Chapter 3 Probability Basics PDF

Chapter 3 Probability Basics: Basic rules, normal and binomial distributions. 2-1- General Definitions and Concepts: Probability: 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: A...

Chapter 3 Probability Basics: Basic rules, normal and binomial distributions. 2-1- General Definitions and Concepts: Probability: 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: An experiment is some procedure (or process) that we do. Sample Space: The sample space of an experiment is the set of all possible outcomes of an experiment. Also, it is called the universal set, and is denoted by Ω.  An Event: Any subset of the sample space Ω is called an event. -ɸ ⊆Ω is an event (impossible event). - Ω ⊆Ω is an event (sure event). Example: Experiment: Selecting a ball from a box containing 6 balls numbered from 1 to 6 and observing the number on the selected ball. This experiment has 6 possible outcomes. The sample space is: Ω = 1; 2; 3; 4; 5; 6. Consider the following events: 𝐸1 = 𝑔𝑒𝑡𝑡𝑖𝑛𝑔 𝑎𝑛 𝑒𝑣𝑒𝑛 𝑛𝑢𝑚𝑏𝑒𝑟 = 2; 4; 6 ⊆ Ω 𝐸2 = 𝑔𝑒𝑡𝑡𝑖𝑛𝑔 𝑎 𝑛𝑢𝑚𝑏𝑒𝑟 𝑙𝑒𝑠𝑠 𝑡ℎ𝑎𝑛 4 = 1; 2; 3 ⊆ Ω 𝐸3 = 𝑔𝑒𝑡𝑡𝑖𝑛𝑔 1 𝑜𝑟 3 = 1; 3 ⊆ Ω 𝐸4 = 𝑔𝑒𝑡𝑡𝑖𝑛𝑔 𝑎𝑛 𝑜𝑑𝑑 𝑛𝑢𝑚𝑏𝑒𝑟 = 1; 3; 5 ⊆ Ω 𝐸5 = 𝑔𝑒𝑡𝑡𝑖𝑛𝑔 𝑎 𝑛𝑒𝑔𝑎𝑡𝑖𝑣𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 = ɸ ⊆ Ω 𝐸6 = 𝑔𝑒𝑡𝑡𝑖𝑛𝑔 𝑎 𝑛𝑢𝑚𝑏𝑒𝑟 𝑙𝑒𝑠𝑠 𝑡ℎ𝑎𝑛 10 = 1; 2; 3; 4; 5; 6 =Ω⊆Ω Notation: −n(Ω)= number of outcomes (elements) in Ω −n(E)=number of outcomes (elements) in the event E Equally Likely Outcomes: The outcomes of an experiment are equally likely if the outcomes have the same chance of occurrence. Probability of An Event: If the experiment has n(Ω) equally likely outcomes, then the probability of the event E is denoted by P(E) and is defined by: 𝑛(𝐸) 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠 𝑖𝑛 𝐸 P(E)= = 𝑛(Ω) 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠 𝑖𝑛Ω Example: In the ball experiment in the previous example, suppose the ball is selected at random. Determine the probabilities of the following events: 𝐸1 = 𝑔𝑒𝑡𝑡𝑖𝑛𝑔 𝑎𝑛 𝑒𝑣𝑒𝑛 𝑛𝑢𝑚𝑏𝑒𝑟 = 2; 4; 6 ⊆ Ω 𝐸2 = 𝑔𝑒𝑡𝑡𝑖𝑛𝑔 𝑎 𝑛𝑢𝑚𝑏𝑒𝑟 𝑙𝑒𝑠𝑠 𝑡ℎ𝑎𝑛 4 = 1; 2; 3 ⊆ Ω 𝐸3 = 𝑔𝑒𝑡𝑡𝑖𝑛𝑔 1 𝑜𝑟 3 = 1; 3 ⊆ Ω Solution: Ω = 1; 2; 3; 4; 5; 6 , n(Ω)=6 𝐸1 = 2; 4; 6 , 𝑛 𝐸1 = 3 𝐸2 = 1; 2; 3 , n 𝐸2 = 3 𝐸3 = 1; 3 , n 𝐸3 = 2 The outcomes are equally likely. 