{"id":2598,"date":"2016-11-04T00:23:31","date_gmt":"2016-11-04T00:23:31","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/waymakercollegealgebra\/?post_type=chapter&#038;p=2598"},"modified":"2017-04-04T20:06:02","modified_gmt":"2017-04-04T20:06:02","slug":"putting-it-together-probability-and-counting-principles","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/chapter\/putting-it-together-probability-and-counting-principles\/","title":{"raw":"Putting It Together: Probability and Counting Principles","rendered":"Putting It Together: Probability and Counting Principles"},"content":{"raw":"The lottery jackpot has continued to climb as you completed this module. \u00a0Now it is time to determine how likely you are to win.\r\n\r\nLet\u2019s first assume that you not only need to pick six specific numbers from 1 \u2013 49, but you need to pick them in the correct order. \u00a0If this is the case, you know you need to use a permutation to figure out the size of the sample space.\r\n<p style=\"text-align: center;\">[latex]P\\left(n,r\\right)=\\frac{n!}{\\left(n-r\\right)!}[\/latex]<\/p>\r\nIn this case, [latex]n[\/latex] is the possible numbers, which is 49, and [latex]r[\/latex] is the number of choices you make, which is 6.\r\n<p style=\"text-align: center;\">[latex]P\\left(49,6\\right)=\\frac{49!}{\\left(49-6\\right)!}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]P\\left(49,6\\right)=\\frac{49!}{43!}=10,068,347,520[\/latex]<\/p>\r\nThis tells you that there is one way out of about 10 billion to win. \u00a0Your chances are not good at all.\r\n\r\nFortunately, most lottery winnings do not depend on order so you can use a combination instead.\r\n<p style=\"text-align: center;\">[latex]C\\left(n,r\\right)=\\frac{n!}{r!\\left(n-r\\right)!}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]C\\left(49,6\\right)=\\frac{49!}{6!\\left(49-6\\right)!}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]C\\left(49,6\\right)=\\frac{49!}{6!\\left(43\\right)!}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]C\\left(49,6\\right)=\\frac{49!}{6!\\left(43\\right)!}=13,983,816[\/latex]<\/p>\r\nNotice that the sample space has been greatly reduced from about 10 billion to about 14 million. \u00a0So the likelihood of you winning is much greater than before, but still very slim.\r\n\r\nWhat would happen to your chances of winning if you bought more than one ticket? \u00a0Suppose you bought 100 tickets and chose a different combination of six numbers on each ticket. \u00a0You could compare the number of tickets to sample space to determine your probability.\r\n<p style=\"text-align: center;\">[latex]{\\large\\frac{100}{14\\text{ million}}}=0.0000071\\ =\\ 0.00071\\%[\/latex]<\/p>\r\nThat\u2019s much less than a 1% chance of winning. \u00a0Still not very good. \u00a0So suppose you gather up some cash and buy 1,000 tickets.\r\n<p style=\"text-align: center;\">[latex]{\\large\\frac{1,000}{14\\text{ million}}}=0.000071\\ =\\ 0.0071\\%[\/latex]<\/p>\r\nNow you are out $1000, assuming each ticket costs $1, and your chances are still less than a 1% chance.\r\n\r\n&nbsp;\r\n\r\nOkay, maybe you are ready to go for broke. \u00a0You and a group of friends gather your funds to purchase 1 million tickets.\r\n<p style=\"text-align: center;\">[latex]{\\large\\frac{1\\text{ million}}{14\\text{ million}}}=0.071\\ =\\ 7.1\\%[\/latex]<\/p>\r\n&nbsp;\r\n\r\nSo even after purchasing 1 million tickets, which might cost $1 million, your probability of winning the big jackpot is only about 7%. \u00a0To raise your probability to just 50%, you would have to purchase 7 million tickets. \u00a0\u00a0It\u2019s up to you do decide how lucky you feel. Maybe just buy one ticket and see what happens. \u00a0Good luck!\r\n\r\n&nbsp;","rendered":"<p>The lottery jackpot has continued to climb as you completed this module. \u00a0Now it is time to determine how likely you are to win.<\/p>\n<p>Let\u2019s first assume that you not only need to pick six specific numbers from 1 \u2013 49, but you need to pick them in the correct order. \u00a0If this is the case, you know you need to use a permutation to figure out the size of the sample space.<\/p>\n<p style=\"text-align: center;\">[latex]P\\left(n,r\\right)=\\frac{n!