{"id":190,"date":"2019-03-11T10:26:25","date_gmt":"2019-03-11T10:26:25","guid":{"rendered":"http:\/\/esp14.epgpbooks.inflibnet.ac.in\/?post_type=chapter&#038;p=190"},"modified":"2019-03-11T11:05:17","modified_gmt":"2019-03-11T11:05:17","slug":"addition-and-multiplication-properties-of-probability-theory","status":"publish","type":"chapter","link":"https:\/\/ebooks.inflibnet.ac.in\/esp14\/chapter\/addition-and-multiplication-properties-of-probability-theory\/","title":{"rendered":"Addition and Multiplication Properties of Probability Theory"},"content":{"raw":"<div><span style=\"float: right\"><a href=\"https:\/\/youtu.be\/WREm900g1hw\" target=\"_blank\" rel=\"noopener\"><img src=\"http:\/\/epgpbooks.inflibnet.ac.in\/wp-content\/uploads\/2018\/11\/download.png\" alt=\"epgp books\" width=\"75px\" height=\"75px;\" \/><\/a>\r\n<\/span><\/div>\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n<strong style=\"text-align: initial;font-size: 1em\">Learning Objectives<\/strong>\r\n<div>\r\n<ul>\r\n \t<li><strong>Introduction<\/strong><\/li>\r\n \t<li><strong>Addition Probability Theorem<\/strong><\/li>\r\n \t<li><strong>Conditional Probability<\/strong><\/li>\r\n \t<li><strong>Multiplication Rule<\/strong><\/li>\r\n \t<li><strong>Summary<\/strong><\/li>\r\n \t<li><strong>Suggested Readings<\/strong><\/li>\r\n<\/ul>\r\n<strong>\u00a0 \u00a0 1.\u00a0<\/strong><strong>Learning Objectives<\/strong>\r\n\r\n&nbsp;\r\n<p style=\"text-align: justify\">This module is in continuation of the module \"Introduction to Probability\". Basic key concepts of probability were discussed in the above said module. In this module, our main concern is to give brief\u00a0<span style=\"text-align: initial;font-size: 1em\">understanding of some basic and most important rules for probability theory. We will discuss these concepts with examples for better understanding. Additions and multiplication rules are considered as <\/span>an important features<span style=\"text-align: initial;font-size: 1em\"> of probability. Construction of formulas and examples are discussed to understand addition and multiplication probability theorem in detail.<\/span><\/p>\r\n\r\n<\/div>\r\n<div>\r\n\r\n<strong>\u00a0 \u00a0 2.<\/strong>\u00a0<strong>Introduction<\/strong>\r\n\r\n&nbsp;\r\n<p style=\"text-align: justify\">Probability is an important concept for all type of study variable. Theory of probability deals with several type of problems. To settle problems, there are many rules related to probability theory. Addition and multiplication rules are one of them. The addition and multiplication rule are applied to solve the probability problem which has two or more events.<\/p>\r\n&nbsp;\r\n<p style=\"text-align: justify\">Addition rule is used to detect probability of one of the two or more events occurring in the respective random experiment at one task. i.e. at least one event occurred among all events.<\/p>\r\n&nbsp;\r\n<p style=\"text-align: justify\">Multiplication rule is used to detect outcome that occurs from more than one task in respective random experiment.<\/p>\r\n&nbsp;\r\n<p style=\"text-align: justify\">Let us suppose, if there is only one task of tossing a coin two times and find the probability of getting heads on both time. This type of probability is problem solved by addition rule of the probability.<\/p>\r\n&nbsp;\r\n<p style=\"text-align: justify\">Suppose tossing a coin two times one after another. If one get heads on the first toss and also on the second toss. These two outcomes are independent as outcome of first toss is not affecting the outcome of second toss. This type of probability problem is solved by multiplication rule of the probability.<\/p>\r\n&nbsp;\r\n<p style=\"text-align: justify\">Here, some basic features of an event will help one to understand mathematical behavior of the probability problem. These features will be helpful to understand statistical addition and multiplication probability Theorem. These are:<\/p>\r\n? \u222a ? = {? \u2208 ?; ? \u2208 ? ?? ? \u2208 ?};\r\n\r\n? = {? \u2208 ?; ? does not belong to E};\r\n\r\n? \u2229 ? = {? \u2208 ?; ? \u2208 ? ??? ? \u2208 ?};\r\n\r\n? \u2282 ? for every ? \u2208 ?, ? \u2208 ?.\r\n\r\n&nbsp;\r\n\r\nwhere ? \u222a ? (say), E union F, means at least one event happens.\r\n\r\n? \u0305(say), E bar or E compliment, means outcomes does not belong to E.\r\n\r\n? \u2229 ? (say), E intersection F, means both events happens.\r\n\r\n? \u2282 ? (say), E subset of F, means outcome of event belongs to outcome of F.\r\n\r\n&nbsp;\r\n\r\nLet us discuss in detail explanation of the addition and multiplication probability theorem.\r\n\r\n&nbsp;\r\n\r\n<strong style=\"text-align: initial;font-size: 1em\">3.<\/strong><span style=\"text-align: initial;font-size: 1em\">\u00a0\u00a0\u00a0 <\/span><strong style=\"text-align: initial;font-size: 1em\">Addition Probability Theorem<\/strong>\r\n\r\n&nbsp;\r\n<p style=\"text-align: justify\"><span style=\"text-align: initial;font-size: 1em\">A statistical property of probability theory is used to explain happening of one or more event at one task. Suppose a random experiment, selecting a card from a deck of card and event, is selected card jack and queen. Since it is impossible to draw both jack and queen in <\/span>same<span style=\"text-align: initial;font-size: 1em\"> draw. <\/span>Outcome<span style=\"text-align: initial;font-size: 1em\"> will be either jack or queen. This type of probability problem is solved by using <\/span>addition<span style=\"text-align: initial;font-size: 1em\"> rule.<\/span><\/p>\r\n&nbsp;\r\n<p style=\"text-align: justify\"><span style=\"text-align: initial;font-size: 1em\">\u201cThe addition rule is used to find <\/span>compound<span style=\"text-align: initial;font-size: 1em\"> probability of one or more events with the either-or statement format of the problem.\u201d<\/span><\/p>\r\n&nbsp;\r\n<p style=\"text-align: justify\"><span style=\"text-align: initial;font-size: 1em\">In other words, the addition theorem solves the probability of happening at least one of the given events. Union concept is used to find <\/span>probability<span style=\"text-align: initial;font-size: 1em\"> of happening at least one event. <\/span>Union<span style=\"text-align: initial;font-size: 1em\"> is <\/span>key<span style=\"text-align: initial;font-size: 1em\"> concept of mathematical probability theory and set theory. This union representation looks like alphabet U.<\/span><\/p>\r\n&nbsp;\r\n<p style=\"text-align: justify\">Since,<span style=\"text-align: initial;font-size: 1em\"> two events cannot occur simultaneously in one task and cannot have <\/span>common<span style=\"text-align: initial;font-size: 1em\"> outcome, it is known as <\/span>mutually<span style=\"text-align: initial;font-size: 1em\"> exclusive event. If events have <\/span>common<span style=\"text-align: initial;font-size: 1em\"> outcome, <\/span>event<span style=\"text-align: initial;font-size: 1em\"> will not be <\/span>mutually<span style=\"text-align: initial;font-size: 1em\"> exclusive event. It may <\/span>be say<span style=\"text-align: initial;font-size: 1em\"> as <\/span>non-mutually<span style=\"text-align: initial;font-size: 1em\"> exclusive event.<\/span><\/p>\r\n&nbsp;\r\n<p style=\"text-align: justify\"><span style=\"text-align: initial;font-size: 1em\">Based on these two types of events, mutually exclusive and non-mutually exclusive event mathematical formulation are described as:<\/span><\/p>\r\n\r\n<\/div>\r\n<div>\r\n\r\n<strong>\u00a0 \u00a0 3.1 Mathematical Formulation<\/strong>\r\n\r\n&nbsp;\r\n<p style=\"text-align: justify\">Mathematical formulation of addition probability theorem is slightly different from mutually exclusive event than non-mutually exclusive event. There is only one difference i.e. addition formula for mutually exclusive does not have intersection term but addition formula for non-mutually exclusive event has\u00a0<span style=\"text-align: initial;font-size: 1em\">intersection term. Addition probability formulation for non-mutually exclusive is the generalization of the addition probability formulation for mutually exclusive formula.<\/span><\/p>\r\n&nbsp;\r\n<p style=\"text-align: justify\"><span style=\"text-align: initial;font-size: 1em\">Let us discuss addition probability formulation for <\/span>non-mutually<span style=\"text-align: initial;font-size: 1em\"> exclusive event and mutually exclusive event.<\/span><\/p>\r\n\r\n<\/div>\r\n<div>\r\n\r\n<strong>\u00a0 \u00a0 Addition Probability Formulation for Non-mutually Exclusive Events<\/strong>\r\n\r\n&nbsp;\r\n<p style=\"text-align: justify\">Venn diagram is a graphical representation for set theory. Sample space is a set of all possible outcomes and event is a subset of sample space. Event is also a set. Venn diagram is being used here, to show graphical representation of the given event.<\/p>\r\n&nbsp;\r\n\r\nSuppose E and F are two events of a random experiment.\r\n\r\n&nbsp;\r\n\r\n<img class=\"aligncenter size-full wp-image-195\" src=\"http:\/\/esp14.epgpbooks.inflibnet.ac.in\/wp-content\/uploads\/sites\/176\/2019\/03\/Untitled-96.png\" alt=\"\" width=\"179\" height=\"177\" \/>\r\n\r\n&nbsp;\r\n\r\nIn Figure 1 Venn diagram, event E is shown as blue circle and event F is shown as yellow circle. Event E and F have some common area and have common outcomes, so event E and F are non-mutually exclusive event.\r\n\r\n&nbsp;\r\n\r\nAddition Probability Formulation for non-mutually exclusive events E or F is given by:\r\n\r\n&nbsp;\r\n<p style=\"text-align: center\">?(? \u222a ?) = ?(?) + ?(?) \u2212 ?(? \u2229 ?)<\/p>\r\n&nbsp;\r\n\r\nThis formulation can be read as:\r\n\r\n&nbsp;\r\n<p style=\"text-align: justify\">Probability of happening either event E or event F = probability of happening event E + probability of happening event F \u2013 probability of happening events E and F.