3 3 2 P(𝐸1 )= ; P(𝐸2 )= ; P(𝐸3 )= 6 6 6 6 n(∅) 0 P(Ω)= = 1; 𝑃 ∅ = = =0 6 n(Ω) 6 Some Operations on Events: Let A and B be two events defined on the sample space Ω  Union of Two events: 𝐴 ∪ 𝐵 𝑜𝑟 𝐴 + 𝐵 The event 𝐴 ∪ 𝐵 consists of all outcomes in 𝐴 𝐨𝐫 𝐵 or in both 𝐀 𝐚𝐧𝐝 𝐁. The event 𝐴 ∪ 𝐵 occurs if 𝐴 occurs or 𝐵 occurs or both 𝐴 𝑎𝑛𝑑 𝐵 occurs. Intersection of Two events: 𝐴 ∩ 𝐵 The event 𝐴 ∩ 𝐵 consists of all outcomes in both 𝐴 and 𝐵. The event 𝐴 ∩ 𝐵 occurs if both 𝐀 𝐚𝐧𝐝 𝐁 occurs. Complement of an events: 𝐴ҧ 𝑜𝑟 𝐴𝑐 𝑜𝑟 𝐴′ The complement of the event A is denoted by 𝐴.ҧ The event 𝐴ҧ consists of all outcomes of Ω but are not in A. The event 𝐴ҧ occurs if A does not occurs. 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. 𝐸2 = 1; 2; 3 = 𝑔𝑒𝑡𝑡𝑖𝑛𝑔 𝑎 𝑛𝑢𝑚𝑏𝑒𝑟 𝑙𝑒𝑠𝑠 𝑡ℎ𝑎𝑛 4. 𝐸3 = 1; 3. 𝐸4 = 1; 3; 5 = 𝑔𝑒𝑡𝑡𝑖𝑛𝑔 an odd number. 1) 𝐸1 ∪ 𝐸2 = 1; 2; 3; 4; 6 = 𝑔𝑒𝑡𝑡𝑖𝑛𝑔 𝑎𝑛 𝑒𝑣𝑒𝑛 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑟𝑎 𝑛𝑢𝑚𝑏𝑒𝑟 𝑙𝑒𝑠𝑠 𝑡ℎ𝑎𝑛 4. 𝑛 𝐸1 ∪𝐸2 5 P(𝐸1 ∪ 𝐸2 )= = 𝑛 Ω 6 2) 𝐸1 ∪ 𝐸4 = 1; 2; 3; 4; 5; 6 = Ω = 𝑔𝑒𝑡𝑡𝑖𝑛𝑔 𝑎𝑛 𝑒𝑣𝑒𝑛 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑟 𝑎𝑛 𝑜𝑑𝑑 𝑛𝑢𝑚𝑏𝑒𝑟. 𝑛 𝐸1 ∪𝐸4 6 P(𝐸1 ∪ 𝐸4 )= 𝑛 Ω = 6 =1 Note: 𝐸1 ∪ 𝐸4 =Ω. 𝐸1 𝑎𝑛𝑑 𝐸4 are called exhaustive events. The union of these events gives the whole sample space. 3) 𝐸1 ∩ 𝐸2 = 2. = 𝑔𝑒𝑡𝑡𝑖𝑛𝑔 𝑎𝑛 𝑒𝑣𝑒𝑛 𝑛𝑢𝑚𝑏𝑒𝑟 𝑎𝑛𝑑 𝑎 𝑛𝑢𝑚𝑏𝑒𝑟 𝑙𝑒𝑠𝑠 𝑡ℎ𝑎𝑛 4. 𝑛 𝐸1 ∩𝐸2 1 P(𝐸1 ∩ 𝐸2 )= 𝑛 Ω = 6. 4) 𝐸1 ∩ 𝐸4 = ɸ. = 𝑔𝑒𝑡𝑡𝑖𝑛𝑔 𝑎𝑛 𝑒𝑣𝑒𝑛 𝑛𝑢𝑚𝑏𝑒𝑟 𝑎𝑛𝑑 𝑎𝑛 𝑜𝑑𝑑 𝑛𝑢𝑚𝑏𝑒𝑟. 𝑛 𝐸1 ∩𝐸4 0 P(𝐸1 ∩ 𝐸4 )= 𝑛 Ω = 6 = 0.  Note: 𝐸1 ∩ 𝐸4 = ɸ. 𝐸1 𝑎𝑛𝑑 𝐸4 are called disjoint (or mutually exclusive) events. These kinds of events can not occurred simultaneously (together in the same time). 5) The complement of 𝐸1 𝐸1 = 𝑛𝑜𝑡 𝑔𝑒𝑡𝑡𝑖𝑛𝑔 𝑎𝑛 𝑒𝑣𝑒𝑛 𝑛𝑢𝑚𝑏𝑒𝑟 = 2; 4; 6 = 1; 3; 5 = 𝑔𝑒𝑡𝑡𝑖𝑛𝑔 𝑎𝑛 𝑜𝑑𝑑 𝑛𝑢𝑚𝑏𝑒𝑟 = 𝐸4. Mutually exclusive (disjoint) Events: The events A and B are disjoint (or mutually exclusive) if: 𝐴 ∩ 𝐵 = ∅. For this case, it is impossible that both events occur simultaneously (i.e., together in the same time). In this case: (i) 𝑃 𝐴 ∩ 𝐵 = 0. (ii) 𝑃 𝐴 ∪ 𝐵 = 𝑃 𝐴 + 𝑃 𝐵. If 𝐴 ∩ 𝐵 ≠ ∅, then A and B are not mutually exclusive (not disjoint).  