{"id":1030,"date":"2023-06-22T01:45:11","date_gmt":"2023-06-22T01:45:11","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/introstatstest\/chapter\/binomial-distribution-background-youll-need-1\/"},"modified":"2023-11-01T16:34:20","modified_gmt":"2023-11-01T16:34:20","slug":"binomial-distribution-background-youll-need-1","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/introstatstest\/chapter\/binomial-distribution-background-youll-need-1\/","title":{"raw":"Supplemental Module - Binomial Distribution: Background You'll Need 2","rendered":"Supplemental Module &#8211; Binomial Distribution: Background You&#8217;ll Need 2"},"content":{"raw":"<section class=\"textbox learningGoals\">\r\n<ul>\r\n\t<li><span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;Compute probabilities that involve exponents&quot;}\" data-sheets-userformat=\"{&quot;2&quot;:4609,&quot;3&quot;:{&quot;1&quot;:0},&quot;12&quot;:0,&quot;15&quot;:&quot;Calibri&quot;}\">Compute probabilities that involve exponents<\/span><\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>How Many 1's?<\/h2>\r\n<section class=\"textbox recall\">When events [latex]A[\/latex] and [latex]B[\/latex] are independent, then [latex]P(A \\text{ and } B) = P(A) \\cdot P(B)[\/latex]<\/section>\r\n<p>When using this rule, there are situations in which the probabilities of some events are the same, and that\u2019s when exponents come into play.<\/p>\r\n<section class=\"textbox recall\">\r\n<h3>Exponent Review<\/h3>\r\n<ul>\r\n\t<li>[latex] p^{x} [\/latex] means that [latex] p [\/latex] is multiplied by itself [latex] x [\/latex] times. For example: [latex] (0.5)^{2} = (0.5)\\cdot (0.5) = 0.25, \\mbox{ and } (0.5)^{3} = (0.5)\\cdot (0.5)\\cdot (0.5) = 0.125 [\/latex].<\/li>\r\n\t<li>Multiplication is commutative, which means that the order we multiply numbers in does not matter when multiplication is the only operation present. For example:\u00a0<span style=\"font-size: 1rem; text-align: initial;\">[latex] (0.25)\\cdot (0.5)\\cdot (0.25) = (0.25)\\cdot (0.25)\\cdot (0.5) = (0.25)^{2}(0.5) [\/latex]<\/span><\/li>\r\n<\/ul>\r\n<\/section>\r\n<section class=\"textbox example\">Suppose we are rolling a fair, [latex]6[\/latex]-sided die [latex]4[\/latex] times.Calculate the probability\u00a0that the first roll is not a [latex]6[\/latex], the second roll is a [latex]6[\/latex], and the final two rolls are also not [latex]6[\/latex].[reveal-answer q=\"387070\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"387070\"]The probability that the first roll is not a [latex]6[\/latex], the second roll is a [latex]6[\/latex], and the final two rolls are also not [latex]6[\/latex] is:\r\n\r\n<p style=\"text-align: center;\"><span style=\"font-size: 1rem; text-align: initial;\">[latex]P([\/latex]not [latex]6[\/latex] AND [latex]6[\/latex] AND not [latex]6[\/latex] AND not [latex]6)[\/latex]<\/span> =\u00a0<span style=\"font-size: 1rem; text-align: initial;\">[latex]P(\\text{not} 6)[\/latex]<\/span><span style=\"font-size: 1rem; text-align: initial;\"> [latex]\\cdot[\/latex] [latex]P(6)[\/latex] [latex]\\cdot[\/latex] [latex]P(\\text{not} 6)[\/latex] [latex]\\cdot[\/latex] [latex]P(\\text{not} 6)[\/latex]<\/span><\/p>\r\n<p style=\"text-align: center;\"><span style=\"font-size: 1rem; text-align: initial;\">=\u00a0 [latex]P(6) \\cdot (P(\\text{not} 6))^3 = (\\frac{1}{6}) \\cdot (\\frac{5}{6})^3 \\approx 0.0965 [\/latex]<\/span><\/p>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>\r\n<section class=\"textbox tryIt\">[ohm2_question hide_question_numbers=1]1476[\/ohm2_question]<\/section>","rendered":"<section class=\"textbox learningGoals\">\n<ul>\n<li><span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;Compute probabilities that involve exponents&quot;}\" data-sheets-userformat=\"{&quot;2&quot;:4609,&quot;3&quot;:{&quot;1&quot;:0},&quot;12&quot;:0,&quot;15&quot;:&quot;Calibri&quot;}\">Compute probabilities that involve exponents<\/span><\/li>\n<\/ul>\n<\/section>\n<h2>How Many 1&#8217;s?