Toss 3 coins and observe the H & T sequence S={HHH,HHT,HTH,HTT,THH,THT,TTH,TTT} Let A be the two consecutive heads event, B same outcome event. A={HHH,HHT,THH} B={HHH,TTT} A B = {HHH} is the elementary event with only heads.

7 Probability Spaces A probability space is a triple (S, A, P), where: S sample space, all possible outcomes. A event space, sample events/outcomes P is a probability measure. We also have a set of probability axioms e.g. the probability of an event is a nonnegative real number.

8 Probability Spaces A probability space consists of a sample space together with a positive, additive measure, called a probability measure, which sums to one; the points of the sample space represent the different possible outcomes of the phenomenon, and the probability measure assigns probabilities to sets of outcomes.

9 Finite Probability Spaces Let S be a finite sample space S={a 1,a 2,a 3...a n }. A finite probability space is obtained by assigning to each sample point a i S a real number p i, called the probability of a i satisfying the following conditions: Each p i is non-negative. The sum of p i is one. We write P(A) for the sum of the probabilities sample points in A.

10 Finite Probability Spaces Three runners A,B,C; A is twice a likely to win as B, and B is twice as likely to win as C. What is P(A),P(B),P(C) winning? Let P(C) = p P(B) = 2p P(A)=4p p+2p+4p=1 therefore p = 1/7 P(A)=4/7, P(B)=2/7, P(C)=1/7 P({B,C}) = P(B)+P(C)=3/7

11 Equiprobable Spaces If all the sample points within a given finite probability space are equal to each other, then it is known as an equiprobable space. An example would be a fair die, where each number is equally possible P(1) = P(2) = P(3) = P(4) =P(5) = P(6) = 1/6

12 Equiprobable Spaces If S contains n points, then the probability of each point is 1/n. If an event A contains r points then its probability is: r 1/n = r/n P(A) = number of elements in A number of elements in S

13 Equiprobable Spaces S = cards in the deck = 52 A = card is spade B = card is a face P(A) = 13/52 P(B) = 12/52 P(A B) = 3/52

14 Axioms of Finite Probability Spaces 1. For every event A, 0 P(A) 1 2. P(S)=1, where S is sample space, 3. If events A and B are mutually exclusive (or disjoint), then P(A B) = P(A) + P(B)

15 Theorems of Finite Probability 1. P( ) = 0 Spaces 2. P(A c )= 1 P(A) 3. P(A\B)= P(A) - P(A B) 4. A B implies P(A) P(B) 5. P(A) 1 6. P(A B) = P(A) + P(B) - P(A B) 7. P(A B) = P(A) P(B A) Where P(B A) reads the probability B given A

16 Addition P(A B) = P(A) + P(B) - P(A B) Sums are used when we have two events, and we want to know the probability that either event occurs (Event A union Event B). In the Addition Rule, A and B may or may not be disjoint. Mutually exclusive or disjoint events cannot occur together, so we have: P(A B) = 0. Then the addition rule reduces to: P(A U B) = P(A) + P(B)

17 Addition Rule Example Suppose a student is selected at random from 100 students where 30 are taking maths, 20 are taking chemistry, and 10 are taking maths and chemistry. Find the probability p that the student is taking maths or chemistry. P(M) = 30/100, P(C)=20/100 P(M C) = 10/100 P(M C)=P(M) + P(C) - P(M C) P(M C)= 30/100+20/100 10/100=2/5

18 Rule of Multiplication Is used when we want to know the probability that two events occur (Event A intersection Event B). Rule of Multiplication The probability that Events A and B both occur is equal to the probability that Event A occurs times the probability that Event B occurs, given that A has occurred. P(A B) = P(A) P(B A)

19 Rule of Multiplication A bag contains 6 red marbles and 4 blue marbles. Two marbles are drawn without replacement from the bag. What is the probability that both of the marbles are blue? A = first marble is blue, B = second marble is blue. Therefore, P(A) = 4/10, P(B A) = 3/9. Using P(A B) = P(A) P(B A) P(A B) = (4/10) * (3/9) = 12/90 = 2/15

20 Rule of Multiplication A bag contains 6 red marbles and 4 blue marbles. Two marbles are drawn with replacement from the bag. What is the probability that both of the marbles are blue? A = first marble is blue, B = second marble is blue. Therefore, P(A) = 4/10, P(B A) = 4/10. Using P(A B) = P(A) P(B A) P(A B) = (4/10) * (4/10) = 16/100 = 4/25

21 Conditional Probability E is an event in S with P(E)>0. Conditional probability of A is defined as the probability that A has occurred after E has occurred. We say the conditional probability of A given E: P(A E) = P(A E) P(E) P(A E) = number of elements in A E number of elements in E

