STAT3600

1. STAT3600 Lecture 3 Chapter II Probability

2. Probability • Probability is the study of randomness and uncertainty. • An experiment is any action or process whose outcome is subject to uncertainty. • Tossing a coin, drawing cards from a deck, rolling a die, etc. • The sample space of an experiment, denoted by U, is the set of all possible outcomes of that experiment.

3. Examples: ExperimentSample Space Tossing a coin U={Head, Tail} Roll a die U={1,2,3,4,5,6} Example 2.3 Two gas stations, each with 6 pumps. Observe the number of pumps in use in both stations at a certain time of day. 0 1 2 3 4 5 6 0 (0,0) (0,1) (0,2) (0,3) (0,4) (0,5) (0,6) 1 (1,0) (1,1) (1,2) (1,3) (1,4) (1,5) (1,6) 2 (2,0) (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) 3 (3,0) (3,1) (3,2) (3,3) (3,4) (3,5) (3,6) 4 (4,0) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) 5 (5,0) (5,1) (5,2) (5,3) (5,4) (5,5) (5,6) 6 (6,0) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6) 2nd Station 1st Station

4. An event is any collection (subset) of outcomes contained in the sample space of an experiment. • Simple (or elementary) event: Consists of exactly one outcome. • E.g. say we are observing one gas station only. U={0,1,2,3,4,5,6}. A simple event would be E={2} meaning only two pumps are in use. Similarly E0={0}, E1={1}, …, E6={6} are all simple events. • Compound event: Consists of more than one outcome. • E.g. A compound event would be having at most three gas pumps in use, i.e. E={0,1,2,3}. E is a combination of E0, E1, E2 and E3.

5. Relations with set theory (Let C={2}, D={2,4}) • Union of two events A and B is denoted by A  B and read “A or B”: is the event consisting of all the outcomes that are either in A or in B, or in both. • E.g. The event C  D={2,4}, meaning either 2 or 4 pumps are in use. • Intersection of A and B is denoted by A  B and read “A and B”: is the event consisting of all outcomes in both A and B. • E.g. The event C  D={2}, meaning only 2 pumps are in use. • Complement of an event A is denoted by A: is the event consisting of all the outcomes that are not contained in A. • E.g. The event C={0,1,3,4,5,6}, meaning not 2 pumps are in use, i.e. the event C would occur if there were no pumps in use, or a single pump in use, or three, or four, or five, or six pumps are in use.

6. Mutually exclusive events: If events A and B have no outcomes in common, then they are called “mutually exclusive” or “disjoint” events. • Example: Roll a die. • Define events A, B and C as: • A=Rolling a 6, i.e. A={6} • B=Rolling even, i.e. B={0,2,4,6} • C=Rolling odd, i.e. C={1,3,5} • A and C are mutually exclusive. • A and B are not mutually exclusive. • B and C are mutually exclusive.

7. Properties of probability • Our objective is to assign probability, P, values to events. P(A) denotes the probability that the event A will occur once the experiment is conducted. • For any event A, P(A)  0 • P(U) = 1 • If A1, A2, A3, …, Ak is a finite collection of mutuallyexclusive events, then • P(A1A2A3…Ak) = • If A1, A2, A3, … is an infinite collection of mutuallyexclusive events, then • P(A1A2A3…) =

8. Example • Toss a coin. The sample space U={H,T}, where H represents head, and T represents tail. • What are P(H) and P(T)? • We know that H and T are mutually exclusive. • HT=U  P(HT) = P(U) = 1 •  P(H) + P(T) = 1  P(H) = 1 – P(T) • If fair coin, P(H) = 0.5  P(T) = 0.5

9. How to assign probability values to experiments we are not familiar with? • Consider an experiment that can be repeated in an identical manner and independent fashion, and let A be an event whose probability value is of interest. • If the experiment is conducted n times, on some of the replications the event A will occur, and for the other replications it will not occur. • Let n(A) denote the number of replications with outcome A. Then n(A) / n is called the relative frequency of occurrence of the event A in a sequence of n replications. • As n grows large, n(A) / n ratio converges to a steady number, called the limiting relative frequency, which is used to estimate P(A).