<\/p>\r\n&nbsp;\r\n\r\nSuppose there are three non-mutually exclusive events E, F and G\r\n\r\n<\/div>\r\n<div>\r\n\r\n<img class=\"aligncenter size-full wp-image-197\" src=\"http:\/\/esp14.epgpbooks.inflibnet.ac.in\/wp-content\/uploads\/sites\/176\/2019\/03\/Untitled-97.png\" alt=\"\" width=\"159\" height=\"179\" \/>\r\n\r\n&nbsp;\r\n\r\nAddition probability formulation for non-mutually exclusive events E, F or G are given by:\r\n\r\n&nbsp;\r\n<p style=\"text-align: center\">?(? \u222a ? \u222a ?) = ?(?) + ?(?) + ?(?) \u2212 ?(? \u2229 ?) \u2212 ?(? \u2229 ?) \u2212 ?(? \u2229 ?) + ?(? \u2229 ? \u2229 ?)<\/p>\r\n&nbsp;\r\n\r\nSimilarly, addition probability formulation can be extented for n events.\r\n\r\n&nbsp;\r\n\r\n<strong>Question 1<\/strong>\r\n\r\n&nbsp;\r\n<p style=\"text-align: justify\">Suppose in a class there are 13 students, 7 are boys and 6 are girls. On unit test 4 boys and 5 girls received A grade. If a student chooses at random in the class, what is the probability of choosing a girl or an A grade student?<\/p>\r\n&nbsp;\r\n\r\n<strong>Answer<\/strong>\r\n\r\n&nbsp;\r\n<p style=\"text-align: justify\">Suppose E is an event of choosing girl and F is an event of choosing A grade student. Hence we are interested in to find out the probability of E or F.<\/p>\r\n&nbsp;\r\n<p style=\"text-align: justify\">Event E and F have 5 common outcomes i.e. 5 girls made A grade. Hence E and F are non-mutually exclusive events.<\/p>\r\n&nbsp;\r\n<p style=\"text-align: justify\">We have to find probability of non-mutually exclusive events by using addition probability formulation.<\/p>\r\n&nbsp;\r\n\r\nProbabilities are given as:\r\n\r\n<img class=\"aligncenter size-full wp-image-198\" src=\"http:\/\/esp14.epgpbooks.inflibnet.ac.in\/wp-content\/uploads\/sites\/176\/2019\/03\/Untitled-98.png\" alt=\"\" width=\"138\" height=\"189\" \/>\r\n\r\n&nbsp;\r\n\r\n<span style=\"text-align: initial;font-size: 1em\">Addition probability formulation for two non-mutually exclusive events is given as:<\/span>\r\n\r\n<\/div>\r\n<div>\r\n\r\n<img class=\"aligncenter size-full wp-image-199\" src=\"http:\/\/esp14.epgpbooks.inflibnet.ac.in\/wp-content\/uploads\/sites\/176\/2019\/03\/Untitled-99.png\" alt=\"\" width=\"311\" height=\"156\" \/>\r\n\r\n&nbsp;\r\n\r\nProbability of choosing a girl or A grade student is 0.7692.\r\n\r\n&nbsp;\r\n\r\n<strong>Question 2<\/strong>\r\n\r\n&nbsp;\r\n\r\nIf the probability of solving a quiz by Rubi and Kiran are 3\/4 and 2\/3 respectively, what is the probability of solving the quiz?\r\n\r\n&nbsp;\r\n\r\n<strong>Answer<\/strong>\r\n<p style=\"text-align: justify\">Let E and F are the event of solving a quiz by Rubi and Kiran respectively. The probabilities of these events are given as:<\/p>\r\n<img class=\"aligncenter size-full wp-image-200\" src=\"http:\/\/esp14.epgpbooks.inflibnet.ac.in\/wp-content\/uploads\/sites\/176\/2019\/03\/Untitled-100.png\" alt=\"\" width=\"188\" height=\"46\" \/>\r\n<p style=\"text-align: justify\">As we are interested in solving the quiz hence there are the following possibilities. First possibility is that either Kiran solve the quiz or Rubi solve the quiz. It means that quiz will be solved if atleast one of them will solve the quiz. Second possibility is that both solve the quiz. As Kiran and Rubi can solve the quiz independently. So we can consider events E and F are independent events.<\/p>\r\n&nbsp;\r\n\r\nTherefore:\r\n<p style=\"text-align: center\">?(? \u2229 ?) = ?(?) \u2217 ?(?).<\/p>\r\n&nbsp;\r\n<p style=\"text-align: justify\">As the probability of solving the quiz by atleast one of two events is represented as union of two events. So, find the probability by using addition probability formulation. The formula is given as<\/p>\r\n&nbsp;\r\n<p style=\"text-align: center\">?(? \u222a ?) = ?(?) + ?(?) \u2212 ?(? \u2229 ?);<\/p>\r\n&nbsp;\r\n\r\n<img class=\"aligncenter size-full wp-image-201\" src=\"http:\/\/esp14.epgpbooks.inflibnet.ac.in\/wp-content\/uploads\/sites\/176\/2019\/03\/Untitled-101.png\" alt=\"\" width=\"297\" height=\"57\" \/>\r\n\r\n<\/div>\r\n<div>\r\n\r\n\u00a0 \u00a0 Hence the probability of solving the quiz is 0.9166.\r\n\r\n&nbsp;\r\n\r\n<strong>Mutually exclusive addition formula:<\/strong>\r\n\r\n&nbsp;\r\n\r\nSuppose E and F are two events of a random experiment respectively.\r\n\r\n&nbsp;\r\n\r\n<img class=\"aligncenter size-full wp-image-202\" src=\"http:\/\/esp14.epgpbooks.inflibnet.ac.in\/wp-content\/uploads\/sites\/176\/2019\/03\/Untitled-102.png\" alt=\"\" width=\"176\" height=\"185\" \/>\r\n\r\n&nbsp;\r\n<p style=\"text-align: justify\">In Figure 3 of Venn diagram, event E is shown as blue circle and event F is shown as yellow circle. Events E and F do not have common area and same outcomes, so event E and F are mutually exclusive event.<\/p>\r\n&nbsp;\r\n\r\nAddition probability formulation for mutually exclusive events E and F is given by:\r\n\r\n&nbsp;\r\n<p style=\"text-align: center\">?(? \u222a ?) = ?(?) + ?(?).<\/p>\r\n&nbsp;\r\n\r\nThis formulation can be read as:\r\n\r\n&nbsp;\r\n\r\nProbability of happening either event E or event F = probability of happening event E +probability of happening event F.\r\n\r\n&nbsp;\r\n\r\nIf there are three mutually exclusive events E, F and G.\r\n\r\n<img class=\"aligncenter size-full wp-image-203\" src=\"http:\/\/esp14.epgpbooks.inflibnet.ac.in\/wp-content\/uploads\/sites\/176\/2019\/03\/Untitled-103.png\" alt=\"\" width=\"272\" height=\"214\" \/>\r\n<p style=\"text-align: center\">Figure 4<\/p>\r\n\r\n<\/div>\r\nAddition probability formulation for three mutually exclusive events is given as follows\r\n\r\n&nbsp;\r\n<p style=\"text-align: center\">?(? \u222a ? \u222a ?) = ?(?) + ?(?) + ?(?)<\/p>\r\n&nbsp;\r\n\r\nSimilarly, Addition probability formulation can be extended to n events or outcomes.\r\n\r\n&nbsp;\r\n\r\n<strong>Question 3<\/strong>\r\n\r\n&nbsp;\r\n<p style=\"text-align: justify\">Suppose in a group of 50 members, there are 15 junior students , 20 senior students and 15 instructor . What is the probability that a person randomly picked from this group is either junior or senior student?<\/p>\r\n&nbsp;\r\n\r\n<strong>Answer<\/strong>\r\n\r\n&nbsp;\r\n\r\nLet E and F be the event of selecting junior and senior student respectively from a group of 50 members.\r\n\r\n&nbsp;\r\n<p style=\"text-align: justify\">As events E and F do not have any common outcomes i.e. student can either be junior and senior. So E and F are mutually exclusive events.<\/p>\r\n&nbsp;\r\n<p style=\"text-align: justify\">Addition probability formulation is used to find probability of picking a member from a group of 50 members that can be either junior or senior student.<\/p>\r\n&nbsp;\r\n\r\nProbabilities of events E and F are given as:\r\n\r\n&nbsp;\r\n\r\nP(E) = 15\/50.\r\n\r\n&nbsp;\r\n\r\nP(F) = 20\/50.\r\n\r\n&nbsp;\r\n\r\nAddition probability formulation for two mutually exclusive events is given as:\r\n\r\n&nbsp;\r\n\r\n<img class=\"aligncenter size-full wp-image-204\" src=\"http:\/\/esp14.epgpbooks.inflibnet.ac.in\/wp-content\/uploads\/sites\/176\/2019\/03\/Untitled-104.png\" alt=\"\" width=\"209\" height=\"158\" \/>\r\n\r\n&nbsp;\r\n\r\nProbability of selectinga junior or senior student randomly from this group is 0.7.\r\n\r\n&nbsp;\r\n\r\n<strong>Question 4<\/strong>\r\n\r\n&nbsp;\r\n\r\n<span style=\"text-align: initial;font-size: 1em\">Suppose a card selected from standard deck of cards. What is the probability of selecting either ace or queen card from a deck of card?<\/span>\r\n\r\n&nbsp;\r\n\r\n<strong style=\"text-align: initial;font-size: 1em\">Answer<\/strong>\r\n<div>\r\n\r\n&nbsp;\r\n\r\nLet us consider two events E and F\r\n\r\n&nbsp;\r\n\r\nE:\u00a0\u00a0 selected card is ace.\r\n\r\n&nbsp;\r\n\r\nF:\u00a0 selected card is queen.\r\n\r\n&nbsp;\r\n\r\nAs events E and F do not have common outcomes. So, E and F are mutually exclusive event.\r\n\r\n&nbsp;\r\n\r\nAddition probability formulation is used to find the probability of selecting card either ace and queen.\r\n\r\n&nbsp;\r\n\r\nProbabilities of events E and F are given as:\r\n\r\n<img class=\"size-full wp-image-206 alignleft\" src=\"http:\/\/esp14.epgpbooks.inflibnet.ac.in\/wp-content\/uploads\/sites\/176\/2019\/03\/Untitled-105.png\" alt=\"\" width=\"592\" height=\"256\" \/>\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\nHence the probability of selecting an ace and a queen from a deck of card is 2\/13.\r\n\r\n&nbsp;\r\n\r\n<strong>4.\u00a0\u00a0 Conditional Probability<\/strong>\r\n\r\n&nbsp;\r\n<p style=\"text-align: justify\">When one event has a relationship with other respective events. Suppose event E is the event that is related with raining, there is 0.7% chance of raining. Conditional probability is used to find out probability of these types of events.<\/p>\r\n&nbsp;\r\n<p style=\"text-align: justify\">\u201cConditional probability measured the probability of an event given that one event occurred.\u201d<\/p>\r\n&nbsp;\r\n\r\nSuppose E and F are two dependent events of random experiment.\r\n\r\n&nbsp;\r\n\r\n<span style=\"text-align: initial;font-size: 1em\">Conditional Probability formulation for dependent event is given as:<\/span>\r\n\r\n<\/div>\r\n<div>\r\n\r\n<img class=\"aligncenter size-full wp-image-207\" src=\"http:\/\/esp14.epgpbooks.inflibnet.ac.in\/wp-content\/uploads\/sites\/176\/2019\/03\/Untitled-106.png\" alt=\"\" width=\"166\" height=\"55\" \/>\r\n\r\nThis formulation can be read as:\r\n\r\n&nbsp;\r\n\r\nProbability of event E given event F = probability of events E and F \/ probability of F.\r\n\r\n&nbsp;\r\n\r\nSuppose E and F are two independent events of random experiment.