Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Chapter 3: Basic Probability Concepts • Probability: • is a measure (or number) used to measure the chance of the occurrence of some event. This number is between 0 and 1. • An experiment: • is some procedure (or process) that we do. • Sample Space: • The set of all possible outcomes of an experiment is called the sample space (or Universal set) and is denoted by An Event: . is a subset of the sample space · is an event (impossible event) · is an event (sure event) Example: Experiment: Selecting a ball from a box containing 6 balls numbered 1,2,3,4,5 and 6. This experiment has 6 possible outcomes 1, 2, 3, 4, 5, 6 · Consider the following events: E1: getting an event number 2, 4, 6 E2: getting a number less than 4 1, 2, 3 E3 =getting 1 or 3 1, 3 E4 =getting an odd number 1, 3, 5 n no. of outcomes (elements) in nEi no. of outcomes (elements) in Ei Notation Equally likely outcomes: The outcomes of an experiment are equally likely if the occurrences of the outcomes have the same chance. Probability of an event: · If the experiment has N equally likely outcomes, then the probability of the event E is: n E n E no. of outcomes in E PE n N no. of outcomes in Example: In the ball experiment in the previous example, suppose the ball is selected randomly. 1, 2, 3, 4, 5, 6 ; n 6 E1 2, 4, 6 ; nE1 3 E 2 1, 2, 3 E3 1, 3 ; ; n E 2 3 n E3 2 The outcomes are equally likely. 3 PE1 , 6 3 PE 2 , 6 2 PE3 , 6 Some Operations on Events: Let A and B be two events defined on Union: A B A· B Consists of all outcomes in A or in B or in both A and B. A · B Occurs if A occurs, or B occurs, or both A and B occur. Intersection: A B A ·B Consists of all outcomes in both A and B. A B Occurs if both A and B occur . Complement: Ac · Ac is the complement of A. · Ac consists of all outcomes of but are not in A. · Ac occurs if A does not. Ac A Example: Experiment: Selecting a ball from a box containing 6 balls numbered 1, 2, 3, 4, 5, and 6 randomly. Define the following events: E1 2, 4, 6 = getting an even number. E 2 1, 2, 3 = getting a number < 4. E3 1, 3 = getting 1 or 3. E4 1, 3, 5 = getting an odd number. (1) E1 E2 1, 2, 3, 4, 6 = getting an even no. or a no. less than 4. nE1 E2 5 PE1 E2 n 6 ( 2 ) E1 E4 1, 2, 3, 4, 5, 6 = getting an even no. or an odd no. nE1 E4 6 PE1 E4 1 n 6 Note: E1 E 4 E1 and E4 are called exhaustive events. ( 3 ) E1 E2 2 = getting an even no. and a no.less than 4. nE1 E2 1 PE1 E2 n 6 ( 4 ) E1 E4 = getting an even no. and an odd no. nE1 E4 n 0 PE1 E4 0 n 6 6 Note: E1 E 4 E1 and E4 are called disjoint (or mutually exclusive) events. ( 5 ) E1c = not getting an even no. = {1, 3, 5} = getting an odd no. = E4 Notes: 1. The event A1 , A2 , …, An are exhaustive events if A1 A2 An 2. The events A and B are disjoint (or mutually exclusive) if A B In this case : (i) P A B 0 (ii) P A B P A PB 3. A Ac , A and Ac are exhaustive events. A Ac , A and Ac are disjoint events. 4. nAc n n A PAc 1 P A General Probability Rules:0 P A 1 1. 2. P 1 3. P 0 4. PA 1 P A 5. For any events A and B c P A B P A PB P A B 6. 7. For disjoint events A and B P A B P A PB For disjoint events E1 , E2 , … , En PE1 E 2 E n PE1 PE 2 PE n 2.3. Probability applied to health data:Example 3.1: 630 patients are classified as follows : (Simple frequency table) Blood Type O (E1) A (E2) B (E3) AB (E4) Total No. of patient s 284 258 63 25 630 · Experiment: Selecting a patient at random and observe his/her blood type. · This experiment has 630 equally likely outcomes n 630 Define the events : E1 = The blood type of the selected patient is O E2 = The blood type of the selected patient is A E3 = The blood type of the selected patient is B E4 = The blood type of the selected patient is AB n(E1) = 284, n(E2) = 258, n(E3) = 63 , n(E4) = 25 . 