{"id":467,"date":"2019-07-15T22:45:22","date_gmt":"2019-07-15T22:45:22","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/waymakercollegealgebracorequisite\/chapter\/putting-it-together-probability-and-counting-principles\/"},"modified":"2019-07-15T22:45:22","modified_gmt":"2019-07-15T22:45:22","slug":"putting-it-together-probability-and-counting-principles","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/ntcc-collegealgebracorequisite\/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":"\nThe lottery jackpot has continued to climb as you completed this module. &nbsp;Now it is time to determine how likely you are to win.\n\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. &nbsp;If this is the case, you know you need to use a permutation to figure out the size of the sample space.\n<p style=\"text-align: center\">[latex]P\\left(n,r\\right)=\\dfrac{n!}{\\left(n-r\\right)!}[\/latex]<\/p>\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.\n<p style=\"text-align: center\">[latex]P\\left(49,6\\right)=\\dfrac{49!}{\\left(49-6\\right)!}[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]P\\left(49,6\\right)=\\dfrac{49!}{43!}=10,068,347,520[\/latex]<\/p>\nThis tells you that there is one way out of about 10 billion to win. &nbsp;Your chances are not good at all.\n\nFortunately, most lottery winnings do not depend on order so you can use a combination instead.\n<p style=\"text-align: center\">[latex]C\\left(n,r\\right)=\\dfrac{n!}{r!\\left(n-r\\right)!}[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]C\\left(49,6\\right)=\\dfrac{49!}{6!\\left(49-6\\right)!}[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]C\\left(49,6\\right)=\\dfrac{49!}{6!\\left(43\\right)!}[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]C\\left(49,6\\right)=\\dfrac{49!}{6!\\left(43\\right)!}=13,983,816[\/latex]<\/p>\nNotice that the sample space has been greatly reduced from about 10 billion to about 14 million. &nbsp;So the likelihood of you winning is much greater than before, but still very slim.\n\nWhat would happen to your chances of winning if you bought more than one ticket? &nbsp;Suppose you bought 100 tickets and chose a different combination of six numbers on each ticket. &nbsp;You could compare the number of tickets to sample space to determine your probability.\n<p style=\"text-align: center\">[latex]{\\large\\frac{100}{14\\text{ million}}}=0.0000071\\ =\\ 0.00071\\%[\/latex]<\/p>\nThat\u2019s much less than a 1% chance of winning. &nbsp;Still not very good. &nbsp;So suppose you gather up some cash and buy 1,000 tickets.\n<p style=\"text-align: center\">[latex]{\\large\\frac{1,000}{14\\text{ million}}}=0.000071\\ =\\ 0.0071\\%[\/latex]<\/p>\nNow you are out $1000, assuming each ticket costs $1, and your chances are still less than a 1% chance.\n\n&nbsp;\n\nOkay, maybe you are ready to go for broke. &nbsp;You and a group of friends gather your funds to purchase 1 million tickets.\n<p style=\"text-align: center\">[latex]{\\large\\frac{1\\text{ million}}{14\\text{ million}}}=0.071\\ =\\ 7.1\\%[\/latex]<\/p>\n&nbsp;\n\nSo even after purchasing 1 million tickets, which might cost $1 million, your probability of winning the big jackpot is only about 7%. &nbsp;To raise your probability to just 50%, you would have to purchase 7 million tickets. &nbsp;&nbsp;It\u2019s up to you do decide how lucky you feel. Maybe just buy one ticket and see what happens. &nbsp;Good luck!\n\n&nbsp;\n","rendered":"<p>The lottery jackpot has continued to climb as you completed this module. &nbsp;Now 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. &nbsp;If 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)=\\dfrac{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)=\\dfrac{49!}{\\left(49-6\\right)!}[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]P\\left(49,6\\right)=\\dfrac{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. &nbsp;Your 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)=\\dfrac{n!}{r!\\left(n-r\\right)!}[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]C\\left(49,6\\right)=\\dfrac{49!}{6!\\left(49-6\\right)!}[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]C\\left(49,6\\right)=\\dfrac{49!}{6!\\left(43\\right)!}[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]C\\left(49,6\\right)=\\dfrac{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. &nbsp;So 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? &nbsp;Suppose you bought 100 tickets and chose a different combination of six numbers on each ticket. &nbsp;You 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. &nbsp;Still not very good. &nbsp;So 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. &nbsp;You 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%. &nbsp;To raise your probability to just 50%, you would have to purchase 7 million tickets. &nbsp;&nbsp;It\u2019s up to you do decide how lucky you feel. Maybe just buy one ticket and see what happens. &nbsp;Good 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-467\">\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":17533,"menu_order":14,"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-467","chapter","type-chapter","status-publish","hentry"],"part":453,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/ntcc-collegealgebracorequisite\/wp-json\/pressbooks\/v2\/chapters\/467","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/ntcc-collegealgebracorequisite\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/ntcc-collegealgebracorequisite\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ntcc-collegealgebracorequisite\/wp-json\/wp\/v2\/users\/17533"}],"version-history":[{"count":0,"href":"https:\/\/courses.lumenlearning.com\/ntcc-collegealgebracorequisite\/wp-json\/pressbooks\/v2\/chapters\/467\/revisions"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/ntcc-collegealgebracorequisite\/wp-json\/pressbooks\/v2\/parts\/453"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/ntcc-collegealgebracorequisite\/wp-json\/pressbooks\/v2\/chapters\/467\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/ntcc-collegealgebracorequisite\/wp-json\/wp\/v2\/media?parent=467"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ntcc-collegealgebracorequisite\/wp-json\/pressbooks\/v2\/chapter-type?post=467"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ntcc-collegealgebracorequisite\/wp-json\/wp\/v2\/contributor?post=467"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ntcc-collegealgebracorequisite\/wp-json\/wp\/v2\/license?post=467"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}