\r\n\r\n&nbsp;\r\n\r\nConditional probability formulation for independent event is given as:\r\n\r\n&nbsp;\r\n\r\n<img class=\"aligncenter size-full wp-image-208\" src=\"http:\/\/esp14.epgpbooks.inflibnet.ac.in\/wp-content\/uploads\/sites\/176\/2019\/03\/Untitled-107.png\" alt=\"\" width=\"239\" height=\"54\" \/>\r\n\r\nThis formulation can be read as:\r\n\r\n&nbsp;\r\n\r\nProbability of event E given event F = probability of E.\r\n\r\n&nbsp;\r\n\r\n<strong>Question 5<\/strong>\r\n\r\n&nbsp;\r\n<p style=\"text-align: justify\">A sports person group buys 60 sports cars, 30 bought alarm systems, 20 bought bucket seats and 10 bought an alarm system and bucket seats. If a car buyer chooses at random bucket seat car, what is the probability they also have alarm system?<\/p>\r\n&nbsp;\r\n\r\n<strong>Answer<\/strong>\r\n\r\n&nbsp;\r\n<p style=\"text-align: justify\">Suppose F is event denoting car with bucket seats, E is event denoting car with alarm system and (E and F) is event denoting car with both bucket seats and alarm system.<\/p>\r\n&nbsp;\r\n<p style=\"text-align: justify\">We have to find out the probability of car having alarm system given that it has bucket seat. This probability can be found by using conditional probability formulation for dependent event.<\/p>\r\n&nbsp;\r\n\r\n<img class=\"aligncenter size-full wp-image-209\" src=\"http:\/\/esp14.epgpbooks.inflibnet.ac.in\/wp-content\/uploads\/sites\/176\/2019\/03\/Untitled-108.png\" alt=\"\" width=\"708\" height=\"258\" \/>\r\n\r\n<span style=\"text-align: initial;font-size: 1em\">\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 Figure 5<\/span>\r\n\r\n<img class=\"size-full wp-image-210 alignleft\" src=\"http:\/\/esp14.epgpbooks.inflibnet.ac.in\/wp-content\/uploads\/sites\/176\/2019\/03\/Untitled-109.png\" alt=\"\" width=\"509\" height=\"354\" \/>\r\n\r\n<\/div>\r\n<div>\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\nProbability of a car having alarm system given that it has bucket seat is 0.5.\r\n\r\n&nbsp;\r\n\r\n<strong>5.<\/strong>\u00a0\u00a0\u00a0 <strong>Multiplication Rule<\/strong>\r\n\r\n&nbsp;\r\n<p style=\"text-align: justify\">A statistical property of probability theory is used to find out the probability of happening events in more than one task. Suppose two dice are rolling followed by one after another and find out probability of rolling 3 on one dice and rolling odd number in second dice. Here two tasks are happening, one is rolling dice first time and second one is rolling dice second time simultaneously and we have to find out probability of happening events. These types of problem can be solved by using multiplication probability theorem.<\/p>\r\n&nbsp;\r\n<p style=\"text-align: justify\">\u201cMultiplication probability theorem is applied to find out compound probability of two and more events of more than one task.\u201d<\/p>\r\n&nbsp;\r\n<p style=\"text-align: justify\">In other words, Multiplication probability theorem is used to find probability of events occurring in different task. Intersection concept is used to find probability of all occurring events. Intersection concept is also a key concept of mathematical probability and set theory similar as union concept. It is denoted as \u2229.<\/p>\r\n&nbsp;\r\n<p style=\"text-align: justify\">Since, two events can occur simultaneously in different two tasks. If an event depends on the event of different trial, it is dependent event otherwise it will be independent event. Mathematical formulation\u00a0<span style=\"text-align: initial;font-size: 1em\">for independent and dependent will be slightly different. It is known as multiplication probability formulation for dependent event and multiplication probability formulation for independent event respectively.<\/span><\/p>\r\n&nbsp;\r\n<p style=\"text-align: justify\"><span style=\"text-align: initial;font-size: 1em\">Let us discuss multiplication probability formulation for <\/span>dependent<span style=\"text-align: initial;font-size: 1em\"> and independent event.<\/span><\/p>\r\n&nbsp;\r\n<p style=\"text-align: justify\"><strong style=\"text-align: initial;font-size: 1em\">5.1 Mathematical Formulation<\/strong><\/p>\r\n&nbsp;\r\n<p style=\"text-align: justify\"><span style=\"text-align: initial;font-size: 1em\">Multiplication probability formulation for <\/span>dependent<span style=\"text-align: initial;font-size: 1em\"> event is slightly different from independent events happening in more than two tasks. Multiplication probability formulation for <\/span>dependent<span style=\"text-align: initial;font-size: 1em\"> event is <\/span>generalization<span style=\"text-align: initial;font-size: 1em\"> of the multiplication probability formulation for <\/span>independent<span style=\"text-align: initial;font-size: 1em\"> event.<\/span><\/p>\r\n&nbsp;\r\n<p style=\"text-align: justify\"><strong style=\"text-align: initial;font-size: 1em\">Multiplication probability formulation for dependent event<\/strong><\/p>\r\n&nbsp;\r\n<p style=\"text-align: justify\"><span style=\"text-align: initial;font-size: 1em\">Suppose E and F are two dependent events occurring in two different tasks. Since E and F are two dependent <\/span>event<span style=\"text-align: initial;font-size: 1em\"> i.e. <\/span>outcome<span style=\"text-align: initial;font-size: 1em\"> of one is affecting the outcome of <\/span>second<span style=\"text-align: initial;font-size: 1em\"> event.<\/span><\/p>\r\n&nbsp;\r\n<p style=\"text-align: justify\"><span style=\"text-align: initial;font-size: 1em\">One more thing is that <\/span>second<span style=\"text-align: initial;font-size: 1em\"> task will happen after <\/span>completed<span style=\"text-align: initial;font-size: 1em\"> first task. So that, conditional probability used in multiplication rule.<\/span><\/p>\r\n&nbsp;\r\n<p style=\"text-align: justify\"><span style=\"text-align: initial;font-size: 1em\">Multiplication probability formulation for two dependent events is given as:<\/span><\/p>\r\n&nbsp;\r\n<p style=\"text-align: justify\"><span style=\"text-align: initial;font-size: 1em\">?(? \u2229 ?) = ?(?|?)?(?).<\/span><\/p>\r\n&nbsp;\r\n<p style=\"text-align: justify\"><span style=\"text-align: initial;font-size: 1em\">This formulation can be read as:<\/span><\/p>\r\n&nbsp;\r\n<p style=\"text-align: justify\">Probability<span style=\"text-align: initial;font-size: 1em\"> of events E and F = probability of event E <\/span>given<span style=\"text-align: initial;font-size: 1em\"> F event has already occured \u00d7 probability of event F.<\/span><\/p>\r\n&nbsp;\r\n<p style=\"text-align: justify\"><span style=\"text-align: initial;font-size: 1em\">Multiplication probability formulation for three dependent event E, F and G is given as:<\/span><\/p>\r\n\r\n<\/div>\r\n<div>\r\n<p style=\"text-align: center\">?(? \u2229 ? \u2229 ?) = ?(?)?(?|?)?(?|? \u2229 ?).<\/p>\r\n&nbsp;\r\n\r\nThis formulation can be read as:\r\n\r\n&nbsp;\r\n\r\nProbabilit<span style=\"text-align: initial;font-size: 1em\">y of event E, F and G = probability of event E \u00d7 probability of event F given event E \u00d7 probability of event G given events E and F.<\/span>\r\n\r\n&nbsp;\r\n\r\n<span style=\"text-align: initial;font-size: 1em\">Similarly, multiplication probability formulation can be extended to n dependent events.<\/span>\r\n\r\n<\/div>\r\n<div>\r\n\r\n&nbsp;\r\n\r\n<strong>Question 6<\/strong>\r\n\r\n&nbsp;\r\n<p style=\"text-align: justify\">Two cards are selected from a deck of card followed by one, without replacement of the first card. What is the probability that the first card is club and the second card is ace?<\/p>\r\n&nbsp;\r\n\r\n<strong>Answer<\/strong>\r\n\r\n&nbsp;\r\n<p style=\"text-align: justify\">Suppose E is an event of selecting first card as a club, F is an event of selecting second card as an ace. E and F events occur in two different trails. Second card is selected after first card in second trail with no placement of first card in the deck of card. So, E and F affecting each other and are dependent event. Multiplication probability formulation for dependent event is used to find out the probability that second card is club ace.<\/p>\r\n&nbsp;\r\n\r\nProbabilities are given as:\r\n\r\n<img class=\"size-full wp-image-211 alignleft\" src=\"http:\/\/esp14.epgpbooks.inflibnet.ac.in\/wp-content\/uploads\/sites\/176\/2019\/03\/Untitled-110.png\" alt=\"\" width=\"508\" height=\"387\" \/>\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\nProbability of the first card club and second card ace is 0.0049.\r\n\r\n&nbsp;\r\n\r\n<strong style=\"text-align: initial;font-size: 1em\">Multiplication Probability Formulation for Independent Events<\/strong>\r\n\r\n&nbsp;\r\n\r\n<span style=\"text-align: initial;font-size: 1em\">Suppose E and F are two independent events of two different task i.e. happening of one event is not affecting the happening of another event. So, this multiplication rule formula does not contain conditional probability. It contains simple multiplication of events.<\/span>\r\n\r\n<\/div>\r\n<div>\r\n\r\nMultiplication probability formulation to calculate probability of two independent events E and F is given as:\r\n\r\n&nbsp;\r\n<p style=\"text-align: center\">?(? \u2229 ?) = ?(?)?(?).<\/p>\r\n&nbsp;\r\n\r\nThis formulation can be read as:\r\n\r\n&nbsp;\r\n\r\nProbability of events E and F = probability of event E \u00d7 probability of event F\r\n\r\n&nbsp;\r\n\r\nMultiplication Probability Formulation for three independent event E, F and G is given as:\r\n\r\n&nbsp;\r\n<p style=\"text-align: center\">?(? \u2229 ? \u2229 ?) = ?(?)?(?)?(?).<\/p>\r\n&nbsp;\r\n\r\nSimilarly, Multiplication probability formulation can be extend to n events as need.\r\n\r\n&nbsp;\r\n\r\n<strong>Question 7<\/strong>\r\n\r\n&nbsp;\r\n<p style=\"text-align: justify\">Suppose a person has BMW, Ferrari and Fortuner luxury cars. He also has Puma and Action shoes. If he wants to choose one luxury car and one shoe at random, what is the probability that person choose Fortuner luxury car and Puma shoes?