{"id":1034,"date":"2023-06-22T01:45:13","date_gmt":"2023-06-22T01:45:13","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/introstatstest\/chapter\/binomial-distribution-learn-it-3\/"},"modified":"2023-10-23T13:05:42","modified_gmt":"2023-10-23T13:05:42","slug":"binomial-distribution-learn-it-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/introstatstest\/chapter\/binomial-distribution-learn-it-3\/","title":{"raw":"Binomial Distribution: Learn It 3","rendered":"Binomial Distribution: Learn It 3"},"content":{"raw":"<section class=\"textbox learningGoals\">\r\n<ul>\r\n\t<li>Use a binomial distribution to calculate probability<\/li>\r\n\t<li>Determine if a probability model meets the conditions for a binomial distribution<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h3>Binomial Distribution<\/h3>\r\n<section class=\"textbox recall\">If the event [latex]A \\text{ and } B[\/latex] are mutually exclusive (meaning they cannot happen at the same time), then [latex] P(A~or~B) =P(A)+P(B). [\/latex]<\/section>\r\n<section class=\"textbox example\">\r\n<p style=\"text-align: left;\">Let's revisit the experiment flipping the coin [latex]3[\/latex] times and counting the number of tails obtained.<\/p>\r\n<p style=\"text-align: left;\">Find\u00a0[latex] P(X = 1). [\/latex]<\/p>\r\n<div align=\"center\">\r\n<p style=\"text-align: left;\">[reveal-answer q=\"326395\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"326395\"]<\/p>\r\n<p style=\"text-align: left;\">Since the outcomes THH, HTH, and HHT are mutually exclusive, the probability of obtaining [latex]1[\/latex] tail in [latex]3[\/latex] coin flips, i.e., [latex] P(X = 1) [\/latex] , is:<\/p>\r\n<p style=\"text-align: center;\">[latex] P(X = 1) = P(THH) + P(HTH) + P(HHT) = 0.125 + 0.125 + 0.125 = 0.375 [\/latex]<\/p>\r\n<p style=\"text-align: left;\">Since all [latex]3[\/latex] of the outcomes yielding [latex]1[\/latex] tail have the same probability, then<\/p>\r\n<p style=\"text-align: center;\">[latex] P(THH) = P(HTH) = P(HHT) = P(T)P(H)P(H)=P(T)(P(H))^{2} [\/latex]<\/p>\r\n<p style=\"text-align: left;\">and we can compute the probability [latex] P(X = 1) [\/latex] as follows:<\/p>\r\n<p style=\"text-align: center;\">[latex] P(X=1) = 3 P(T)P(H)P(H) = 3P(T)(P(H))^{2} = 3(0.5)(0.5)^{2} = 0.375 [\/latex]<\/p>\r\n<p style=\"text-align: left;\">where, as you observed in the table below, [latex]3[\/latex] is the number of ways of obtaining [latex]1[\/latex] tail in [latex]3[\/latex] coin flips.<\/p>\r\n<p style=\"text-align: left;\">Recall that the outcomes of the experiment are as given in the following table:<\/p>\r\n<div align=\"center\">\r\n<table style=\"width: 338px;\">\r\n<tbody>\r\n<tr>\r\n<td style=\"text-align: center; width: 99.5781px;\">Experimental Outcome<\/td>\r\n<td style=\"text-align: center; width: 260.203px;\">\r\n<p>[latex]X[\/latex]<\/p>\r\n<p>Number of Tails in 3 Flips of a Coin<\/p>\r\n<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 99.5781px; text-align: center;\">HHH<\/td>\r\n<td style=\"width: 260.203px; text-align: center;\">[latex]0[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 99.5781px; text-align: center;\">HHT<\/td>\r\n<td style=\"width: 260.203px; text-align: center;\">[latex]1[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 99.5781px; text-align: center;\">HTH<\/td>\r\n<td style=\"width: 260.203px; text-align: center;\">[latex]1[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 99.5781px; text-align: center;\">THH<\/td>\r\n<td style=\"width: 260.203px; text-align: center;\">[latex]1[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 99.5781px; text-align: center;\">TTH<\/td>\r\n<td style=\"width: 260.203px; text-align: center;\">[latex]2[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 99.5781px; text-align: center;\">THT<\/td>\r\n<td style=\"width: 260.203px; text-align: center;\">[latex]2[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 99.5781px; text-align: center;\">HTT<\/td>\r\n<td style=\"width: 260.203px; text-align: center;\">[latex]2[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 99.5781px; text-align: center;\">TTT<\/td>\r\n<td style=\"width: 260.203px; text-align: center;\">[latex]3[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/div>\r\n<\/section>\r\n<p>The example above gives us a glimpse into the formula for the <strong>binomial distribution<\/strong>, which is used to model a binomial experiment.<\/p>\r\n<section class=\"textbox keyTakeaway\">\r\n<h3>binomial distribution formula<\/h3>\r\n<p>In general, the formula for the probability of obtaining [latex]x[\/latex] successes from [latex]n[\/latex] independent trials where the probability of success is [latex]p[\/latex] is:<\/p>\r\n<p style=\"text-align: center;\">[latex] P(X = x) = (\\text{number of ways to obtain }x\\text{ successes in }n\\text{ trials}) \\cdot p^{x} \\cdot (1-p)^{n-x} [\/latex]<\/p>\r\n<p>where [latex] p^{x} [\/latex] occurs because there are [latex]x[\/latex] successes, and [latex] (1-p)^{n-x} [\/latex] occurs because if there are [latex]x[\/latex] successes and [latex] n [\/latex] trials total, there must be [latex] n-x [\/latex] failures.<\/p>\r\n<\/section>\r\n<section class=\"textbox tryIt\">[ohm2_question hide_question_numbers=1]1483[\/ohm2_question]<\/section>","rendered":"<section class=\"textbox learningGoals\">\n<ul>\n<li>Use a binomial distribution to calculate probability<\/li>\n<li>Determine if a probability model meets the conditions for a binomial distribution<\/li>\n<\/ul>\n<\/section>\n<h3>Binomial Distribution<\/h3>\n<section class=\"textbox recall\">If the event [latex]A \\text{ and } B[\/latex] are mutually exclusive (meaning they cannot happen at the same time), then [latex]P(A~or~B) =P(A)+P(B).