Chapter5 Probability. Introduction. We will consider random experiments with chance outcomes. Events are outcomes that may or may not occur. Notation: Capital letters like E will denote events Probability of event E=P(E) Frequency interpretation of probability: P(E) can be approximated by the relative frequency of occurrences of E in a very large number of repetitions of our chance experiment. Ex. In a 1024 single baby births in a Phoenix hospital 507 babies were girls. If A=event that a girl is born in that hospital, then P( A) 507 1024 PROBABILITY (classical approach) The Sample Space (S) associated with any experiment is the set of all possible outcomes that can occur as a result of the experiment. So naturally, we will call each element of the sample space an outcome. An event (E) is any subset of the sample space. The probability of an event E (written as P(E)) in a sample space (S) with equally likely outcomes is given by number of outcomes in E P(E) = number of outcomes in S EXAMPLE1 Consider the experiment of rolling a fair die two times (or a pair of dice once) The figure below gives a representation of all the 36 equally likely outcomes of the sample space associated with this experiment. (1,1) (2,1) (3,1) (4,1) (5,1) (6,1) (1,2) (2,2) (3,2) (4,2) (5,2) (6,2) (1,3) (2,3) (3,3) (4,3) (5,3) (6,3) (1,4) (2,4) (3,4) (4,4) (5,4) (6,4) (1,5) (2,5) (3,5) (4,5) (5,5) (6,5) (1,6) (2,6) (3,6) (4,6) (5,6) (6,6) If for example E=sum of both rolls is 7, then E={(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)} P(E)= 6/36=1/6
Outcomes in a sample space do not have to be equally likely, next example will consider that case: EXAMPLE2 Suppose we roll a die that is not balanced and probabilities of each face showing are as follows: P(1)=0.1, P(2)=0.1, P(3)=0.2, P(4)=0.5, P(5)=0.05, P(6)=0.05 Let A=event that we will get a number larger than 4, A={5, 6} P(A)=P(5)+P(6)=0.05+0.05=0.1 In general if sample space S={ e 1, e 2,....e n } and events e i do not have to be equally likely, but sum of the probabilities of all outcomes in S still equals 1. For event E that consists of some outcomes in S, P(E)=sum of the probabilities of these outcomes: P(E)= P(e i ) Some Probability Rules. 1) 0 P (E) 1 P(impossible event)=0 and P(certain event)=1 2) P (E c )=1 P (E), where E c =not E is an opposite event to E 3) Sum of probabilities of all outcomes in a sample space is 1 4) P ( Aor B)=P ( A)+ P( B) P ( Aand B) or another notation: Notation: A B= Aor B A and B= A B= A & B These rules can be illustrated using Venn Diagrams.
Mutually Exclusive Events Two or more events are said to be mutually exclusive if at most one of them can occur when the experiment is performed, that is, if no two of them have outcomes in common. On the following diagram a) represents mutually exclusive events and b) not mutually exclusive events Example1 continue: Consider an experiment with rolling a balanced die twice (or pair of fair dice once) a) What is the probablity that sum of both dice is not 7. That is P(notE)=1-1/6=5/6 b) P(sum=7 and number on the first roll is 2)=P((2,5))=1/36 since F=number on the first roll is 2={(2,1), (2,2), (2,3), (2,4), (2,5), (2,6)} c) P( sum=7 or number on the first roll is 2)=6/36+6/36-1/36=11/36 Example3. Blood type distribution (in %) in the USA is given in the following table: Blood Type A B AB O % 42 10 4 44 Suppose we select randomly one American, Compute the probabilities of the following events: a) P(AB)=.04 b) P(A or B) =.52 c) P( A c )= 1-.42=.58 d) P(A and O)=0 (Events A, O are mutually exclusive) Example4 Consider a random family with 3 children, denote the sex of each child: B=boy, G=girl Let A, B, C be the following events: A=family has exactly 1 boy C=family has no more than 1 girl D= family has at least 2 girls a) Give sample space : S={GGG, GGB, GBG, BGG, BBB, BBG, BGB, GBB}
b) Compute P(A and C)=0 c) Compute P(A or C)=7/8 d) Compute P(A and D)=3/8 e) Compute P(A or D)=4/8=0.5 f) Compute P(C and D)=0 g) Compute P(C or D)=1 Example5 Following table represents distribution of X=number of children in a family for some community in Tempe: X 0 1 2 3 4 Relative frequency 0.03 0.17 0.50 0.25 0.05 Suppose one family is randomly selected from that community, compute following probabilities: P(family has no children)=0.03 P(family has some children)=1-0.03=0.97 P(family has at most 2 or more than 3 children)=1-0.25=0.75 P(family has at most 2 and more than 3 children)=0 P(family has at least 2 and no more than 3 children)=0.5+0.25=0.75 P(family has at least 2 or no more than 3 children)=1 Example 6 In a study of the relationship between health risks and income levels a large group of people reported that they are stresses (which means extremely or quite stressful most days) or not stressed ( which means most days are a bit stressful, not very stressful or not stressful at all). Their income level was also recorded. Summary is presented in the table below: Income Level Low Medium High total stressed 526 274 216 1016 Not stressed 1954 1680 1899 5533 total 2480 1954 2115 6549 If the random person from this study is selected, compute the following: a) probability that person is stressed= 1016/6549=.155 b) probability that person is low income=2480/6549=.3787 Probabilities in a and b are called marginal probabilities, events are related to one variable only. c) probability that person is stressed and low income= 526/6549=.080 d) probability that person is stressed or is low income=.155+.3787-.080=.4537 Probability in c is called joint probability, it involves both variables.
Following problems use conditional probability rules listed below in optional section. e) conditional probability that low income person is stressed= 526/2480=.2121 (we consider only low income people and out of these we count how many are stressed) (or using the formula: 526 /6549 2480/6549 gives the same answer) f) are events person has low income and person is stressed independent? P(stressed)=.155, P(low income)=.3787, independent.155(.3787)=.059.080, so events are not Example 7 Researchers mapped a large Canadian forest marking the species of trees in it and their stage of life. Data is presented in the table below. Life stage Dead Live Sapling total Western red cedar 0.02 0.10 0.08 0.20 Douglas fir 0.16 0.16 0.01 0.33 Western hemlock 0.23 0.21 0.03 0.47 total 0.41 0.47 0.12 a) What is the probability that randomly selected tree is a live Western hemlock? 0.21 b) What is the probability that a randomly selected tree is a sapling or Douglas fir? 0.12+0.33-0.01=0.44 Use conditional probability for parts c and d, follow optional rules listed below. c) What is the probability that randomly selected tree is a Douglas fir given that it is a sapling? 0.01/0.12=0.083 d) Compute conditional probability that a random tree is live given that it is a Western cedar. 0.10/0.20=0.50
Following is OPTIONAL : Other Probability Rules: P ( Aand B) 5) P ( A/B)= P (B), provided P(B) 0 (Conditional probability) 6) P ( Aand B)=P ( A)P (B) and P ( A/B)=P ( A) for independent events only We can use these rules to answer following questions using Example 6 e) conditional probability that low income person is stressed= 526/2480=.2121 (we consider only low income people and out of these we count how many are stressed) (or using the formula: 526 /6549 2480 /6549 gives the same answer) f) are events person has low income and person is stressed independent? P(stressed)=.155, P(low income)=.3787, independent.155(.3787)=.059.080, so events are not In general: To check independence of two events A, B you either check if P(A B)=P(A) or if P( A B)=P (A )P(B) In case of equality you have independence, otherwise events are not independent.