<\/h2>\n<section class=\"textbox recall\">When events [latex]A[\/latex] and [latex]B[\/latex] are independent, then [latex]P(A \\text{ and } B) = P(A) \\cdot P(B)[\/latex]<\/section>\n<p>When using this rule, there are situations in which the probabilities of some events are the same, and that\u2019s when exponents come into play.<\/p>\n<section class=\"textbox recall\">\n<h3>Exponent Review<\/h3>\n<ul>\n<li>[latex]p^{x}[\/latex] means that [latex]p[\/latex] is multiplied by itself [latex]x[\/latex] times. For example: [latex](0.5)^{2} = (0.5)\\cdot (0.5) = 0.25, \\mbox{ and } (0.5)^{3} = (0.5)\\cdot (0.5)\\cdot (0.5) = 0.125[\/latex].<\/li>\n<li>Multiplication is commutative, which means that the order we multiply numbers in does not matter when multiplication is the only operation present. For example:\u00a0<span style=\"font-size: 1rem; text-align: initial;\">[latex](0.25)\\cdot (0.5)\\cdot (0.25) = (0.25)\\cdot (0.25)\\cdot (0.5) = (0.25)^{2}(0.5)[\/latex]<\/span><\/li>\n<\/ul>\n<\/section>\n<section class=\"textbox example\">Suppose we are rolling a fair, [latex]6[\/latex]-sided die [latex]4[\/latex] times.Calculate the probability\u00a0that the first roll is not a [latex]6[\/latex], the second roll is a [latex]6[\/latex], and the final two rolls are also not [latex]6[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q387070\">Show Solution<\/button><\/p>\n<div id=\"q387070\" class=\"hidden-answer\" style=\"display: none\">The probability that the first roll is not a [latex]6[\/latex], the second roll is a [latex]6[\/latex], and the final two rolls are also not [latex]6[\/latex] is:<\/p>\n<p style=\"text-align: center;\"><span style=\"font-size: 1rem; text-align: initial;\">[latex]P([\/latex]not [latex]6[\/latex] AND [latex]6[\/latex] AND not [latex]6[\/latex] AND not [latex]6)[\/latex]<\/span> =\u00a0<span style=\"font-size: 1rem; text-align: initial;\">[latex]P(\\text{not} 6)[\/latex]<\/span><span style=\"font-size: 1rem; text-align: initial;\"> [latex]\\cdot[\/latex] [latex]P(6)[\/latex] [latex]\\cdot[\/latex] [latex]P(\\text{not} 6)[\/latex] [latex]\\cdot[\/latex] [latex]P(\\text{not} 6)[\/latex]<\/span><\/p>\n<p style=\"text-align: center;\"><span style=\"font-size: 1rem; text-align: initial;\">=\u00a0 [latex]P(6) \\cdot (P(\\text{not} 6))^3 = (\\frac{1}{6}) \\cdot (\\frac{5}{6})^3 \\approx 0.0965[\/latex]<\/span><\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm1476\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=1476&theme=lumen&iframe_resize_id=ohm1476&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n","protected":false},"author":8,"menu_order":3,"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":"background_you_need","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\/1030"}],"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\/1030\/revisions"}],"predecessor-version":[{"id":4071,"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/pressbooks\/v2\/chapters\/1030\/revisions\/4071"}],"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\/1030\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/wp\/v2\/media?parent=1030"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/pressbooks\/v2\/chapter-type?post=1030"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/wp\/v2\/contributor?post=1030"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/wp\/v2\/license?post=1030"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}