{"id":1755,"date":"2025-07-28T19:23:19","date_gmt":"2025-07-28T19:23:19","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&#038;p=1755"},"modified":"2026-03-26T21:12:29","modified_gmt":"2026-03-26T21:12:29","slug":"probability-learn-it-6","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/probability-learn-it-6\/","title":{"raw":"Probability: Learn It 6","rendered":"Probability: Learn It 6"},"content":{"raw":"<h2>Computing Probability Using Counting Theory<\/h2>\r\nMany interesting probability problems involve counting principles, permutations, and combinations. In these problems, we will use permutations and combinations to find the number of elements in events and sample spaces. These problems can be complicated, but they can be made easier by breaking them down into smaller counting problems.\r\n\r\n<section class=\"textbox example\" aria-label=\"Example\">A store has [latex]8[\/latex] cellular phones and that [latex]3[\/latex] of those are defective. Find the probability that a couple purchasing [latex]2[\/latex] phones receives [latex]2[\/latex] phones that are not defective.\r\n\r\n\r\n[reveal-answer q=\"782213\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"782213\"]\r\nTo solve this problem, we need to calculate all of the ways to select [latex]2[\/latex] phones that are not defective as well as all of the ways to select [latex]2[\/latex] phones.\r\n<ul>\r\n \t<li>There are [latex]5[\/latex] phones that are not defective, so there are [latex]C\\left(5,2\\right)[\/latex] ways to select [latex]2[\/latex] phones that are not defective.<\/li>\r\n \t<li>There are [latex]8[\/latex] phones, so there are [latex]C\\left(8,2\\right)[\/latex] ways to select [latex]2[\/latex] phones.<\/li>\r\n<\/ul>\r\nThe probability of selecting [latex]2[\/latex] phones that are not defective is:\r\n<p style=\"text-align: center;\">[latex]\\begin{align}\\dfrac{\\text{Number of ways to select 2 phones that are not defective}}{\\text{Number of ways to select 2 phones}}&amp;=\\dfrac{C\\left(5,2\\right)}{C\\left(8,2\\right)} \\\\[1mm] &amp;=\\dfrac{10}{28} \\\\[1mm] &amp;=\\dfrac{5}{14} \\end{align}[\/latex]<\/p>\r\n\r\n\r\n[\/hidden-answer]<\/section><section class=\"textbox example\" aria-label=\"Example\">A child randomly selects [latex]5[\/latex] toys from a bin containing [latex]3[\/latex] bunnies, [latex]5[\/latex] dogs, and [latex]6[\/latex] bears.\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>Find the probability that only bears are chosen.<\/li>\r\n \t<li>Find the probability that [latex]2[\/latex] bears and [latex]3[\/latex] dogs are chosen.<\/li>\r\n \t<li>Find the probability that at least [latex]2[\/latex] dogs are chosen.<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"14948\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"14948\"]\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>We need to count the number of ways to choose only bears and the total number of possible ways to select [latex]5[\/latex] toys. There are [latex]6[\/latex] bears, so there are [latex]C\\left(6,5\\right)[\/latex] ways to choose [latex]5[\/latex] bears. There are [latex]14[\/latex] toys, so there are [latex]C\\left(14,5\\right)[\/latex] ways to choose any [latex]5[\/latex] toys.\r\n<div style=\"text-align: center;\">[latex]\\dfrac{C\\left(6\\text{,}5\\right)}{C\\left(14\\text{,}5\\right)}=\\dfrac{6}{2\\text{,}002}=\\dfrac{3}{1\\text{,}001}[\/latex]<\/div><\/li>\r\n \t<li>We need to count the number of ways to choose [latex]2[\/latex] bears and [latex]3[\/latex] dogs and the total number of possible ways to select [latex]5[\/latex] toys. There are [latex]6[\/latex] bears, so there are [latex]C\\left(6,2\\right)[\/latex] ways to choose 2 bears. There are 5 dogs, so there are [latex]C\\left(5,3\\right)[\/latex] ways to choose 3 dogs. Since we are choosing both bears and dogs at the same time, we will use the Multiplication Principle. There are [latex]C\\left(6,2\\right)\\cdot C\\left(5,3\\right)[\/latex] ways to choose 2 bears and 3 dogs. We can use this result to find the probability.\r\n<div style=\"text-align: center;\">[latex]\\dfrac{C\\left(6\\text{,}2\\right)C\\left(5\\text{,}3\\right)}{C\\left(14\\text{,}5\\right)}=\\dfrac{15\\cdot 10}{2\\text{,}002}=\\dfrac{75}{1\\text{,}001}[\/latex]<\/div><\/li>\r\n \t<li>It is often easiest to solve \"at least\" problems using the Complement Rule. We will begin by finding the probability that fewer than [latex]2[\/latex] dogs are chosen. If less than [latex]2[\/latex] dogs are chosen, then either no dogs could be chosen, or [latex]1[\/latex] dog could be chosen. When no dogs are chosen, all [latex]5[\/latex] toys come from the [latex]9[\/latex] toys that are not dogs. There are [latex]C\\left(9,5\\right)[\/latex] ways to choose toys from the [latex]9[\/latex] toys that are not dogs. Since there are [latex]14[\/latex] toys, there are [latex]C\\left(14,5\\right)[\/latex] ways to choose the [latex]5[\/latex] toys from all of the toys.\r\n<div style=\"text-align: center;\">[latex]\\dfrac{C\\left(9\\text{,}5\\right)}{C\\left(14\\text{,}5\\right)}=\\dfrac{63}{1\\text{,}001}[\/latex]<\/div>\r\nIf there is [latex]1[\/latex] dog chosen, then [latex]4[\/latex] toys must come from the [latex]9[\/latex] toys that are not dogs, and [latex]1[\/latex] must come from the [latex]5[\/latex] dogs. Since we are choosing both dogs and other toys at the same time, we will use the Multiplication Principle. There are [latex]C\\left(5,1\\right)\\cdot C\\left(9,4\\right)[\/latex] ways to choose [latex]1[\/latex] dog and [latex]1[\/latex] other toy.\r\n<div style=\"text-align: center;\">[latex]\\dfrac{C\\left(5\\text{,}1\\right)C\\left(9\\text{,}4\\right)}{C\\left(14\\text{,}5\\right)}=\\dfrac{5\\cdot 126}{2\\text{,}002}=\\dfrac{315}{1\\text{,}001}[\/latex]<\/div>\r\nBecause these events would not occur together and are therefore mutually exclusive, we add the probabilities to find the probability that fewer than [latex]2[\/latex] dogs are chosen.\r\n<div style=\"text-align: center;\">[latex]\\dfrac{63}{1\\text{,}001}+\\dfrac{315}{1\\text{,}001}=\\dfrac{378}{1\\text{,}001}[\/latex]<\/div>\r\nWe then subtract that probability from [latex]1[\/latex] to find the probability that at least [latex]2[\/latex] dogs are chosen.\r\n<div style=\"text-align: center;\">[latex]1-\\dfrac{378}{1\\text{,}001}=\\dfrac{623}{1\\text{,}001}[\/latex]<\/div><\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm_question hide_question_numbers=1]322253[\/ohm_question]<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm_question hide_question_numbers=1]322254[\/ohm_question]<\/section>","rendered":"<h2>Computing Probability Using Counting Theory<\/h2>\n<p>Many interesting probability problems involve counting principles, permutations, and combinations. In these problems, we will use permutations and combinations to find the number of elements in events and sample spaces. These problems can be complicated, but they can be made easier by breaking them down into smaller counting problems.<\/p>\n<section class=\"textbox example\" aria-label=\"Example\">A store has [latex]8[\/latex] cellular phones and that [latex]3[\/latex] of those are defective. Find the probability that a couple purchasing [latex]2[\/latex] phones receives [latex]2[\/latex] phones that are not defective.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q782213\">Show Solution<\/button><\/p>\n<div id=\"q782213\" class=\"hidden-answer\" style=\"display: none\">\nTo solve this problem, we need to calculate all of the ways to select [latex]2[\/latex] phones that are not defective as well as all of the ways to select [latex]2[\/latex] phones.<\/p>\n<ul>\n<li>There are [latex]5[\/latex] phones that are not defective, so there are [latex]C\\left(5,2\\right)[\/latex] ways to select [latex]2[\/latex] phones that are not defective.<\/li>\n<li>There are [latex]8[\/latex] phones, so there are [latex]C\\left(8,2\\right)[\/latex] ways to select [latex]2[\/latex] phones.<\/li>\n<\/ul>\n<p>The probability of selecting [latex]2[\/latex] phones that are not defective is:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}\\dfrac{\\text{Number of ways to select 2 phones that are not defective}}{\\text{Number of ways to select 2 phones}}&=\\dfrac{C\\left(5,2\\right)}{C\\left(8,2\\right)} \\\\[1mm] &=\\dfrac{10}{28} \\\\[1mm] &=\\dfrac{5}{14} \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">A child randomly selects [latex]5[\/latex] toys from a bin containing [latex]3[\/latex] bunnies, [latex]5[\/latex] dogs, and [latex]6[\/latex] bears.<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>Find the probability that only bears are chosen.<\/li>\n<li>Find the probability that [latex]2[\/latex] bears and [latex]3[\/latex] dogs are chosen.<\/li>\n<li>Find the probability that at least [latex]2[\/latex] dogs are chosen.<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q14948\">Show Solution<\/button><\/p>\n<div id=\"q14948\" class=\"hidden-answer\" style=\"display: none\">\n<ol style=\"list-style-type: lower-alpha;\">\n<li>We need to count the number of ways to choose only bears and the total number of possible ways to select [latex]5[\/latex] toys. There are [latex]6[\/latex] bears, so there are [latex]C\\left(6,5\\right)[\/latex] ways to choose [latex]5[\/latex] bears. There are [latex]14[\/latex] toys, so there are [latex]C\\left(14,5\\right)[\/latex] ways to choose any [latex]5[\/latex] toys.