258 63 , 25 284 , , P E 2 PE1 PE 3 PE 4 630 630 630 630 E 2 E 4 the blood type of the selected patients is A or AB nE2 E4 258 25 283 0.4492 n 630 630 or P( E2 E4 ) PE PE 258 25 283 0.4492 2 4 630 630 630 (since E 2 E 4 ) Notes: 1. E1 ,E2 ,E3 ,E4 are mutually disjoint, Ei E j i j 2. E1 ,E2 ,E3 ,E4 are exhaustive events, .E1 E 2 E3 E 4 Example 3.2: Smoking Habit Daily (B1) Occasionally (B2) Not at all (B3) Total 20 - 29 (A1) 31 9 7 47 30 - 39 (A2) 110 30 49 189 40 - 49 (A3) 29 21 29 79 50+ 6 0 18 24 176 60 103 339 Total (A4) Experiment: Selecting a physician at random n 339 equally likely outcomes Events: · A3 = the selected physician n A 79 is aged 40 - 49 3 P A3 0.2330 n 339 · B2 = the selected physician smokes occasionally nB2 60 PB2 0.1770 n 339 ·A3 B2 the selected physician is aged 40-49 and smokes occasionally. n A3 B2 21 P A3 B2 0.06195 n 339 the selected physician is aged A3 B2 · 40-49 or smokes occasionally (or both) P A3 B2 P A3 PB2 P A3 B2 79 60 21 339 339 339 0.233 0.177 0.06195 0.3481 ·A4c the selected physician is not 50 years or older. P A4c 1 P A4 n A4 24 1 1 0.9292 n 339 A2 · A3 the selected physician is aged 30-49 or is aged 40-49 = the selected physician is aged 30-49 n A2 A3 189 79 286 0.7906 P A2 A3 n 339 339 or 189 79 0.7906 P A2 A3 P A2 P A3 339 339 ( Since A2 A3 ) 3.3. (Percentage/100) as probabilities and the use of venn diagrams: nE P E n n ?? unknown nE ?? unknown %E Percentage of elements of E relative ,, to the elements of , n nE %E 100% n , is known. Example 3.3: (p.72) A population of pregnant women with: · 10% of the pregnant women delivered prematurely. · 25% of the pregnant women used some sort of medication. · 5% of the pregnant women delivered prematurely and used some sort of medication. Experiment : Selecting a woman randomly from this population. Define the events: · D = The selected woman delivered prematurely. · M = The selected women used some sort of medication. D M The selected woman delivered prematurely and used some sort of medication. %D 10% %M 25% PD %D 10% 0.1 100% 100% PM %M 25% 0.25 100% 100% PD M %D M 5% 0.05 100% 100% %D M 5% A Venn diagram: PD 0.1 PM 0.25 PD M 0.05 PD c M 0.2 PD M c 0.05 PD c M c 0.70 PD M 0.30 Probability given by a Venn diagram A Two-way table: M Mc Total D 0.05 0.05 0.10 Dc 0.20 0.70 0.90 Total 0.25 0.75 1.00 Probabilities given by a two-way table. Calculating probabilities of some events: Mc = The selected woman did not use medication PM c 1 PM 1 0.25 0.75 Dc M c the selected woman din prematurely and did not use medication. not deliver P D c M c 1 P D M ?? D M the selected woman delivered prematurely or used some medication. PD M PD M PD M 0.1 0.25 0.05 0.3 PDc M c 1 PD M 1 0.3 0.7 Note: From the Venn diagram, it is clear that: PD PD M PD M c PM PD M PD c M PD M c PD PD M PDc M PM PD M PD c M c 1 PD M 3.4. Conditional Probability: · The conditional probability of the event A given the event B is defined by: P A B P A | B ; P B 0 P B · P(A | B) = the probability of the event A if we know that the event B has occurred. P A B Note: P A | B P B n A B / n nB / n P A | B n A B nB P A B PB P A | B P A B P APB | A Smoking Habbit Example: Age 20-29 A1) multiplication rules Daily Occasionally Not at all (B1) (B2) (B3) 31 9 7 Total 47 30-39 (A2) 110 30 49 189 40-49 (A3) 29 21 29 79 50+ (A4) 6 0 18 24 176 60 103 339 For calculating P A | B , we can use P A | B P A B n A B or P A | B PB nB Using the restricted table directly 176 PB1 0.519 339 PB1 A2 PB1 | A2 P A2 0.324484 0.5820 0.557522 n( B1 A2 ) 110 P B1 A2 0.324484 = n( ) 339 n( A2 ) 189 0.557522 P A2 339 n() OR nB1 A2 PB1 | A2 n A2 110 0.5820 189 Notice that P B1 < PB1 | A2 !! … What does this mean? Independent Events There are 3 cases (1) P A | B P A which means that knowing B increases the probability of occurrence of A . (2) P A | B P A which means that knowing B decreases the probability of occurrence of A . (3) P A | B P A which means that knowing B has no effect on the probability of occurrence of A . B A In this case A is independent of B. Independent Events: Two events A and B are independent if one of the following conditions is satisfied: ( i ) P A | B P A (ii ) P B | A P B (iii ) P B A P A P B (multiplication rule) Example: In the previous , A2 and B1 are not independent because: also PB1 0.5192 PB1 | A2 0.5820 PB1 A2 0.32448 PB1 P A2 0.28945 Combinations: •Notation: n factorial is denoted by n! and is defined by: n! nn 1n 221 for n 1 0! 1 Example: 5! = (5) (4) (3) (2) (1) = 120 • Combinations: The number of different ways for selecting r objects from n is given by: n distinct objects is denoted by and r n n! ; r r ! n r ! r 0, 1, 2, , n n is read as “ n “ choose “ r ”. r n 1 n n 1 0 n n r n r n r n-r Example 3.9: If we have 10 equal–priority operations and only 4 operating rooms, in how many ways can we choose the 4 patients to be operated on first? Answer: The number of different ways for selecting 4 patients from 10 patients is 10 10! 10! 10 98 21 4 4! 10 4! 4! 6! 4321 654321 210 (different ways)