{"id":360,"date":"2020-07-08T22:02:28","date_gmt":"2020-07-09T02:02:28","guid":{"rendered":"https:\/\/openbooks.macewan.ca\/rcommander\/?post_type=chapter&#038;p=360"},"modified":"2025-04-25T18:44:59","modified_gmt":"2025-04-25T22:44:59","slug":"3-4-probability-rules","status":"publish","type":"chapter","link":"https:\/\/openbooks.macewan.ca\/introstats\/chapter\/3-4-probability-rules\/","title":{"raw":"3.4 Probability Rules","rendered":"3.4 Probability Rules"},"content":{"raw":"The following are some basic rules of probability which can be easily verified with a Venn diagram:\r\n<ul>\r\n \t<li><strong>Complement Rule<\/strong>: [latex] P(\\text{not }E) = 1 - P(E)[\/latex] or [latex] P(E) = 1 - P(\\text{not }E)[\/latex].<\/li>\r\n \t<li><strong>General Addition Rule<\/strong>: [latex] P(A \\text{ or } B) = P(A) + P(B) - P(A \\: \\&amp; \\: B)[\/latex].<\/li>\r\n \t<li><strong>Special Addition Rule<\/strong>: when two events <strong>A and B are mutually exclusive<\/strong>,\u00a0<span style=\"text-align: initial; font-size: 1em;\">[latex] P(A \\text{ or } B) = P(A) + P(B)[\/latex].<\/span><\/li>\r\n<\/ul>\r\n<p style=\"padding-left: 40px;\">This is due to the fact that [latex] P(A \\: \\&amp; \\: B) = 0[\/latex], because it is impossible to observe both A and B.<\/p>\r\n<p style=\"padding-left: 40px;\">More generally, if events [latex] A, B, C, \\cdots[\/latex] are mutually exclusive, then [latex] P(A \\text{ or } B \\text{ or } C \\text{ or } \\cdots) = P(A) + P(B) + P(C) + \\cdots.[\/latex]<\/p>\r\n\r\n<div class=\"textbox textbox--examples\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Examples: Probability Rules<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nSuppose we roll a fair die, so that the sample space is S = {1, 2, 3, 4, 5, 6}. Consider the following events:\r\n<ul>\r\n \t<li>Observing a six, E = {6}.<\/li>\r\n \t<li>Observing an even number, A = {2, 4, 6}.<\/li>\r\n \t<li>Observing an outcome that is less than 3, B = {1, 2}.<\/li>\r\n \t<li>Observing an odd number, C = {1, 3, 5}.<\/li>\r\n<\/ul>\r\nSince the die is fair, each outcome is equally likely. Therefore we can use the f\/N rule to find the probabilities.\r\n<p align=\"center\">[latex] P(E) = \\frac{f}{N} = \\frac{1}{6}; \\quad P(A) = \\frac{f}{N} = \\frac{3}{6}, [\/latex]<\/p>\r\n<p align=\"center\">[latex] P(B) = \\frac{f}{N} = \\frac{2}{6}; \\quad P(C) = \\frac{f}{N} = \\frac{3}{6}.[\/latex]<\/p>\r\nFind the probabilities of the following events:\r\n<ol>\r\n \t<li>(not E). Using the complement rule, [latex] P(\\text{not }E) = 1 - P(E) = 1 - \\frac{1}{6} = \\frac{5}{6}[\/latex].<\/li>\r\n \t<li>(A &amp; C). Events A and C are mutually exclusive, since it is impossible to observe a number that is both even and odd. Therefore, [latex]A \\mbox{ \\&amp; } C=\\varnothing[\/latex] and [latex] P(A \\: \\&amp; \\: C) = 0[\/latex].<\/li>\r\n \t<li>(A or B). Since events A and B are NOT mutually exclusive, we use the general addition rule. The overlap of events A and B is {2}, that is, A &amp; B = {2} and therefore, [latex] P(A \\: \\&amp; \\: B) = \\frac{f}{N} = \\frac{1}{6}[\/latex].\r\nBy the general addition rule,\r\n[latex]P(A \\text{ or }B) = P(A) + P(B) - P(A \\: \\&amp; \\: B) = \\frac{3}{6} + \\frac{2}{6} - \\frac{1}{6} = \u00a0\\frac{4}{6}[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<div style=\"height: 55px; margin-top: 2.1428571429em;\">\r\n\r\n<img class=\"size-full wp-image-99 alignleft\" src=\"https:\/\/openbooks.macewan.ca\/rcommander\/wp-content\/uploads\/sites\/8\/2020\/06\/instructornote.png\" alt=\"\" width=\"250\" height=\"50\" \/>\r\n\r\n<\/div>\r\nSince we are rolling a fair die, we can also use the [latex] \\frac{f}{N}[\/latex]\u00a0rule to find the probabilities in the previous example.\r\n<ul>\r\n \t<li style=\"list-style-type: none;\">\r\n<ul>\r\n \t<li style=\"list-style-type: none;\">\r\n<ol>\r\n \t<li>(not E) = not observing a six={1, 2, 3, 4, 5}. [latex] P(\\text{not }E)= \\frac{f}{N} = \\frac{5}{6}[\/latex].<\/li>\r\n \t<li>(A or B) = observing an even number or a number not more than 3 = {1, 2, 4, 6} [latex] P(A \\text{ or }B)= \\frac{f}{N} = \\frac{4}{6}[\/latex].<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ul>\r\nThe results are identical to those using the probability rules.","rendered":"<p>The following are some basic rules of probability which can be easily verified with a Venn diagram:<\/p>\n<ul>\n<li><strong>Complement Rule<\/strong>: [latex]P(\\text{not }E) = 1 - P(E)[\/latex] or [latex]P(E) = 1 - P(\\text{not }E)[\/latex].<\/li>\n<li><strong>General Addition Rule<\/strong>: [latex]P(A \\text{ or } B) = P(A) + P(B) - P(A \\: \\& \\: B)[\/latex].