<\/p>\r\n&nbsp;\r\n\r\n<strong>Answer<\/strong>\r\n\r\n&nbsp;\r\n<p style=\"text-align: justify\">Suppose E is an event of selecting Fortuner luxury car and F is an event of selecting Puma shoes. E and F are independent events because selection of one does not affect other. Multiplication probability formulation for independent event is used to find out the probability of E and F events.<\/p>\r\n&nbsp;\r\n\r\nProbabilities of these events are given as:\r\n\r\n<img class=\"aligncenter size-full wp-image-212\" src=\"http:\/\/esp14.epgpbooks.inflibnet.ac.in\/wp-content\/uploads\/sites\/176\/2019\/03\/Untitled-111.png\" alt=\"\" width=\"101\" height=\"115\" \/>\r\n\r\n&nbsp;\r\n\r\nMultiplication probability formulation for two independent events is given as:\r\n\r\n&nbsp;\r\n\r\n<img class=\"aligncenter size-full wp-image-213\" src=\"http:\/\/esp14.epgpbooks.inflibnet.ac.in\/wp-content\/uploads\/sites\/176\/2019\/03\/Untitled-112.png\" alt=\"\" width=\"200\" height=\"148\" \/>\r\n\r\n<\/div>\r\n<div>\r\n\r\n\u00a0 \u00a0\u00a0Probability of choosing Fortuner and Puma shoe is 0.1666.\r\n\r\n&nbsp;\r\n<p style=\"text-align: justify\">In this module, we discussed about the addition and multiplication property of probability. The main purpose of this module is to gain an understanding about the application of addition and multiplication theorem. If one can understand how to apply these rules based upon the problem then the concept of probability is very easy to understand. Many questions and their answers are discussed in this modules regarding this.<\/p>\r\n&nbsp;\r\n\r\n<strong>6.\u00a0\u00a0 Summary<\/strong>\r\n\r\n&nbsp;\r\n<p style=\"text-align: justify\">Probability problem cannot be solved only by using definition and approach of probability. To solve probability problem, one needs to discuss about some probability theorem like addition probability theorem and multiplication probability theorem. Conditional probability is the key concept of probability theory. Conditional probability is used to find probability of an event when prior knowledge of that event is given. Conditional probability is applied for those event that are related to each other. To estimate prior probabilities in terms of posterior probabilities by using external new information, Bayes Theorem is used. Bayes Theorem will be discussed in the module \u201cIntroduction to Bayes Theorem\u201d.<\/p>\r\n&nbsp;\r\n\r\n<strong>7.\u00a0\u00a0 Suggested Readings<\/strong>\r\n\r\n&nbsp;\r\n\r\nAgresti, A. and B. Finlay, <em>Statistical Methods for the Social Science, 3<\/em><em>rd<\/em> <em>Edition, Prentice Hall<\/em>, 1997.\r\n\r\n&nbsp;\r\n<p style=\"text-align: justify\">Daniel, W. W. and C. L. Cross, C. L., <em>Biostatistics<\/em>: <em>A Foundation for Analysis in the Health Sciences,<\/em> 10th Edition<em>,<\/em> John Wiley &amp; Sons, 2013.<\/p>\r\n&nbsp;\r\n\r\nHogg, R. V., J. Mckean and A. Craig, <em>Introduction to Mathematical Statistics<\/em>, Macmillan Pub. Co. Inc., 1978.\r\n\r\n&nbsp;\r\n\r\nMeyer, P. L., <em>Introductory Probability and Statistical Applications<\/em>, Oxford &amp; IBH Pub, 1975.\r\n\r\n<\/div>\r\nStephens, L. J., <em>Schaum\u2019s Series Outline: Beginning Statistics,<\/em> 2nd Edition<em>,<\/em> McGraw Hill, 2006.\r\n\r\n&nbsp;\r\n\r\nTriola, M. F., <em>Elementary Statistics<\/em>, 13th\u00a0 Edition, Pearson, 2017.\r\n\r\n&nbsp;\r\n\r\nWeiss, N. A., <em>Introductory Statistics<\/em>, 10th Edition, Pearson, 2017.\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td><strong>you can view video on Addition and Multiplication Properties of Probability Theory<\/strong><\/td>\r\n<td><a href=\"https:\/\/youtu.be\/WREm900g1hw\" target=\"_blank\" rel=\"noopener\"><img class=\"alignnone wp-image-120\" src=\"http:\/\/epgpbooks.inflibnet.ac.in\/wp-content\/uploads\/2018\/11\/download.png\" alt=\"\" width=\"36\" height=\"36\" \/><\/a><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n&nbsp;\r\n\r\nOne can refer to the following links for further understanding of the statistics terms.\r\n\r\n&nbsp;\r\n\r\n<a href=\"http:\/\/biostat.mc.vanderbilt.edu\/wiki\/pub\/Main\/ClinStat\/glossary.pdf\">http:\/\/biostat.mc.vanderbilt.edu\/wiki\/pub\/Main\/ClinStat\/glossary.pdf<\/a>\r\n\r\n&nbsp;\r\n\r\n<a href=\"http:\/\/www.stats.gla.ac.uk\/steps\/glossary\/alphabet.html\">http:\/\/www.stats.gla.ac.uk\/steps\/glossary\/alphabet.html<\/a>\r\n\r\n&nbsp;\r\n\r\n<a href=\"http:\/\/www.reading.ac.uk\/ssc\/resources\/Docs\/Statistical_Glossary.pdf\">http:\/\/www.reading.ac.uk\/ssc\/resources\/Docs\/Statistical_Glossary.pdf<\/a>\r\n\r\n&nbsp;\r\n\r\n<a href=\"https:\/\/stats.oecd.org\/glossary\/\">https:\/\/stats.oecd.org\/glossary\/<\/a>\r\n\r\n&nbsp;\r\n\r\n<a href=\"http:\/\/www.statsoft.com\/Textbook\/Statistics-Glossary\">http:\/\/www.statsoft.com\/Textbook\/Statistics-Glossary<\/a>\r\n\r\n&nbsp;\r\n\r\n<a href=\"https:\/\/www.stat.berkeley.edu\/~stark\/SticiGui\/Text\/gloss.htm\">https:\/\/www.stat.berkeley.edu\/~stark\/SticiGui\/Text\/gloss.htm<\/a>\r\n\r\n&nbsp;\r\n\r\n<a href=\"https:\/\/stats.oecd.org\/glossary\/alpha.asp?Let=A\">https:\/\/stats.oecd.org\/glossary\/alpha.asp?Let=A<\/a>","rendered":"<div><span style=\"float: right\"><a href=\"https:\/\/youtu.be\/WREm900g1hw\" target=\"_blank\" rel=\"noopener\"><img decoding=\"async\" src=\"http:\/\/epgpbooks.inflibnet.ac.in\/wp-content\/uploads\/2018\/11\/download.png\" alt=\"epgp books\" width=\"75px\" height=\"75px;\" \/><\/a><br \/>\n<\/span><\/div>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p><strong style=\"text-align: initial;font-size: 1em\">Learning Objectives<\/strong><\/p>\n<div>\n<ul>\n<li><strong>Introduction<\/strong><\/li>\n<li><strong>Addition Probability Theorem<\/strong><\/li>\n<li><strong>Conditional Probability<\/strong><\/li>\n<li><strong>Multiplication Rule<\/strong><\/li>\n<li><strong>Summary<\/strong><\/li>\n<li><strong>Suggested Readings<\/strong><\/li>\n<\/ul>\n<p><strong>\u00a0 \u00a0 1.\u00a0<\/strong><strong>Learning Objectives<\/strong><\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: justify\">This module is in continuation of the module &#8220;Introduction to Probability&#8221;. Basic key concepts of probability were discussed in the above said module. In this module, our main concern is to give brief\u00a0<span style=\"text-align: initial;font-size: 1em\">understanding of some basic and most important rules for probability theory. We will discuss these concepts with examples for better understanding. Additions and multiplication rules are considered as <\/span>an important features<span style=\"text-align: initial;font-size: 1em\"> of probability. Construction of formulas and examples are discussed to understand addition and multiplication probability theorem in detail.<\/span><\/p>\n<\/div>\n<div>\n<p><strong>\u00a0 \u00a0 2.<\/strong>\u00a0<strong>Introduction<\/strong><\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: justify\">Probability is an important concept for all type of study variable. Theory of probability deals with several type of problems. To settle problems, there are many rules related to probability theory. Addition and multiplication rules are one of them. The addition and multiplication rule are applied to solve the probability problem which has two or more events.<\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: justify\">Addition rule is used to detect probability of one of the two or more events occurring in the respective random experiment at one task. i.e. at least one event occurred among all events.<\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: justify\">Multiplication rule is used to detect outcome that occurs from more than one task in respective random experiment.<\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: justify\">Let us suppose, if there is only one task of tossing a coin two times and find the probability of getting heads on both time. This type of probability is problem solved by addition rule of the probability.<\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: justify\">Suppose tossing a coin two times one after another. If one get heads on the first toss and also on the second toss. These two outcomes are independent as outcome of first toss is not affecting the outcome of second toss. This type of probability problem is solved by multiplication rule of the probability.<\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: justify\">Here, some basic features of an event will help one to understand mathematical behavior of the probability problem. These features will be helpful to understand statistical addition and multiplication probability Theorem. These are:<\/p>\n<p>? \u222a ? = {? \u2208 ?; ? \u2208 ? ?? ? \u2208 ?};<\/p>\n<p>? = {? \u2208 ?; ? does not belong to E};<\/p>\n<p>? \u2229 ? = {? \u2208 ?; ? \u2208 ? ??? ? \u2208 ?};<\/p>\n<p>? \u2282 ? for every ? \u2208 ?, ? \u2208 ?.<\/p>\n<p>&nbsp;<\/p>\n<p>where ? \u222a ? (say), E union F, means at least one event happens.<\/p>\n<p>? \u0305(say), E bar or E compliment, means outcomes does not belong to E.<\/p>\n<p>? \u2229 ? (say), E intersection F, means both events happens.<\/p>\n<p>? \u2282 ? (say), E subset of F, means outcome of event belongs to outcome of F.<\/p>\n<p>&nbsp;<\/p>\n<p>Let us discuss in detail explanation of the addition and multiplication probability theorem.<\/p>\n<p>&nbsp;<\/p>\n<p><strong style=\"text-align: initial;font-size: 1em\">3.