A and B are not mutually A and B are mutually exclusive exclusive (It is possible that (disjoint) (It is impossible that both events occur in the both events occur in the same same time) time) Exhaustive Events: The events 𝐴1 , 𝐴2 , … , 𝐴𝑛 are exhaustive events if: 𝐴1 ∪ 𝐴2 ∪ ⋯ ∪ 𝐴𝑛 = Ω. For this case 𝑃 𝐴1 ∪ 𝐴2 ∪ ⋯ ∪ 𝐴𝑛 = 𝑃 Ω = 1. Note: ഥ = Ω, (A and A 1) A ∪ A ഥ are exhaustive events). ഥ = ∅, (A 𝑎𝑛𝑑 𝐴ҧ are mutually exclusive 2) A ∩ A (disjoint) events). 3) 𝑛 𝐴ҧ = 𝑛 Ω − 𝑛 𝐴. 4) 𝑃 𝐴ҧ = 1 − 𝑃 𝐴. General Probability Rules: 1) 0 ≤ 𝑃(𝐴) ≤ 1. 2)𝑃(Ω) = 1. 3)𝑃(∅) = 0. 4) 𝑃 𝐴ҧ = 1 − 𝑃 𝐴. The Addition Rules: For any two events A and B: 𝑃 𝐴 ∪ 𝐵 = 𝑃(𝐴)+ 𝑃 𝐵 − 𝑃(𝐴 ∩ 𝐵). Special cases: 1) For mutually exclusive (disjoint) events A and B 𝑃 𝐴 ∪ 𝐵 = 𝑃(𝐴)+ 𝑃 𝐵. 2) For mutually exclusive (disjoint) events 𝐸1 , 𝐸2 ,…, 𝐸𝑛 : 𝑃(𝐸1 ∪ 𝐸2 ∪…∪ 𝐸𝑛 ) = 𝑃(𝐸1 )+𝑃(𝐸2 )+…+𝑃(𝐸𝑛 ) If the Note: events 𝐴1 , 𝐴2 ,…, 𝐴𝑛 are exhaustive and mutually exclusive (disjoint) events, then: 𝑃(𝐴1 ∪ 𝐴2 ∪…∪ 𝐴𝑛 ) = 𝑃(𝐴1 )+𝑃(𝐴2 )+…+𝑃(𝐴𝑛 ) = P Ω = 1. Marginal Probability: Given some variable that can be broken down into (m) categories designated by 𝐴1 , 𝐴2 ,…, 𝐴m and another jointly occurring variable that is broken down into (n) categories designated by 𝐵1 , 𝐵2 ,…, 𝐵𝑛. ∩ 𝐵1 𝐵2 … 𝐵𝐧 Total 𝐴1 n(𝐴1 ∩ B1 ) n(𝐴1 ∩ B2 ) … n(𝐴1 ∩ Bn ) n(𝐴1 ) 𝐴2 n(𝐴2 ∩ B1 ) n(𝐴2 ∩ B2 ) … n(𝐴2 ∩ Bn ) n(𝐴2 )... 𝐴m n(𝐴m ∩ B1 ) n(𝐴m ∩ B2 ) … n(𝐴m ∩ Bn ) n(𝐴m ) Total n(B1 ) n(B2 ) … n(Bn ) n(Ω) (This table contains the number of elements in each event). 𝐵1 𝐵2 … 𝐵𝐧 Marginal probabilit y 𝐴1 P(𝐴1 ∩ B1 ) P(𝐴1 ∩ B2 ) … P(𝐴1 ∩ Bn ) P(𝐴1 ) 𝐴2 P(𝐴2 ∩ B1 ) P(𝐴2 ∩ B2 ) … P(𝐴2 ∩ Bn ) P(𝐴2 )... 𝐴m P(𝐴m ∩ B1 ) P(𝐴m ∩ B2 ) … P(𝐴m ∩ Bn ) P(𝐴m ) Marginal P(B1 ) P(B2 ) … P(Bn ) 1 probabili ty (This table contains the probability of each event)  The marginal probability of 𝐴i , P(𝐴i ) is equal to the sum of the joint probabilities of 𝐴i. with all categories of B. That is: P(𝐴i ) = P(𝐴i ∩ B1 ) + P(𝐴i ∩ B2 ) + ⋯ + P(𝐴i ∩ Bn ) 𝑛 = ෍ P(𝐴i ∩ Bj ) 𝑗=1 For example, P(𝐴2 ) = P(𝐴2 ∩ B1 ) + P(𝐴2 ∩ B2 ) + ⋯ + P(𝐴2 ∩ Bn ) 𝑛 = ෍ P(𝐴2 ∩ Bj ) 𝑗=1 We define the marginal probability Bj , P(Bj ), in a similar of way. Example: Table of number of elements in each event: 𝑩𝟏 𝑩𝟐 𝑩𝟑 Total 𝑨𝟏 50 30 70 150 𝑨𝟐 20 70 10 100 𝑨𝟑 30 100 120 250 Total 100 200 200 500 Table of probability of each event: 𝑩𝟏 𝑩𝟐 𝑩𝟑 Marginal probability 𝑨𝟏 0.1 0.06 0.14 0.3 𝑨𝟐 0.04 0.14 0.02 0.2 𝑨𝟑 0.06 0.2 0.24 0.5 Marginal probability 0.2 0.4 0.4 1 For example: P(𝐴2 ) = P(𝐴2 ∩ B1 ) + P(𝐴2 ∩ B2 ) + P(𝐴2 ∩ B3 ) = 0.04 + 0.14 + 0.02 = 0.2 Applications: Example: 630 patients are classified as follows: Blood type O A B AB Total (𝑬𝟏 ) (𝑬𝟐 ) (𝑬𝟑 ) (𝑬𝟒 ) Number of patients 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 Ω = 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" Number of elements in each event: 𝑛(𝐸1 ) = 284; 𝑛(𝐸2 ) = 258; 𝑛(𝐸3 ) = 63; 𝑛(𝐸4 ) = 25. Probabilities of the events: 284 258 P(𝐸1 ) = = 0,4508; P(𝐸2 ) = = 0,4095; 630 630 63 25 P(𝐸3 ) = = 0,1; P(𝐸4 ) = = 0,0397. 630 630  Some operations on the events: 1-𝐸2 ∩ 𝐸4 : the blood type of the selected patients is "A" and "AB". 𝐸2 ∩ 𝐸4 = ∅, (disjoint events / mutually exclusive events) P(𝐸2 ∩ 𝐸4 ) = 𝑃(∅) = 0 2-𝐸2 ∪ 𝐸4 : the blood type of the selected patients is "A" or "AB". 𝑛(𝐸2 ∪ 𝐸4 ) 258 + 25 = = 0,4492 𝑛(Ω) 630 P(𝐸2 ∪ 𝐸4 ) = 258 25 P(𝐸2 ) + P(𝐸4 ) = + = 0,4492 630 630 (𝑠𝑖𝑛𝑐𝑒 𝐸2 ∩ 𝐸4 = ∅) 3- 𝐸1 = the blood type of the selected patients is not "O". 𝑛 𝐸1 = 𝑛 Ω − 𝑛 𝐸1 = 630 − 284 = 346. 𝑛 𝐸1 346 𝑃 𝐸1 = = = 0,5492 𝑛 Ω 630 another solution: 𝑃 𝐸1 = 𝑃 𝐸1𝑐 = 1 − 𝑃 𝐸1 = 1 − 0,4508 = 0,5492 Notes: 1. 𝐸1 , 𝐸2 , 𝐸3 , 𝐸4 are mutually disjoint , 𝐸i ∩ 𝐸j = ∅; i≠𝑗. 12. 𝐸1 , 𝐸2 , 𝐸3 , 𝐸4 are exhaustive events,𝐸1 ∪ 𝐸2 ∪ 𝐸3 ∪ 𝐸4 = Ω.. Example: 339 physicians are classified based on their ages and smoking habits as follows. Smoking Daily Occa Not at all Total (𝐵1 ) siona (𝐵3 ) lly (𝐵2 ) Age 20 − 29 (𝐴1 ) 31 9 7 47 30 − 39 (𝐴2 ) 110 30 49 189 40 − 49 (𝐴3 ) 29 21 29 79 50 + (𝐴4 ) 6 0 18 24 Total 176 60 103 339 Experiment: Selecting a physician at random The number