<\/li>\n<li><strong>Special Addition Rule<\/strong>: when two events <strong>A and B are mutually exclusive<\/strong>,\u00a0<span style=\"text-align: initial; font-size: 1em;\">[latex]P(A \\text{ or } B) = P(A) + P(B)[\/latex].<\/span><\/li>\n<\/ul>\n<p style=\"padding-left: 40px;\">This is due to the fact that [latex]P(A \\: \\& \\: B) = 0[\/latex], because it is impossible to observe both A and B.<\/p>\n<p style=\"padding-left: 40px;\">More generally, if events [latex]A, B, C, \\cdots[\/latex] are mutually exclusive, then [latex]P(A \\text{ or } B \\text{ or } C \\text{ or } \\cdots) = P(A) + P(B) + P(C) + \\cdots.[\/latex]<\/p>\n<div class=\"textbox textbox--examples\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Examples: Probability Rules<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>Suppose we roll a fair die, so that the sample space is S = {1, 2, 3, 4, 5, 6}. Consider the following events:<\/p>\n<ul>\n<li>Observing a six, E = {6}.<\/li>\n<li>Observing an even number, A = {2, 4, 6}.<\/li>\n<li>Observing an outcome that is less than 3, B = {1, 2}.<\/li>\n<li>Observing an odd number, C = {1, 3, 5}.<\/li>\n<\/ul>\n<p>Since the die is fair, each outcome is equally likely. Therefore we can use the f\/N rule to find the probabilities.<\/p>\n<p style=\"text-align: center;\">[latex]P(E) = \\frac{f}{N} = \\frac{1}{6}; \\quad P(A) = \\frac{f}{N} = \\frac{3}{6},[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]P(B) = \\frac{f}{N} = \\frac{2}{6}; \\quad P(C) = \\frac{f}{N} = \\frac{3}{6}.[\/latex]<\/p>\n<p>Find the probabilities of the following events:<\/p>\n<ol>\n<li>(not E). Using the complement rule, [latex]P(\\text{not }E) = 1 - P(E) = 1 - \\frac{1}{6} = \\frac{5}{6}[\/latex].<\/li>\n<li>(A &amp; C). Events A and C are mutually exclusive, since it is impossible to observe a number that is both even and odd. Therefore, [latex]A \\mbox{ \\& } C=\\varnothing[\/latex] and [latex]P(A \\: \\& \\: C) = 0[\/latex].<\/li>\n<li>(A or B). Since events A and B are NOT mutually exclusive, we use the general addition rule. The overlap of events A and B is {2}, that is, A &amp; B = {2} and therefore, [latex]P(A \\: \\& \\: B) = \\frac{f}{N} = \\frac{1}{6}[\/latex].<br \/>\nBy the general addition rule,<br \/>\n[latex]P(A \\text{ or }B) = P(A) + P(B) - P(A \\: \\& \\: B) = \\frac{3}{6} + \\frac{2}{6} - \\frac{1}{6} = \u00a0\\frac{4}{6}[\/latex].<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<div style=\"height: 55px; margin-top: 2.1428571429em;\">\n<p><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-99 alignleft\" src=\"https:\/\/openbooks.macewan.ca\/rcommander\/wp-content\/uploads\/sites\/8\/2020\/06\/instructornote.png\" alt=\"\" width=\"250\" height=\"50\" \/><\/p>\n<\/div>\n<p>Since we are rolling a fair die, we can also use the [latex]\\frac{f}{N}[\/latex]\u00a0rule to find the probabilities in the previous example.<\/p>\n<ul>\n<li style=\"list-style-type: none;\">\n<ul>\n<li style=\"list-style-type: none;\">\n<ol>\n<li>(not E) = not observing a six={1, 2, 3, 4, 5}. [latex]P(\\text{not }E)= \\frac{f}{N} = \\frac{5}{6}[\/latex].<\/li>\n<li>(A or B) = observing an even number or a number not more than 3 = {1, 2, 4, 6} [latex]P(A \\text{ or }B)= \\frac{f}{N} = \\frac{4}{6}[\/latex].<\/li>\n<\/ol>\n<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<p>The results are identical to those using the probability rules.<\/p>\n","protected":false},"author":19,"menu_order":4,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-360","chapter","type-chapter","status-publish","hentry"],"part":327,"_links":{"self":[{"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/pressbooks\/v2\/chapters\/360","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/wp\/v2\/users\/19"}],"version-history":[{"count":59,"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/pressbooks\/v2\/chapters\/360\/revisions"}],"predecessor-version":[{"id":5450,"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/pressbooks\/v2\/chapters\/360\/revisions\/5450"}],"part":[{"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/pressbooks\/v2\/parts\/327"}],"metadata":[{"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/pressbooks\/v2\/chapters\/360\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/wp\/v2\/media?parent=360"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/pressbooks\/v2\/chapter-type?post=360"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/wp\/v2\/contributor?post=360"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/wp\/v2\/license?post=360"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}