\n<div style=\"text-align: center;\">[latex]\\dfrac{C\\left(6\\text{,}5\\right)}{C\\left(14\\text{,}5\\right)}=\\dfrac{6}{2\\text{,}002}=\\dfrac{3}{1\\text{,}001}[\/latex]<\/div>\n<\/li>\n<li>We need to count the number of ways to choose [latex]2[\/latex] bears and [latex]3[\/latex] dogs and the total number of possible ways to select [latex]5[\/latex] toys. There are [latex]6[\/latex] bears, so there are [latex]C\\left(6,2\\right)[\/latex] ways to choose 2 bears. There are 5 dogs, so there are [latex]C\\left(5,3\\right)[\/latex] ways to choose 3 dogs. Since we are choosing both bears and dogs at the same time, we will use the Multiplication Principle. There are [latex]C\\left(6,2\\right)\\cdot C\\left(5,3\\right)[\/latex] ways to choose 2 bears and 3 dogs. We can use this result to find the probability.\n<div style=\"text-align: center;\">[latex]\\dfrac{C\\left(6\\text{,}2\\right)C\\left(5\\text{,}3\\right)}{C\\left(14\\text{,}5\\right)}=\\dfrac{15\\cdot 10}{2\\text{,}002}=\\dfrac{75}{1\\text{,}001}[\/latex]<\/div>\n<\/li>\n<li>It is often easiest to solve &#8220;at least&#8221; problems using the Complement Rule. We will begin by finding the probability that fewer than [latex]2[\/latex] dogs are chosen. If less than [latex]2[\/latex] dogs are chosen, then either no dogs could be chosen, or [latex]1[\/latex] dog could be chosen. When no dogs are chosen, all [latex]5[\/latex] toys come from the [latex]9[\/latex] toys that are not dogs. There are [latex]C\\left(9,5\\right)[\/latex] ways to choose toys from the [latex]9[\/latex] toys that are not dogs. Since there are [latex]14[\/latex] toys, there are [latex]C\\left(14,5\\right)[\/latex] ways to choose the [latex]5[\/latex] toys from all of the toys.\n<div style=\"text-align: center;\">[latex]\\dfrac{C\\left(9\\text{,}5\\right)}{C\\left(14\\text{,}5\\right)}=\\dfrac{63}{1\\text{,}001}[\/latex]<\/div>\n<p>If there is [latex]1[\/latex] dog chosen, then [latex]4[\/latex] toys must come from the [latex]9[\/latex] toys that are not dogs, and [latex]1[\/latex] must come from the [latex]5[\/latex] dogs. Since we are choosing both dogs and other toys at the same time, we will use the Multiplication Principle. There are [latex]C\\left(5,1\\right)\\cdot C\\left(9,4\\right)[\/latex] ways to choose [latex]1[\/latex] dog and [latex]1[\/latex] other toy.<\/p>\n<div style=\"text-align: center;\">[latex]\\dfrac{C\\left(5\\text{,}1\\right)C\\left(9\\text{,}4\\right)}{C\\left(14\\text{,}5\\right)}=\\dfrac{5\\cdot 126}{2\\text{,}002}=\\dfrac{315}{1\\text{,}001}[\/latex]<\/div>\n<p>Because these events would not occur together and are therefore mutually exclusive, we add the probabilities to find the probability that fewer than [latex]2[\/latex] dogs are chosen.<\/p>\n<div style=\"text-align: center;\">[latex]\\dfrac{63}{1\\text{,}001}+\\dfrac{315}{1\\text{,}001}=\\dfrac{378}{1\\text{,}001}[\/latex]<\/div>\n<p>We then subtract that probability from [latex]1[\/latex] to find the probability that at least [latex]2[\/latex] dogs are chosen.<\/p>\n<div style=\"text-align: center;\">[latex]1-\\dfrac{378}{1\\text{,}001}=\\dfrac{623}{1\\text{,}001}[\/latex]<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm322253\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=322253&theme=lumen&iframe_resize_id=ohm322253&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm322254\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=322254&theme=lumen&iframe_resize_id=ohm322254&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n","protected":false},"author":13,"menu_order":23,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":513,"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\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/1755"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/users\/13"}],"version-history":[{"count":5,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/1755\/revisions"}],"predecessor-version":[{"id":6068,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/1755\/revisions\/6068"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/parts\/513"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/1755\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/media?parent=1755"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapter-type?post=1755"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/contributor?post=1755"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/license?post=1755"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}