<\/strong><span style=\"text-align: initial;font-size: 1em\">\u00a0\u00a0\u00a0 <\/span><strong style=\"text-align: initial;font-size: 1em\">Addition Probability Theorem<\/strong><\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: justify\"><span style=\"text-align: initial;font-size: 1em\">A statistical property of probability theory is used to explain happening of one or more event at one task. Suppose a random experiment, selecting a card from a deck of card and event, is selected card jack and queen. Since it is impossible to draw both jack and queen in <\/span>same<span style=\"text-align: initial;font-size: 1em\"> draw. <\/span>Outcome<span style=\"text-align: initial;font-size: 1em\"> will be either jack or queen. This type of probability problem is solved by using <\/span>addition<span style=\"text-align: initial;font-size: 1em\"> rule.<\/span><\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: justify\"><span style=\"text-align: initial;font-size: 1em\">\u201cThe addition rule is used to find <\/span>compound<span style=\"text-align: initial;font-size: 1em\"> probability of one or more events with the either-or statement format of the problem.\u201d<\/span><\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: justify\"><span style=\"text-align: initial;font-size: 1em\">In other words, the addition theorem solves the probability of happening at least one of the given events. Union concept is used to find <\/span>probability<span style=\"text-align: initial;font-size: 1em\"> of happening at least one event. <\/span>Union<span style=\"text-align: initial;font-size: 1em\"> is <\/span>key<span style=\"text-align: initial;font-size: 1em\"> concept of mathematical probability theory and set theory. This union representation looks like alphabet U.<\/span><\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: justify\">Since,<span style=\"text-align: initial;font-size: 1em\"> two events cannot occur simultaneously in one task and cannot have <\/span>common<span style=\"text-align: initial;font-size: 1em\"> outcome, it is known as <\/span>mutually<span style=\"text-align: initial;font-size: 1em\"> exclusive event. If events have <\/span>common<span style=\"text-align: initial;font-size: 1em\"> outcome, <\/span>event<span style=\"text-align: initial;font-size: 1em\"> will not be <\/span>mutually<span style=\"text-align: initial;font-size: 1em\"> exclusive event. It may <\/span>be say<span style=\"text-align: initial;font-size: 1em\"> as <\/span>non-mutually<span style=\"text-align: initial;font-size: 1em\"> exclusive event.<\/span><\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: justify\"><span style=\"text-align: initial;font-size: 1em\">Based on these two types of events, mutually exclusive and non-mutually exclusive event mathematical formulation are described as:<\/span><\/p>\n<\/div>\n<div>\n<p><strong>\u00a0 \u00a0 3.1 Mathematical Formulation<\/strong><\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: justify\">Mathematical formulation of addition probability theorem is slightly different from mutually exclusive event than non-mutually exclusive event. There is only one difference i.e. addition formula for mutually exclusive does not have intersection term but addition formula for non-mutually exclusive event has\u00a0<span style=\"text-align: initial;font-size: 1em\">intersection term. Addition probability formulation for non-mutually exclusive is the generalization of the addition probability formulation for mutually exclusive formula.<\/span><\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: justify\"><span style=\"text-align: initial;font-size: 1em\">Let us discuss addition probability formulation for <\/span>non-mutually<span style=\"text-align: initial;font-size: 1em\"> exclusive event and mutually exclusive event.<\/span><\/p>\n<\/div>\n<div>\n<p><strong>\u00a0 \u00a0 Addition Probability Formulation for Non-mutually Exclusive Events<\/strong><\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: justify\">Venn diagram is a graphical representation for set theory. Sample space is a set of all possible outcomes and event is a subset of sample space. Event is also a set. Venn diagram is being used here, to show graphical representation of the given event.<\/p>\n<p>&nbsp;<\/p>\n<p>Suppose E and F are two events of a random experiment.<\/p>\n<p>&nbsp;<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-195\" src=\"http:\/\/esp14.epgpbooks.inflibnet.ac.in\/wp-content\/uploads\/sites\/176\/2019\/03\/Untitled-96.png\" alt=\"\" width=\"179\" height=\"177\" srcset=\"https:\/\/ebooks.inflibnet.ac.in\/esp14\/wp-content\/uploads\/sites\/176\/2019\/03\/Untitled-96.png 179w, https:\/\/ebooks.inflibnet.ac.in\/esp14\/wp-content\/uploads\/sites\/176\/2019\/03\/Untitled-96-65x64.png 65w\" sizes=\"auto, (max-width: 179px) 100vw, 179px\" \/><\/p>\n<p>&nbsp;<\/p>\n<p>In Figure 1 Venn diagram, event E is shown as blue circle and event F is shown as yellow circle. Event E and F have some common area and have common outcomes, so event E and F are non-mutually exclusive event.<\/p>\n<p>&nbsp;<\/p>\n<p>Addition Probability Formulation for non-mutually exclusive events E or F is given by:<\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: center\">?(? \u222a ?) = ?(?) + ?(?) \u2212 ?(? \u2229 ?)<\/p>\n<p>&nbsp;<\/p>\n<p>This formulation can be read as:<\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: justify\">Probability of happening either event E or event F = probability of happening event E + probability of happening event F \u2013 probability of happening events E and F.<\/p>\n<p>&nbsp;<\/p>\n<p>Suppose there are three non-mutually exclusive events E, F and G<\/p>\n<\/div>\n<div>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-197\" src=\"http:\/\/esp14.epgpbooks.inflibnet.ac.in\/wp-content\/uploads\/sites\/176\/2019\/03\/Untitled-97.png\" alt=\"\" width=\"159\" height=\"179\" srcset=\"https:\/\/ebooks.inflibnet.ac.in\/esp14\/wp-content\/uploads\/sites\/176\/2019\/03\/Untitled-97.png 159w, https:\/\/ebooks.inflibnet.ac.in\/esp14\/wp-content\/uploads\/sites\/176\/2019\/03\/Untitled-97-65x73.png 65w\" sizes=\"auto, (max-width: 159px) 100vw, 159px\" \/><\/p>\n<p>&nbsp;<\/p>\n<p>Addition probability formulation for non-mutually exclusive events E, F or G are given by:<\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: center\">?(? \u222a ? \u222a ?) = ?(?) + ?(?) + ?(?) \u2212 ?(? \u2229 ?) \u2212 ?(? \u2229 ?) \u2212 ?(? \u2229 ?) + ?(? \u2229 ? \u2229 ?)<\/p>\n<p>&nbsp;<\/p>\n<p>Similarly, addition probability formulation can be extented for n events.<\/p>\n<p>&nbsp;<\/p>\n<p><strong>Question 1<\/strong><\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: justify\">Suppose in a class there are 13 students, 7 are boys and 6 are girls. On unit test 4 boys and 5 girls received A grade. If a student chooses at random in the class, what is the probability of choosing a girl or an A grade student?<\/p>\n<p>&nbsp;<\/p>\n<p><strong>Answer<\/strong><\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: justify\">Suppose E is an event of choosing girl and F is an event of choosing A grade student. Hence we are interested in to find out the probability of E or F.<\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: justify\">Event E and F have 5 common outcomes i.e. 5 girls made A grade. Hence E and F are non-mutually exclusive events.<\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: justify\">We have to find probability of non-mutually exclusive events by using addition probability formulation.<\/p>\n<p>&nbsp;<\/p>\n<p>Probabilities are given as:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-198\" src=\"http:\/\/esp14.epgpbooks.inflibnet.ac.in\/wp-content\/uploads\/sites\/176\/2019\/03\/Untitled-98.png\" alt=\"\" width=\"138\" height=\"189\" srcset=\"https:\/\/ebooks.inflibnet.ac.in\/esp14\/wp-content\/uploads\/sites\/176\/2019\/03\/Untitled-98.png 138w, https:\/\/ebooks.inflibnet.ac.in\/esp14\/wp-content\/uploads\/sites\/176\/2019\/03\/Untitled-98-65x89.png 65w\" sizes=\"auto, (max-width: 138px) 100vw, 138px\" \/><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"text-align: initial;font-size: 1em\">Addition probability formulation for two non-mutually exclusive events is given as:<\/span><\/p>\n<\/div>\n<div>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-199\" src=\"http:\/\/esp14.epgpbooks.inflibnet.ac.in\/wp-content\/uploads\/sites\/176\/2019\/03\/Untitled-99.png\" alt=\"\" width=\"311\" height=\"156\" srcset=\"https:\/\/ebooks.inflibnet.ac.in\/esp14\/wp-content\/uploads\/sites\/176\/2019\/03\/Untitled-99.png 311w, https:\/\/ebooks.inflibnet.ac.in\/esp14\/wp-content\/uploads\/sites\/176\/2019\/03\/Untitled-99-300x150.png 300w, https:\/\/ebooks.inflibnet.ac.in\/esp14\/wp-content\/uploads\/sites\/176\/2019\/03\/Untitled-99-65x33.png 65w, https:\/\/ebooks.inflibnet.ac.in\/esp14\/wp-content\/uploads\/sites\/176\/2019\/03\/Untitled-99-225x113.png 225w\" sizes=\"auto, (max-width: 311px) 100vw, 311px\" \/><\/p>\n<p>&nbsp;<\/p>\n<p>Probability of choosing a girl or A grade student is 0.7692.<\/p>\n<p>&nbsp;<\/p>\n<p><strong>Question 2<\/strong><\/p>\n<p>&nbsp;<\/p>\n<p>If the probability of solving a quiz by Rubi and Kiran are 3\/4 and 2\/3 respectively, what is the probability of solving the quiz?<\/p>\n<p>&nbsp;<\/p>\n<p><strong>Answer<\/strong><\/p>\n<p style=\"text-align: justify\">Let E and F are the event of solving a quiz by Rubi and Kiran respectively. The probabilities of these events are given as:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-200\" src=\"http:\/\/esp14.epgpbooks.inflibnet.ac.in\/wp-content\/uploads\/sites\/176\/2019\/03\/Untitled-100.png\" alt=\"\" width=\"188\" height=\"46\" srcset=\"https:\/\/ebooks.inflibnet.ac.in\/esp14\/wp-content\/uploads\/sites\/176\/2019\/03\/Untitled-100.png 188w, https:\/\/ebooks.inflibnet.ac.in\/esp14\/wp-content\/uploads\/sites\/176\/2019\/03\/Untitled-100-65x16.png 65w\" sizes=\"auto, (max-width: 188px) 100vw, 188px\" \/><\/p>\n<p style=\"text-align: justify\">As we are interested in solving the quiz hence there are the following possibilities. First possibility is that either Kiran solve the quiz or Rubi solve the quiz. It means that quiz will be solved if atleast one of them will solve the quiz. Second possibility is that both solve the quiz. As Kiran and Rubi can solve the quiz independently. So we can consider events E and F are independent events.<\/p>\n<p>&nbsp;<\/p>\n<p>Therefore:<\/p>\n<p style=\"text-align: center\">?(? \u2229 ?) = ?(?) \u2217 ?(?).<\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: justify\">As the probability of solving the quiz by atleast one of two events is represented as union of two events. So, find the probability by using addition probability formulation. The formula is given as<\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: center\">?(? \u222a ?) = ?(?) + ?(?) \u2212 ?(? \u2229 ?);<\/p>\n<p>&nbsp;<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-201\" src=\"http:\/\/esp14.epgpbooks.inflibnet.ac.in\/wp-content\/uploads\/sites\/176\/2019\/03\/Untitled-101.png\" alt=\"\" width=\"297\" height=\"57\" srcset=\"https:\/\/ebooks.inflibnet.ac.in\/esp14\/wp-content\/uploads\/sites\/176\/2019\/03\/Untitled-101.png 297w, https:\/\/ebooks.inflibnet.ac.in\/esp14\/wp-content\/uploads\/sites\/176\/2019\/03\/Untitled-101-65x12.png 65w, https:\/\/ebooks.inflibnet.ac.in\/esp14\/wp-content\/uploads\/sites\/176\/2019\/03\/Untitled-101-225x43.png 225w\" sizes=\"auto, (max-width: 297px) 100vw, 297px\" \/><\/p>\n<\/div>\n<div>\n<p>\u00a0 \u00a0 Hence the probability of solving the quiz is 0.9166.<\/p>\n<p>&nbsp;<\/p>\n<p><strong>Mutually exclusive addition formula:<\/strong><\/p>\n<p>&nbsp;<\/p>\n<p>Suppose E and F are two events of a random experiment respectively.<\/p>\n<p>&nbsp;<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-202\" src=\"http:\/\/esp14.epgpbooks.inflibnet.ac.in\/wp-content\/uploads\/sites\/176\/2019\/03\/Untitled-102.png\" alt=\"\" width=\"176\" height=\"185\" srcset=\"https:\/\/ebooks.inflibnet.ac.in\/esp14\/wp-content\/uploads\/sites\/176\/2019\/03\/Untitled-102.png 176w, https:\/\/ebooks.inflibnet.ac.in\/esp14\/wp-content\/uploads\/sites\/176\/2019\/03\/Untitled-102-65x68.png 65w\" sizes=\"auto, (max-width: 176px) 100vw, 176px\" \/><\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: justify\">In Figure 3 of Venn diagram, event E is shown as blue circle and event F is shown as yellow circle. Events E and F do not have common area and same outcomes, so event E and F are mutually exclusive event.<\/p>\n<p>&nbsp;<\/p>\n<p>Addition probability formulation for mutually exclusive events E and F is given by:<\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: center\">?(? \u222a ?) = ?(?) + ?(?).<\/p>\n<p>&nbsp;<\/p>\n<p>This formulation can be read as:<\/p>\n<p>&nbsp;<\/p>\n<p>Probability of happening either event E or event F = probability of happening event E +probability of happening event F.<\/p>\n<p>&nbsp;<\/p>\n<p>If there are three mutually exclusive events E, F and G.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-203\" src=\"http:\/\/esp14.epgpbooks.inflibnet.ac.in\/wp-content\/uploads\/sites\/176\/2019\/03\/Untitled-103.png\" alt=\"\" width=\"272\" height=\"214\" srcset=\"https:\/\/ebooks.inflibnet.ac.in\/esp14\/wp-content\/uploads\/sites\/176\/2019\/03\/Untitled-103.png 272w, https:\/\/ebooks.inflibnet.ac.in\/esp14\/wp-content\/uploads\/sites\/176\/2019\/03\/Untitled-103-65x51.png 65w, https:\/\/ebooks.inflibnet.ac.in\/esp14\/wp-content\/uploads\/sites\/176\/2019\/03\/Untitled-103-225x177.png 225w\" sizes=\"auto, (max-width: 272px) 100vw, 272px\" \/><\/p>\n<p style=\"text-align: center\">Figure 4<\/p>\n<\/div>\n<p>Addition probability formulation for three mutually exclusive events is given as follows<\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: center\">?(? \u222a ? \u222a ?) = ?(?) + ?(?) + ?(?)<\/p>\n<p>&nbsp;<\/p>\n<p>Similarly, Addition probability formulation can be extended to n events or outcomes.<\/p>\n<p>&nbsp;<\/p>\n<p><strong>Question 3<\/strong><\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: justify\">Suppose in a group of 50 members, there are 15 junior students , 20 senior students and 15 instructor . What is the probability that a person randomly picked from this group is either junior or senior student?<\/p>\n<p>&nbsp;<\/p>\n<p><strong>Answer<\/strong><\/p>\n<p>&nbsp;<\/p>\n<p>Let E and F be the event of selecting junior and senior student respectively from a group of 50 members.<\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: justify\">As events E and F do not have any common outcomes i.e. student can either be junior and senior. So E and F are mutually exclusive events.<\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: justify\">Addition probability formulation is used to find probability of picking a member from a group of 50 members that can be either junior or senior student.<\/p>\n<p>&nbsp;<\/p>\n<p>Probabilities of events E and F are given as:<\/p>\n<p>&nbsp;<\/p>\n<p>P(E) = 15\/50.<\/p>\n<p>&nbsp;<\/p>\n<p>P(F) = 20\/50.<\/p>\n<p>&nbsp;<\/p>\n<p>Addition probability formulation for two mutually exclusive events is given as:<\/p>\n<p>&nbsp;<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-204\" src=\"http:\/\/esp14.epgpbooks.inflibnet.ac.in\/wp-content\/uploads\/sites\/176\/2019\/03\/Untitled-104.png\" alt=\"\" width=\"209\" height=\"158\" srcset=\"https:\/\/ebooks.inflibnet.ac.in\/esp14\/wp-content\/uploads\/sites\/176\/2019\/03\/Untitled-104.png 209w, https:\/\/ebooks.inflibnet.ac.in\/esp14\/wp-content\/uploads\/sites\/176\/2019\/03\/Untitled-104-65x49.png 65w\" sizes=\"auto, (max-width: 209px) 100vw, 209px\" \/><\/p>\n<p>&nbsp;<\/p>\n<p>Probability of selectinga junior or senior student randomly from this group is 0.7.<\/p>\n<p>&nbsp;<\/p>\n<p><strong>Question 4<\/strong><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"text-align: initial;font-size: 1em\">Suppose a card selected from standard deck of cards. What is the probability of selecting either ace or queen card from a deck of card?<\/span><\/p>\n<p>&nbsp;<\/p>\n<p><strong style=\"text-align: initial;font-size: 1em\">Answer<\/strong><\/p>\n<div>\n<p>&nbsp;<\/p>\n<p>Let us consider two events E and F<\/p>\n<p>&nbsp;<\/p>\n<p>E:\u00a0\u00a0 selected card is ace.<\/p>\n<p>&nbsp;<\/p>\n<p>F:\u00a0 selected card is queen.<\/p>\n<p>&nbsp;<\/p>\n<p>As events E and F do not have common outcomes. So, E and F are mutually exclusive event.<\/p>\n<p>&nbsp;<\/p>\n<p>Addition probability formulation is used to find the probability of selecting card either ace and queen.<\/p>\n<p>&nbsp;<\/p>\n<p>Probabilities of events E and F are given as:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-206 alignleft\" src=\"http:\/\/esp14.epgpbooks.inflibnet.ac.in\/wp-content\/uploads\/sites\/176\/2019\/03\/Untitled-105.png\" alt=\"\" width=\"592\" height=\"256\" srcset=\"https:\/\/ebooks.inflibnet.ac.in\/esp14\/wp-content\/uploads\/sites\/176\/2019\/03\/Untitled-105.png 592w, https:\/\/ebooks.inflibnet.ac.in\/esp14\/wp-content\/uploads\/sites\/176\/2019\/03\/Untitled-105-300x130.png 300w, https:\/\/ebooks.inflibnet.ac.in\/esp14\/wp-content\/uploads\/sites\/176\/2019\/03\/Untitled-105-65x28.png 65w, https:\/\/ebooks.inflibnet.ac.in\/esp14\/wp-content\/uploads\/sites\/176\/2019\/03\/Untitled-105-225x97.png 225w, https:\/\/ebooks.inflibnet.ac.in\/esp14\/wp-content\/uploads\/sites\/176\/2019\/03\/Untitled-105-350x151.png 350w\" sizes=\"auto, (max-width: 592px) 100vw, 592px\" \/><\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>Hence the probability of selecting an ace and a queen from a deck of card is 2\/13.<\/p>\n<p>&nbsp;<\/p>\n<p><strong>4.\u00a0\u00a0 Conditional Probability<\/strong><\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: justify\">When one event has a relationship with other respective events. Suppose event E is the event that is related with raining, there is 0.7% chance of raining. Conditional probability is used to find out probability of these types of events.<\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: justify\">\u201cConditional probability measured the probability of an event given that one event occurred.\u201d<\/p>\n<p>&nbsp;<\/p>\n<p>Suppose E and F are two dependent events of random experiment.<\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"text-align: initial;font-size: 1em\">Conditional Probability formulation for dependent event is given as:<\/span><\/p>\n<\/div>\n<div>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-207\" src=\"http:\/\/esp14.epgpbooks.inflibnet.ac.in\/wp-content\/uploads\/sites\/176\/2019\/03\/Untitled-106.png\" alt=\"\" width=\"166\" height=\"55\" srcset=\"https:\/\/ebooks.inflibnet.ac.in\/esp14\/wp-content\/uploads\/sites\/176\/2019\/03\/Untitled-106.png 166w, https:\/\/ebooks.inflibnet.ac.in\/esp14\/wp-content\/uploads\/sites\/176\/2019\/03\/Untitled-106-65x22.png 65w\" sizes=\"auto, (max-width: 166px) 100vw, 166px\" \/><\/p>\n<p>This formulation can be read as:<\/p>\n<p>&nbsp;<\/p>\n<p>Probability of event E given event F = probability of events E and F \/ probability of F.<\/p>\n<p>&nbsp;<\/p>\n<p>Suppose E and F are two independent events of random experiment.<\/p>\n<p>&nbsp;<\/p>\n<p>Conditional probability formulation for independent event is given as:<\/p>\n<p>&nbsp;<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-208\" src=\"http:\/\/esp14.epgpbooks.inflibnet.ac.in\/wp-content\/uploads\/sites\/176\/2019\/03\/Untitled-107.png\" alt=\"\" width=\"239\" height=\"54\" srcset=\"https:\/\/ebooks.inflibnet.ac.in\/esp14\/wp-content\/uploads\/sites\/176\/2019\/03\/Untitled-107.png 239w, https:\/\/ebooks.inflibnet.ac.in\/esp14\/wp-content\/uploads\/sites\/176\/2019\/03\/Untitled-107-65x15.png 65w, https:\/\/ebooks.inflibnet.ac.in\/esp14\/wp-content\/uploads\/sites\/176\/2019\/03\/Untitled-107-225x51.png 225w\" sizes=\"auto, (max-width: 239px) 100vw, 239px\" \/><\/p>\n<p>This formulation can be read as:<\/p>\n<p>&nbsp;<\/p>\n<p>Probability of event E given event F = probability of E.