of elements of the sample space is 𝑛 Ω = 339 The outcomes of the experiment are equally likely. Some events: - 𝐴3 =the selected physician is aged 40 – 49 𝑛(𝐴3 ) 79 𝑃 𝐴3 = = = 0,2330 𝑛(Ω) 339 - 𝐵2 =the selected physician smokes occasionally 𝑛(𝐵2 ) 60 𝑃 𝐵2 = = = 0,1770 𝑛(Ω) 339 -𝐴3 ∩ 𝐵2 =the selected physician is aged 40-49 and smokes occasionally. 𝑛(𝐴3 ∩ 𝐵2 ) 21 𝑃 𝐴3 ∩ 𝐵2 = = = 0, 06195 𝑛(Ω) 339 -𝐴3 ∪ 𝐵2 =the selected physician is aged 40-49 or smokes occasionally (or both). 𝑃 𝐴3 ∪ 𝐵2 = 𝑃 𝐴3 +𝑃 𝐵2 -𝑃 𝐴3 ∩ 𝐵2 79 60 21 = + - = 0.3481 339 339 339 -𝐴4 =the selected physician is not 50 years or older =𝐴1 ∪ 𝐴2 ∪ 𝐴3. 𝑛 𝐴4 24 𝑃 𝐴4 = 1 − 𝑃 𝐴4 = 1 − =1− = 0.9292. 𝑛 Ω 339 -𝐴2 ∪ 𝐴3 =the selected physician is aged 30-39 or is aged 40-49 =the selected physician is aged 30-49  𝑛(𝐴2 ∪ 𝐴3 ) 189 + 79 = = 0,7906 𝑛(Ω) 339 𝑃 𝐴2 ∪ 𝐴 3 = 𝑆𝑖𝑛𝑐𝑒 189 79 𝑃 𝐴2 + 𝑃 𝐴3 −= + = 0,7906 339 339 (since𝐴2 ∩ 𝐴3 = ∅) Example: Suppose that there is 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 medication. 𝐷 ∩ 𝑀 = The selected woman delivered prematurely and used some sort of medication. Percentages: % 𝐷 = 10%; % 𝑀 = 25%; % 𝐷 ∩ 𝑀 = 5% The complement events: ഥ The selected woman did not deliver prematurely. 𝐷: ഥ The selected women did not use medication. 𝑀: A Two-way table: (Percentages given by a two-way table): The probabilities of the given events are: %(𝐷) 10% 𝑃 𝐷 = = = 0,1. 100% 100% %(𝑀) 25% 𝑃 𝑀 = = = 0,25. 100% 100% %(𝐷 ∩ 𝑀) 5% 𝑃 𝐷∩𝑀 = = = 0,05. 100% 100% Calculating probabilities of some events: 𝐷 ∪ 𝑀: the selected woman delivered prematurely or used medication. 𝑃 𝐷 ∪ 𝑀 = 𝑃 𝐷 + 𝑃 𝑀 − 𝑃 𝐷 ∩ 𝑀 = 0.1 + 0.25 − 0.05 = 0.3 (by the rule). ഥ The selected woman did not use medication 𝑀: 𝑃 𝑀ഥ = 1 − 𝑃 𝑀 = 1 − 0.25 = 0.75 (by the rule). 75 ഥ 𝑃 𝑀 = = 0.75 (from the table). 100 ഥ : The selected woman did not deliver prematurely 𝐷 𝑃 𝐷 ഥ = 1 − 𝑃 𝐷 = 1 − 0.1 = 0.9 (by the rule). 𝑃 𝑀 ഥ = 90 = 0.9 (from the table). 100 𝐷ഥ ∩ 𝑀: ഥ The selected woman did not deliver prematurely and did not use medication. 70 ഥ∩𝑀 𝑃 𝐷 ഥ = = 0.7 (from the table). 100  ഥ ∩ 𝑀: the selected woman did not deliver prematurely and 𝐷 used medication. 