}{\\left(n-r\\right)!}[\/latex]<\/p>\n<p>In this case, [latex]n[\/latex] is the possible numbers, which is 49, and [latex]r[\/latex] is the number of choices you make, which is 6.<\/p>\n<p style=\"text-align: center;\">[latex]P\\left(49,6\\right)=\\frac{49!}{\\left(49-6\\right)!}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]P\\left(49,6\\right)=\\frac{49!}{43!}=10,068,347,520[\/latex]<\/p>\n<p>This tells you that there is one way out of about 10 billion to win. \u00a0Your chances are not good at all.<\/p>\n<p>Fortunately, most lottery winnings do not depend on order so you can use a combination instead.<\/p>\n<p style=\"text-align: center;\">[latex]C\\left(n,r\\right)=\\frac{n!}{r!\\left(n-r\\right)!}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]C\\left(49,6\\right)=\\frac{49!}{6!\\left(49-6\\right)!}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]C\\left(49,6\\right)=\\frac{49!}{6!\\left(43\\right)!}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]C\\left(49,6\\right)=\\frac{49!}{6!\\left(43\\right)!}=13,983,816[\/latex]<\/p>\n<p>Notice that the sample space has been greatly reduced from about 10 billion to about 14 million. \u00a0So the likelihood of you winning is much greater than before, but still very slim.<\/p>\n<p>What would happen to your chances of winning if you bought more than one ticket? \u00a0Suppose you bought 100 tickets and chose a different combination of six numbers on each ticket. \u00a0You could compare the number of tickets to sample space to determine your probability.<\/p>\n<p style=\"text-align: center;\">[latex]{\\large\\frac{100}{14\\text{ million}}}=0.0000071\\ =\\ 0.00071\\%[\/latex]<\/p>\n<p>That\u2019s much less than a 1% chance of winning. \u00a0Still not very good. \u00a0So suppose you gather up some cash and buy 1,000 tickets.<\/p>\n<p style=\"text-align: center;\">[latex]{\\large\\frac{1,000}{14\\text{ million}}}=0.000071\\ =\\ 0.0071\\%[\/latex]<\/p>\n<p>Now you are out $1000, assuming each ticket costs $1, and your chances are still less than a 1% chance.<\/p>\n<p>&nbsp;<\/p>\n<p>Okay, maybe you are ready to go for broke. \u00a0You and a group of friends gather your funds to purchase 1 million tickets.<\/p>\n<p style=\"text-align: center;\">[latex]{\\large\\frac{1\\text{ million}}{14\\text{ million}}}=0.071\\ =\\ 7.1\\%[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>So even after purchasing 1 million tickets, which might cost $1 million, your probability of winning the big jackpot is only about 7%. \u00a0To raise your probability to just 50%, you would have to purchase 7 million tickets. \u00a0\u00a0It\u2019s up to you do decide how lucky you feel. Maybe just buy one ticket and see what happens. \u00a0Good luck!<\/p>\n<p>&nbsp;<\/p>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-2598\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>Putting It Together: Probability and Counting Principles. <strong>Authored by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":21,"menu_order":11,"template":"","meta":{"_candela_citation":"[{\"type\":\"original\",\"description\":\"Putting It Together: Probability and Counting Principles\",\"author\":\"Lumen Learning\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"3c94fb2b-2db1-4da8-bb48-c23b3b91c151","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-2598","chapter","type-chapter","status-publish","hentry"],"part":2518,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/2598","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/wp\/v2\/users\/21"}],"version-history":[{"count":3,"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/2598\/revisions"}],"predecessor-version":[{"id":3824,"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/2598\/revisions\/3824"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/pressbooks\/v2\/parts\/2518"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/2598\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/wp\/v2\/media?parent=2598"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=2598"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/wp\/v2\/contributor?post=2598"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/wp\/v2\/license?post=2598"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}