[\/latex]<\/section>\n<section class=\"textbox example\">\n<p style=\"text-align: left;\">Let&#8217;s revisit the experiment flipping the coin [latex]3[\/latex] times and counting the number of tails obtained.<\/p>\n<p style=\"text-align: left;\">Find\u00a0[latex]P(X = 1).[\/latex]<\/p>\n<div style=\"margin: auto;\">\n<p style=\"text-align: left;\">\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q326395\">Show Solution<\/button><\/p>\n<div id=\"q326395\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: left;\">Since the outcomes THH, HTH, and HHT are mutually exclusive, the probability of obtaining [latex]1[\/latex] tail in [latex]3[\/latex] coin flips, i.e., [latex]P(X = 1)[\/latex] , is:<\/p>\n<p style=\"text-align: center;\">[latex]P(X = 1) = P(THH) + P(HTH) + P(HHT) = 0.125 + 0.125 + 0.125 = 0.375[\/latex]<\/p>\n<p style=\"text-align: left;\">Since all [latex]3[\/latex] of the outcomes yielding [latex]1[\/latex] tail have the same probability, then<\/p>\n<p style=\"text-align: center;\">[latex]P(THH) = P(HTH) = P(HHT) = P(T)P(H)P(H)=P(T)(P(H))^{2}[\/latex]<\/p>\n<p style=\"text-align: left;\">and we can compute the probability [latex]P(X = 1)[\/latex] as follows:<\/p>\n<p style=\"text-align: center;\">[latex]P(X=1) = 3 P(T)P(H)P(H) = 3P(T)(P(H))^{2} = 3(0.5)(0.5)^{2} = 0.375[\/latex]<\/p>\n<p style=\"text-align: left;\">where, as you observed in the table below, [latex]3[\/latex] is the number of ways of obtaining [latex]1[\/latex] tail in [latex]3[\/latex] coin flips.<\/p>\n<p style=\"text-align: left;\">Recall that the outcomes of the experiment are as given in the following table:<\/p>\n<div style=\"margin: auto;\">\n<table style=\"width: 338px;\">\n<tbody>\n<tr>\n<td style=\"text-align: center; width: 99.5781px;\">Experimental Outcome<\/td>\n<td style=\"text-align: center; width: 260.203px;\">\n[latex]X[\/latex]<\/p>\n<p>Number of Tails in 3 Flips of a Coin<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 99.5781px; text-align: center;\">HHH<\/td>\n<td style=\"width: 260.203px; text-align: center;\">[latex]0[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 99.5781px; text-align: center;\">HHT<\/td>\n<td style=\"width: 260.203px; text-align: center;\">[latex]1[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 99.5781px; text-align: center;\">HTH<\/td>\n<td style=\"width: 260.203px; text-align: center;\">[latex]1[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 99.5781px; text-align: center;\">THH<\/td>\n<td style=\"width: 260.203px; text-align: center;\">[latex]1[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 99.5781px; text-align: center;\">TTH<\/td>\n<td style=\"width: 260.203px; text-align: center;\">[latex]2[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 99.5781px; text-align: center;\">THT<\/td>\n<td style=\"width: 260.203px; text-align: center;\">[latex]2[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 99.5781px; text-align: center;\">HTT<\/td>\n<td style=\"width: 260.203px; text-align: center;\">[latex]2[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 99.5781px; text-align: center;\">TTT<\/td>\n<td style=\"width: 260.203px; text-align: center;\">[latex]3[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<p>The example above gives us a glimpse into the formula for the <strong>binomial distribution<\/strong>, which is used to model a binomial experiment.<\/p>\n<section class=\"textbox keyTakeaway\">\n<h3>binomial distribution formula<\/h3>\n<p>In general, the formula for the probability of obtaining [latex]x[\/latex] successes from [latex]n[\/latex] independent trials where the probability of success is [latex]p[\/latex] is:<\/p>\n<p style=\"text-align: center;\">[latex]P(X = x) = (\\text{number of ways to obtain }x\\text{ successes in }n\\text{ trials}) \\cdot p^{x} \\cdot (1-p)^{n-x}[\/latex]<\/p>\n<p>where [latex]p^{x}[\/latex] occurs because there are [latex]x[\/latex] successes, and [latex](1-p)^{n-x}[\/latex] occurs because if there are [latex]x[\/latex] successes and [latex]n[\/latex] trials total, there must be [latex]n-x[\/latex] failures.<\/p>\n<\/section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm1483\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=1483&theme=lumen&iframe_resize_id=ohm1483&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n","protected":false},"author":8,"menu_order":13,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":2912,"module-header":"learn_it","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/pressbooks\/v2\/chapters\/1034"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/wp\/v2\/users\/8"}],"version-history":[{"count":5,"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/pressbooks\/v2\/chapters\/1034\/revisions"}],"predecessor-version":[{"id":4081,"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/pressbooks\/v2\/chapters\/1034\/revisions\/4081"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/pressbooks\/v2\/parts\/2912"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/pressbooks\/v2\/chapters\/1034\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/wp\/v2\/media?parent=1034"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/pressbooks\/v2\/chapter-type?post=1034"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/wp\/v2\/contributor?post=1034"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/wp\/v2\/license?post=1034"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}