<\/p>\n<p>&nbsp;<\/p>\n<p><strong>Question 5<\/strong><\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: justify\">A sports person group buys 60 sports cars, 30 bought alarm systems, 20 bought bucket seats and 10 bought an alarm system and bucket seats. If a car buyer chooses at random bucket seat car, what is the probability they also have alarm system?<\/p>\n<p>&nbsp;<\/p>\n<p><strong>Answer<\/strong><\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: justify\">Suppose F is event denoting car with bucket seats, E is event denoting car with alarm system and (E and F) is event denoting car with both bucket seats and alarm system.<\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: justify\">We have to find out the probability of car having alarm system given that it has bucket seat. This probability can be found by using conditional probability formulation for dependent event.<\/p>\n<p>&nbsp;<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-209\" src=\"http:\/\/esp14.epgpbooks.inflibnet.ac.in\/wp-content\/uploads\/sites\/176\/2019\/03\/Untitled-108.png\" alt=\"\" width=\"708\" height=\"258\" srcset=\"https:\/\/ebooks.inflibnet.ac.in\/esp14\/wp-content\/uploads\/sites\/176\/2019\/03\/Untitled-108.png 708w, https:\/\/ebooks.inflibnet.ac.in\/esp14\/wp-content\/uploads\/sites\/176\/2019\/03\/Untitled-108-300x109.png 300w, https:\/\/ebooks.inflibnet.ac.in\/esp14\/wp-content\/uploads\/sites\/176\/2019\/03\/Untitled-108-65x24.png 65w, https:\/\/ebooks.inflibnet.ac.in\/esp14\/wp-content\/uploads\/sites\/176\/2019\/03\/Untitled-108-225x82.png 225w, https:\/\/ebooks.inflibnet.ac.in\/esp14\/wp-content\/uploads\/sites\/176\/2019\/03\/Untitled-108-350x128.png 350w\" sizes=\"auto, (max-width: 708px) 100vw, 708px\" \/><\/p>\n<p><span style=\"text-align: initial;font-size: 1em\">\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 Figure 5<\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-210 alignleft\" src=\"http:\/\/esp14.epgpbooks.inflibnet.ac.in\/wp-content\/uploads\/sites\/176\/2019\/03\/Untitled-109.png\" alt=\"\" width=\"509\" height=\"354\" srcset=\"https:\/\/ebooks.inflibnet.ac.in\/esp14\/wp-content\/uploads\/sites\/176\/2019\/03\/Untitled-109.png 509w, https:\/\/ebooks.inflibnet.ac.in\/esp14\/wp-content\/uploads\/sites\/176\/2019\/03\/Untitled-109-300x209.png 300w, https:\/\/ebooks.inflibnet.ac.in\/esp14\/wp-content\/uploads\/sites\/176\/2019\/03\/Untitled-109-65x45.png 65w, https:\/\/ebooks.inflibnet.ac.in\/esp14\/wp-content\/uploads\/sites\/176\/2019\/03\/Untitled-109-225x156.png 225w, https:\/\/ebooks.inflibnet.ac.in\/esp14\/wp-content\/uploads\/sites\/176\/2019\/03\/Untitled-109-350x243.png 350w\" sizes=\"auto, (max-width: 509px) 100vw, 509px\" \/><\/p>\n<\/div>\n<div>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>Probability of a car having alarm system given that it has bucket seat is 0.5.<\/p>\n<p>&nbsp;<\/p>\n<p><strong>5.<\/strong>\u00a0\u00a0\u00a0 <strong>Multiplication Rule<\/strong><\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: justify\">A statistical property of probability theory is used to find out the probability of happening events in more than one task. Suppose two dice are rolling followed by one after another and find out probability of rolling 3 on one dice and rolling odd number in second dice. Here two tasks are happening, one is rolling dice first time and second one is rolling dice second time simultaneously and we have to find out probability of happening events. These types of problem can be solved by using multiplication probability theorem.<\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: justify\">\u201cMultiplication probability theorem is applied to find out compound probability of two and more events of more than one task.\u201d<\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: justify\">In other words, Multiplication probability theorem is used to find probability of events occurring in different task. Intersection concept is used to find probability of all occurring events. Intersection concept is also a key concept of mathematical probability and set theory similar as union concept. It is denoted as \u2229.<\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: justify\">Since, two events can occur simultaneously in different two tasks. If an event depends on the event of different trial, it is dependent event otherwise it will be independent event. Mathematical formulation\u00a0<span style=\"text-align: initial;font-size: 1em\">for independent and dependent will be slightly different. It is known as multiplication probability formulation for dependent event and multiplication probability formulation for independent event respectively.<\/span><\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: justify\"><span style=\"text-align: initial;font-size: 1em\">Let us discuss multiplication probability formulation for <\/span>dependent<span style=\"text-align: initial;font-size: 1em\"> and independent event.<\/span><\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: justify\"><strong style=\"text-align: initial;font-size: 1em\">5.1 Mathematical Formulation<\/strong><\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: justify\"><span style=\"text-align: initial;font-size: 1em\">Multiplication probability formulation for <\/span>dependent<span style=\"text-align: initial;font-size: 1em\"> event is slightly different from independent events happening in more than two tasks. Multiplication probability formulation for <\/span>dependent<span style=\"text-align: initial;font-size: 1em\"> event is <\/span>generalization<span style=\"text-align: initial;font-size: 1em\"> of the multiplication probability formulation for <\/span>independent<span style=\"text-align: initial;font-size: 1em\"> event.<\/span><\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: justify\"><strong style=\"text-align: initial;font-size: 1em\">Multiplication probability formulation for dependent event<\/strong><\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: justify\"><span style=\"text-align: initial;font-size: 1em\">Suppose E and F are two dependent events occurring in two different tasks. Since E and F are two dependent <\/span>event<span style=\"text-align: initial;font-size: 1em\"> i.e. <\/span>outcome<span style=\"text-align: initial;font-size: 1em\"> of one is affecting the outcome of <\/span>second<span style=\"text-align: initial;font-size: 1em\"> event.<\/span><\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: justify\"><span style=\"text-align: initial;font-size: 1em\">One more thing is that <\/span>second<span style=\"text-align: initial;font-size: 1em\"> task will happen after <\/span>completed<span style=\"text-align: initial;font-size: 1em\"> first task. So that, conditional probability used in multiplication rule.<\/span><\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: justify\"><span style=\"text-align: initial;font-size: 1em\">Multiplication probability formulation for two dependent events is given as:<\/span><\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: justify\"><span style=\"text-align: initial;font-size: 1em\">?(? \u2229 ?) = ?(?|?)?(?).<\/span><\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: justify\"><span style=\"text-align: initial;font-size: 1em\">This formulation can be read as:<\/span><\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: justify\">Probability<span style=\"text-align: initial;font-size: 1em\"> of events E and F = probability of event E <\/span>given<span style=\"text-align: initial;font-size: 1em\"> F event has already occured \u00d7 probability of event F.<\/span><\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: justify\"><span style=\"text-align: initial;font-size: 1em\">Multiplication probability formulation for three dependent event E, F and G is given as:<\/span><\/p>\n<\/div>\n<div>\n<p style=\"text-align: center\">?(? \u2229 ? \u2229 ?) = ?(?)?(?|?)?(?|? \u2229 ?).<\/p>\n<p>&nbsp;<\/p>\n<p>This formulation can be read as:<\/p>\n<p>&nbsp;<\/p>\n<p>Probabilit<span style=\"text-align: initial;font-size: 1em\">y of event E, F and G = probability of event E \u00d7 probability of event F given event E \u00d7 probability of event G given events E and F.<\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"text-align: initial;font-size: 1em\">Similarly, multiplication probability formulation can be extended to n dependent events.<\/span><\/p>\n<\/div>\n<div>\n<p>&nbsp;<\/p>\n<p><strong>Question 6<\/strong><\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: justify\">Two cards are selected from a deck of card followed by one, without replacement of the first card. What is the probability that the first card is club and the second card is ace?<\/p>\n<p>&nbsp;<\/p>\n<p><strong>Answer<\/strong><\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: justify\">Suppose E is an event of selecting first card as a club, F is an event of selecting second card as an ace. E and F events occur in two different trails. Second card is selected after first card in second trail with no placement of first card in the deck of card. So, E and F affecting each other and are dependent event. Multiplication probability formulation for dependent event is used to find out the probability that second card is club ace.