10. Example: • Roll a fair dice 600 times. Let event A={2}. • n(A)=94 • n(A) / n = 94 / 600 = 0.15666  P(A)=0.15666 • Theoretically, for a fair dice, we would expect P(A) = 1/6 = 0.16666. • If we continue to do the experiment, n(A)/n will get closer and closer to 0.16666 as the number of experiments increases.

11. 1 2 3 4 5 • For any event, P(A) = 1 – P(A) • Example: A series system of five identical components. Each component can either be failed (F) or successful (S). If any of the components fail, the system fails. • For each component, P(F)=0.10, P(S)=0.90 • Let A be the event that the system fails. There are 31 different ways that the system will fail. • E.g. FSSSS, SFSSS, SFSFS, etc. • P(A) calculation will be hard if we calculate and add the probabilities of all possible failure modes. • There is only one way the system can be successful, • SSSSS = A • P(A)=1-P(A)  P(A) = 1  0.900.900.900.900.90 = 1  0.905  P(A) = 1  0.59 = 0.41, i.e. 40% of these systems will not be successful.

12. If A and B are mutually exclusive, then P(AB)=0 • For any two events A and B, P(AB)=P(A)+P(B)  P(AB) Example 2.14 of Devore (page 62): 60% of households subscribe to metropolitan newspaper. 80% subscribed to the local paper. 50% subscribed to both. If a household is selected at random, what is the probability that it subscribes to (1) at least one newspaper and (2) exactly one of the two? A={subscribes to metropolitan}, B={subscribes to local} P(A)=0.60, P(B)=0.80, P(AB)=0.50 P(subscribes to at least one)=P(AB) = 0.60+0.80-0.50 = 0.90 P(exactly one) = P(“subscr. to local only” or “subscr. to metropolitan only”) P(exactly one) = P(“Not A and B” or “Not B and A”) = P(AB)+P(BA) P(AB) = P(A)+P(AB) = 0.60 + P(AB) = 0.90  P(AB) = 0.30 P(AB) = P(B)+P(AB) = 0.80 + P(AB) = 0.90  P(AB) = 0.10  P(exactly one) = 0.30 + 0.10 = 0.40 A B 0.5 0.1 0.3 0.1

13. Number of all elementary outcomes in the event set • Determining probabilities systematically: • Given equally likely outcomes for elementary events. • Example • Three coins are tossed. • U={HHH,THH,HTH,HHT,TTH,THT,HTT,TTT}, N(U)=8 • What is the probability that all are heads? • A1={HHH}, N(A)=1  P(A) = 1/8 • What is the probability that at least 2 are heads? • A2={HHH,THH,HTH,HHT}, N(A)=4  P(A) = 4/8 = 0.50 Number of all possible outcomes

14. Counting techniques • If the first element of an ordered pair can be selected in n1 ways, and the second element can be selected in n2 ways, the number of possible pairs is then equal to n1n2. • Example 1 • 8 Possible colors to paint the living room. • 5 Possible colors to paint the kitchen. • Each of them require one can of paint. • A total of 85=40 different ways to select and purchase two cans of paint. • Example 2 • In how many different ways can a local union with 40 members select a president, followed by a VP and then a secretary treasurer? ANS : 40  39 38 = 59280 different ways.

15. Permutations • Any ordered sequence of k objects taken from a set of n distinct objects is called a permutation of size k of the objects. The number of permutations of size k that can be constructed from the n objects is denoted by Pk,n. • Example • Find the total number of permutations of size 3 that can be formed by using the elements of {a,d,f,g}. • P3,4 = 4!/(4-3)! = 4!/1! = 4! = 24

16. Combinations • Given a set of n distinct objects, any unordered subset of size k of the objects is called a combination. The number of combinations of size k that can be formed from n distinct objects is denoted by . • Example • Consider the set {A,B,C,D,E} • There are 5!/(5-3)! = 60 permutations of size 3. • There are 5!/[3!(5-3)!] = 10 combinations of size 3. • Any ordered subset of size 3 can be written in 6 different ways: • {A,C,E}, {C,E,A}, {A,E,C}, {C,A,E}, {E,A,C}, {E,C,A} • Above subsets are different from permutation viewpoint but same from combination viewpoint!