22 Example: Conditional Probability Alternatively P(A E) = number of ways A and E can occur number of ways E can occur Given the sum of a pair of tossed die is 6. E={sum is 6},5 ways = {(1,5),(2,4),(3,3),(4,2),(5,1)} A= {has at least one two},2 ways= {(2,4),(4,2)} P(A E)=2/5

23 Example 2: Conditional Probability P(A E) = number of ways A and E can occur number of ways E can occur From a class has 12 boys and 4 girls, 3 students are selected. What is the probability that they are all boys? P=Comb(12,3)/Comb(16,3)=11/28 Alternatively P=(12/16)(11/15)(10/14) = 11/28

24 Independence Two events are independent if the occurrence of one of the events gives us no information about whether or not the other event will occur; that is, the events have no influence on each other. We say that two events, A and B, are independent if the probability that they both occur is equal to the product of the probabilities of the two individual events, i.e. P(A B) = P(A) P(B) If two events are independent then they cannot be mutually exclusive (disjoint) and vice versa.

25 Example: Independence Events A and B are independent if P(A B) = P(A) P(B) otherwise they are dependent. A coin tossed three times: S={HHH,HHT,HTH,HTT,THH,THT,TTH,TTT} A={first toss head} B={second toss head} C={exactly 2 heads tossed in a row}

26 Example: Independence Continuing, coin tossed three times: P(A)={HHH,HHT,HTH,HTT}=4/8 (1 st head) P(B)={HHH,HHT,THH,THT}=4/8 (2 nd head) P(C)={HHT,THH} = 1/4 (2 heads in row) P(A B)=P({HHH,HHT})= 1/4 P(A C)=P({HHT})= 1/8 P(B C)=P({HHT,THH})= 1/4

27 Example: Independence Continuing, coin tossed three times: P(A B)=P({HHH,HHT})= 1/4 P(A C)=P({HHT})= 1/8 P(B C)=P({HHT,THH})= 1/4 P(A)P(B)=(1/2) (1/2)=(1/4)= P(A B) P(A)P(C)=(1/2) (1/4)=(1/8)= P(A C) P(B)P(C)=(1/2) (1/4)=(1/8) P(B C) Not independent, B and C are dependent.

28 Repeated Trials The Law of Averages states, in the long run, over repeated trials, random fluctuations eventually average out and the average of our observations will approach the expected value. But at the same time with increasing numbers of observations, the number of observations that differ from what we expect will be larger.

29 Repeated Trials Let S* be a finite probability space. By n independent or repeated trials we mean the probability space S consisting of all ordered n-tuples of elements of S*, with the probability of n-tuple defined to be the product of the probabilities of its components. P(s 1,s 2,s 3...s n )=P(s 1 ) P(s 2 ) P(s n )

30 Repeated Trials Let probability space S*={P(a),p(b),P(c)} represents probabilities three runners winning a race. Their probabilities of winning are P(a)=1/2, P(b)=1/3, P(c)=1/6. If there are two races then the sample space S consisting of two repeated trials is: S={aa,ab,ac,ba,bb,bc,ca,cb,cc}

31 Repeated Trials S={aa,ab,ac,ba,bb,bc,ca,cb,cc} The probability of the sample points of S are: P(aa)=(1/2) (1/2)=1/4 P(ab)=(1/2) (1/3)=1/6 P(ac)=(1/2) (1/6)=1/12 P(ba)=(1/3) (1/2)=1/6 P(bb)=(1/3) (1/3)=1/9 P(bc)=(1/3) (1/6)=1/18 P(ca)=(1/6) (1/2)=1/12 P(cb)=(1/6) (1/3)=1/18 P(cc)=(1/6) (1/6)=1/36 The probability of c winning first race and a the second is P(ca)=1/12 EXCEL =(1/4)+(1/6)+(1/12)+(1/6)+(1/9)+(1/18)+(1/12)+(1/18)+(1/36)

32 Bernoulli Trials with 2 possible outcomes. A Bernoulli trial is a random experiment in which there are only two possible outcomes - success and failure. If p is the probability of success, then q=1-p is the probability of failure. Often we are interested in the number of successes without considering their order. The probability of exactly k successes in n repeated trials is: b(k,n,p)= n p k q n-k k

33 Example: Trials with 2 possible outcomes. A coin is tossed 6 times, H=success,T=failure. n=6, p=q=1/2 The probability of two heads, (k=2) Binomial coefficient 6 b(2,6,1/2)= (1/2) 2 (1/2) 4 =15/64 2

34 Identically Distributed variable Same probability distributions