<\/p>\n<p>&nbsp;<\/p>\n<p>Probabilities are given as:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-211 alignleft\" src=\"http:\/\/esp14.epgpbooks.inflibnet.ac.in\/wp-content\/uploads\/sites\/176\/2019\/03\/Untitled-110.png\" alt=\"\" width=\"508\" height=\"387\" srcset=\"https:\/\/ebooks.inflibnet.ac.in\/esp14\/wp-content\/uploads\/sites\/176\/2019\/03\/Untitled-110.png 508w, https:\/\/ebooks.inflibnet.ac.in\/esp14\/wp-content\/uploads\/sites\/176\/2019\/03\/Untitled-110-300x229.png 300w, https:\/\/ebooks.inflibnet.ac.in\/esp14\/wp-content\/uploads\/sites\/176\/2019\/03\/Untitled-110-65x50.png 65w, https:\/\/ebooks.inflibnet.ac.in\/esp14\/wp-content\/uploads\/sites\/176\/2019\/03\/Untitled-110-225x171.png 225w, https:\/\/ebooks.inflibnet.ac.in\/esp14\/wp-content\/uploads\/sites\/176\/2019\/03\/Untitled-110-350x267.png 350w\" sizes=\"auto, (max-width: 508px) 100vw, 508px\" \/><\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>Probability of the first card club and second card ace is 0.0049.<\/p>\n<p>&nbsp;<\/p>\n<p><strong style=\"text-align: initial;font-size: 1em\">Multiplication Probability Formulation for Independent Events<\/strong><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"text-align: initial;font-size: 1em\">Suppose E and F are two independent events of two different task i.e. happening of one event is not affecting the happening of another event. So, this multiplication rule formula does not contain conditional probability. It contains simple multiplication of events.<\/span><\/p>\n<\/div>\n<div>\n<p>Multiplication probability formulation to calculate probability of two independent events E and F is given as:<\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: center\">?(? \u2229 ?) = ?(?)?(?).<\/p>\n<p>&nbsp;<\/p>\n<p>This formulation can be read as:<\/p>\n<p>&nbsp;<\/p>\n<p>Probability of events E and F = probability of event E \u00d7 probability of event F<\/p>\n<p>&nbsp;<\/p>\n<p>Multiplication Probability Formulation for three independent event E, F and G is given as:<\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: center\">?(? \u2229 ? \u2229 ?) = ?(?)?(?)?(?).<\/p>\n<p>&nbsp;<\/p>\n<p>Similarly, Multiplication probability formulation can be extend to n events as need.<\/p>\n<p>&nbsp;<\/p>\n<p><strong>Question 7<\/strong><\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: justify\">Suppose a person has BMW, Ferrari and Fortuner luxury cars. He also has Puma and Action shoes. If he wants to choose one luxury car and one shoe at random, what is the probability that person choose Fortuner luxury car and Puma shoes?<\/p>\n<p>&nbsp;<\/p>\n<p><strong>Answer<\/strong><\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: justify\">Suppose E is an event of selecting Fortuner luxury car and F is an event of selecting Puma shoes. E and F are independent events because selection of one does not affect other. Multiplication probability formulation for independent event is used to find out the probability of E and F events.<\/p>\n<p>&nbsp;<\/p>\n<p>Probabilities of these events are given as:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-212\" src=\"http:\/\/esp14.epgpbooks.inflibnet.ac.in\/wp-content\/uploads\/sites\/176\/2019\/03\/Untitled-111.png\" alt=\"\" width=\"101\" height=\"115\" srcset=\"https:\/\/ebooks.inflibnet.ac.in\/esp14\/wp-content\/uploads\/sites\/176\/2019\/03\/Untitled-111.png 101w, https:\/\/ebooks.inflibnet.ac.in\/esp14\/wp-content\/uploads\/sites\/176\/2019\/03\/Untitled-111-65x74.png 65w\" sizes=\"auto, (max-width: 101px) 100vw, 101px\" \/><\/p>\n<p>&nbsp;<\/p>\n<p>Multiplication probability formulation for two independent events is given as:<\/p>\n<p>&nbsp;<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-213\" src=\"http:\/\/esp14.epgpbooks.inflibnet.ac.in\/wp-content\/uploads\/sites\/176\/2019\/03\/Untitled-112.png\" alt=\"\" width=\"200\" height=\"148\" srcset=\"https:\/\/ebooks.inflibnet.ac.in\/esp14\/wp-content\/uploads\/sites\/176\/2019\/03\/Untitled-112.png 200w, https:\/\/ebooks.inflibnet.ac.in\/esp14\/wp-content\/uploads\/sites\/176\/2019\/03\/Untitled-112-65x48.png 65w\" sizes=\"auto, (max-width: 200px) 100vw, 200px\" \/><\/p>\n<\/div>\n<div>\n<p>\u00a0 \u00a0\u00a0Probability of choosing Fortuner and Puma shoe is 0.1666.<\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: justify\">In this module, we discussed about the addition and multiplication property of probability. The main purpose of this module is to gain an understanding about the application of addition and multiplication theorem. If one can understand how to apply these rules based upon the problem then the concept of probability is very easy to understand. Many questions and their answers are discussed in this modules regarding this.<\/p>\n<p>&nbsp;<\/p>\n<p><strong>6.\u00a0\u00a0 Summary<\/strong><\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: justify\">Probability problem cannot be solved only by using definition and approach of probability. To solve probability problem, one needs to discuss about some probability theorem like addition probability theorem and multiplication probability theorem. Conditional probability is the key concept of probability theory. Conditional probability is used to find probability of an event when prior knowledge of that event is given. Conditional probability is applied for those event that are related to each other. To estimate prior probabilities in terms of posterior probabilities by using external new information, Bayes Theorem is used. Bayes Theorem will be discussed in the module \u201cIntroduction to Bayes Theorem\u201d.<\/p>\n<p>&nbsp;<\/p>\n<p><strong>7.\u00a0\u00a0 Suggested Readings<\/strong><\/p>\n<p>&nbsp;<\/p>\n<p>Agresti, A. and B. Finlay, <em>Statistical Methods for the Social Science, 3<\/em><em>rd<\/em> <em>Edition, Prentice Hall<\/em>, 1997.<\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: justify\">Daniel, W. W. and C. L. Cross, C. L., <em>Biostatistics<\/em>: <em>A Foundation for Analysis in the Health Sciences,<\/em> 10th Edition<em>,<\/em> John Wiley &amp; Sons, 2013.<\/p>\n<p>&nbsp;<\/p>\n<p>Hogg, R. V., J. Mckean and A. Craig, <em>Introduction to Mathematical Statistics<\/em>, Macmillan Pub. Co. Inc., 1978.<\/p>\n<p>&nbsp;<\/p>\n<p>Meyer, P. L., <em>Introductory Probability and Statistical Applications<\/em>, Oxford &amp; IBH Pub, 1975.<\/p>\n<\/div>\n<p>Stephens, L. J., <em>Schaum\u2019s Series Outline: Beginning Statistics,<\/em> 2nd Edition<em>,<\/em> McGraw Hill, 2006.<\/p>\n<p>&nbsp;<\/p>\n<p>Triola, M. F., <em>Elementary Statistics<\/em>, 13th\u00a0 Edition, Pearson, 2017.<\/p>\n<p>&nbsp;<\/p>\n<p>Weiss, N. A., <em>Introductory Statistics<\/em>, 10th Edition, Pearson, 2017.<\/p>\n<table>\n<tbody>\n<tr>\n<td><strong>you can view video on Addition and Multiplication Properties of Probability Theory<\/strong><\/td>\n<td><a href=\"https:\/\/youtu.be\/WREm900g1hw\" target=\"_blank\" rel=\"noopener\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-120\" src=\"http:\/\/epgpbooks.inflibnet.ac.in\/wp-content\/uploads\/2018\/11\/download.png\" alt=\"\" width=\"36\" height=\"36\" \/><\/a><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>&nbsp;<\/p>\n<p>One can refer to the following links for further understanding of the statistics terms.<\/p>\n<p>&nbsp;<\/p>\n<p><a href=\"http:\/\/biostat.mc.vanderbilt.edu\/wiki\/pub\/Main\/ClinStat\/glossary.pdf\">http:\/\/biostat.mc.vanderbilt.edu\/wiki\/pub\/Main\/ClinStat\/glossary.pdf<\/a><\/p>\n<p>&nbsp;<\/p>\n<p><a href=\"http:\/\/www.stats.gla.ac.uk\/steps\/glossary\/alphabet.html\">http:\/\/www.stats.gla.ac.uk\/steps\/glossary\/alphabet.html<\/a><\/p>\n<p>&nbsp;<\/p>\n<p><a href=\"http:\/\/www.reading.ac.uk\/ssc\/resources\/Docs\/Statistical_Glossary.pdf\">http:\/\/www.reading.ac.uk\/ssc\/resources\/Docs\/Statistical_Glossary.pdf<\/a><\/p>\n<p>&nbsp;<\/p>\n<p><a href=\"https:\/\/stats.oecd.org\/glossary\/\">https:\/\/stats.oecd.org\/glossary\/<\/a><\/p>\n<p>&nbsp;<\/p>\n<p><a href=\"http:\/\/www.statsoft.com\/Textbook\/Statistics-Glossary\">http:\/\/www.statsoft.com\/Textbook\/Statistics-Glossary<\/a><\/p>\n<p>&nbsp;<\/p>\n<p><a href=\"https:\/\/www.stat.berkeley.edu\/~stark\/SticiGui\/Text\/gloss.htm\">https:\/\/www.stat.berkeley.edu\/~stark\/SticiGui\/Text\/gloss.htm<\/a><\/p>\n<p>&nbsp;<\/p>\n<p><a href=\"https:\/\/stats.oecd.org\/glossary\/alpha.asp?Let=A\">https:\/\/stats.oecd.org\/glossary\/alpha.asp?Let=A<\/a><\/p>\n","protected":false},"author":3,"menu_order":11,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":["dr-harmanpreet-singh-kapoor"],"pb_section_license":""},"chapter-type":[],"contributor":[58],"license":[],"class_list":["post-190","chapter","type-chapter","status-publish","hentry","contributor-dr-harmanpreet-singh-kapoor"],"part":3,"_links":{"self":[{"href":"https:\/\/ebooks.inflibnet.ac.in\/esp14\/wp-json\/pressbooks\/v2\/chapters\/190","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/ebooks.inflibnet.ac.in\/esp14\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/ebooks.inflibnet.ac.in\/esp14\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/ebooks.inflibnet.ac.in\/esp14\/wp-json\/wp\/v2\/users\/3"}],"version-history":[{"count":7,"href":"https:\/\/ebooks.inflibnet.ac.in\/esp14\/wp-json\/pressbooks\/v2\/chapters\/190\/revisions"}],"predecessor-version":[{"id":214,"href":"https:\/\/ebooks.inflibnet.ac.in\/esp14\/wp-json\/pressbooks\/v2\/chapters\/190\/revisions\/214"}],"part":[{"href":"https:\/\/ebooks.inflibnet.ac.in\/esp14\/wp-json\/pressbooks\/v2\/parts\/3"}],"metadata":[{"href":"https:\/\/ebooks.inflibnet.ac.in\/esp14\/wp-json\/pressbooks\/v2\/chapters\/190\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/ebooks.inflibnet.ac.in\/esp14\/wp-json\/wp\/v2\/media?parent=190"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/ebooks.inflibnet.ac.in\/esp14\/wp-json\/pressbooks\/v2\/chapter-type?post=190"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/ebooks.inflibnet.ac.in\/esp14\/wp-json\/wp\/v2\/contributor?post=190"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/ebooks.inflibnet.ac.in\/esp14\/wp-json\/wp\/v2\/license?post=190"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}