20 ഥ 𝑃 𝐷∩𝑀 = = 0.2 (from the table). 100 ഥ the selected woman delivered prematurely and did not 𝐷 ∩ 𝑀: use medication. 𝑃 𝐷∩𝑀 ഥ = 5 = 0.05 (from the table). 100 ഥ the selected woman delivered prematurely or did not use 𝐷 ∪ 𝑀: medication. 𝑃 𝐷∪𝑀 ഥ =𝑃 𝐷 +𝑃 𝑀 ഥ − 𝑃(𝐷 ∩ 𝑀) ഥ (by the rule). = 0.1 + 0.75 − 0.05 = 0.8 ഥ ∪ 𝑀: the selected woman did not deliver prematurely or used 𝐷 medication. ഥ∪𝑀 =𝑃 𝐷 𝑃 𝐷 ഥ + 𝑃 𝑀 − 𝑃(𝐷 ഥ ∩ 𝑀) (by the rule). = 0.9 + 0.25 − 0.2 = 0.95 ഥ ∪ 𝑀: 𝐷 ഥ the selected woman did not deliver prematurely or did not use medication. ഥ∪𝑀 𝑃 𝐷 ഥ =𝑃 𝐷 ഥ +𝑃 𝑀 ഥ − 𝑃(𝐷 ഥ ∩ 𝑀) ഥ (by the rule). = 0.9 + 0.75 − 0.7 = 0.95 Conditional Probability: The conditional probability of the event A when we know that the event B has already occurred is defined by: 𝑃(𝐴∩𝐵) 𝑃 𝐴 /𝐵 = ; 𝑃(𝐵) ≠ 0. 𝑃(𝐵) 𝑃 𝐴 /𝐵 :The conditional probability of A given B. Notes: 𝑃(𝐴∩𝐵) 𝑛(𝐴∩𝐵)/𝑛(Ω) 𝑛(𝐴∩𝐵) 1- 𝑃 𝐴 /𝐵 = = =. 𝑃(𝐵) 𝑛(𝐵)/𝑛(Ω) 𝑛(𝐵) 𝑃(𝐴∩𝐵) 2-𝑃 𝐵 /𝐴 =. 𝑃(𝐴) 3-For calculating 𝑃 𝐴 /𝐵 , we may use any one of the following: 𝑃(𝐴∩𝐵) i) 𝑃 𝐴 /𝐵 = 𝑃(𝐵) 𝑛(𝐴∩𝐵) ii) 𝑃 𝐴 /𝐵 =. 𝑛(𝐵) iii) Using the restricted table directly. Multiplication Rules of Probability: For any two events A and B, we have: 𝑃 𝐴 ∩ 𝐵 = 𝑃(𝐵)𝑃 𝐴 /𝐵 𝑃 𝐴 ∩ 𝐵 = 𝑃(𝐴)𝑃 𝐵 /𝐴 Example: Consider the following event: (𝐵1 /𝐴2 ) = the selected physician smokes daily given that his age is between 30 and 39. 𝑛(𝐵1 ) 176 𝑃 𝐵1 = = = 0.519. 𝑛(Ω) 339 𝑃(𝐵1 ∩𝐴2 ) 0.324484 𝑃(𝐵1 /𝐴2 )= = = 0.582. 𝑃(𝐴2 ) 0.557522 Another solution: Notice that: What does this mean? We will answer this question after talking about the concept of independent events. Example: (Multiplication Rule of Probability) A training health program consists of two consecutive parts. To pass this program, the trainee must pass both parts of the program. From the past experience, it is known that 90% of the trainees pass the first part, and 80% of those who pass the first part pass the second part. If you are admitted to this program, what is the probability that you will pass the program? What is the percentage of trainees who pass the program? Solution: Define the following events: 𝐴 = the event of passing the first part B = the event of passing the second part 𝐴 ∩ 𝐵 = the event of passing the first part and the second part = the event of passing both parts = the event of passing both programs. Therefore, the probability of passing the program is P(𝐴 ∩ 𝐵) From the given information: The probability of passing the first part is: The probability of passing the second part given that the trainee has already passed the first part is: Now, we use the multiplication rule to find P(𝐴 ∩ 𝐵) as follows We can conclude that 72% of the trainees pass the program. Independent Events There are 3 cases: (knowing B increases the probability of occurrence of A) (knowing B decreases the probability of occurrence of A) (knowing B has no effect on the probability of occurrence of 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: Note: The third condition is the multiplication rule of independent events. Example: Suppose that A and B are two events such that: Theses two events are not independent (they are dependent) because: Example: (Reading Assignment) Suppose that a dental clinic has 12 nurses classified as follows: The experiment is to randomly choose one of these nurses. Consider the following events: C = the chosen nurse has children N = the chosen nurse works night shift a) Find The probabilities of the following events: 1. the chosen nurse has children. 2. the chosen nurse works night shift. 3. the chosen nurse has children and works night shift. 4. the chosen nurse has children and does not work night shift. b) Find the probability of choosing a nurse who woks at night given that she has children. c) Are the events C and N independent? Why? d) Are the events C and N disjoint? Why? e) Sketch the events C and N with their probabilities using Venn diagram. Solution: We can classify the nurses as follows: 3.5 Bayes' Theorem, Screening Tests, Sensitivity, Specificity, and Predictive Value Positive and Negative: There are two states regarding the disease and two states regarding the result of the screening test: