{"id":999,"date":"2015-08-16T09:53:27","date_gmt":"2015-08-16T08:53:27","guid":{"rendered":"http:\/\/pcool.dyndns.org:8080\/statsbook\/?page_id=999"},"modified":"2025-06-24T20:56:35","modified_gmt":"2025-06-24T19:56:35","slug":"answers-probability","status":"publish","type":"page","link":"https:\/\/pcool.dyndns.org\/index.php\/answers-probability\/","title":{"rendered":"Answers Probability"},"content":{"rendered":"\n<ol class=\"wp-block-list\">\n<li>We need to use:<\/li>\n<\/ol>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\(Probability = {t \\choose n} \\cdot c^n \\cdot{(1-c)^{t-n}}    \\)<script src=\"https:\/\/pcool.dyndns.org\/wp-includes\/js\/dist\/hooks.min.js?ver=dd5603f07f9220ed27f1\" id=\"wp-hooks-js\"><\/script>\n<script src=\"https:\/\/pcool.dyndns.org\/wp-includes\/js\/dist\/i18n.min.js?ver=c26c3dc7bed366793375\" id=\"wp-i18n-js\"><\/script>\n<script id=\"wp-i18n-js-after\">\nwp.i18n.setLocaleData( { 'text direction\\u0004ltr': [ 'ltr' ] } );\n\/\/# sourceURL=wp-i18n-js-after\n<\/script>\n<script  async src=\"https:\/\/cdnjs.cloudflare.com\/ajax\/libs\/mathjax\/2.7.7\/MathJax.js?config=TeX-MML-AM_CHTML\" id=\"mathjax-js\"><\/script>\n<\/div>\n\n\n\n<p>t = 10, n = 3 and c = 1\/10. Therefore<\/p>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\(Probability = {10 \\choose 3} \\cdot {\\frac{1}{10}}^3 \\cdot{(1-{\\frac{1}{10}})^{10-3}}    \\)<\/div>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\(Probability = {10 \\choose 3} \\cdot {\\frac{1}{10}}^3 \\cdot{{\\frac{9}{10}}^{7}}    \\)\n<\/div>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\(Probability =  \\frac{10!}{7!\\cdot3!}\\cdot\\frac{1}{10}^3\\cdot{\\frac{9}{10}^7}   \\)<\/div>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\(Probability =  \\frac{10\\cdot9\\cdot8}{3\\cdot2\\cdot1}\\cdot{\\frac{9^7}{10^{10}}}   \\)\n<\/div>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\"><\/div>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\(Probability = \\frac{3443737680}{60000000000} \\) ~ 0.0574 ~ 5.7% <\/div>\n\n\n\n<p>2. The total number of combinations is: We start by painting 9 of the 20 injects red. There are<\/p>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\(  20 \\choose 9 \\)<\/div>\n\n\n\n<p>possible combinations of doing that. Of the remaining 11 objects, we paint 8 white. That can be done in <\/p>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\(  11 \\choose 8 \\)<\/div>\n\n\n\n<p> possible combinations. The final three objects are painted blue. Obviously, there is only one possible combination of doing that. The total number of possible combinations is therefore:<\/p>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\( {20 \\choose 9}\\cdot{{20-9}\\choose8}\\cdot{{20-9-8}\\choose3}={20 \\choose 9}\\cdot{{11}\\choose8}\\cdot{{3}\\choose3} \\)<\/div>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">= \\( \\frac{20!}{11!\\cdot9!}\\cdot\\frac{11!}{3!\\cdot8!}\\cdot\\frac{3!}{3!}=\\frac{20!}{9!\\cdot3!\\cdot8!} \\) =27713400<\/div>\n\n\n\n<p>One might think that the answer would have been different if we would have started by painting the 3 blue objects. The following formula describes the possible combinations by first painting 3 objects blue, followed by 8 objects white<\/p>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\( {20 \\choose 3}\\cdot{{20-3}\\choose8}\\cdot{{20-3-8}\\choose9}={20 \\choose 3}\\cdot{{17}\\choose8}\\cdot{{9}\\choose9} \\)<\/div>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\( =\\frac{20!}{17!\\cdot3!}\\cdot\\frac{17!}{9!\\cdot8!}\\cdot\\frac{9!}{9!}=\\frac{20!}{3!\\cdot8!\\cdot9!} \\)=27713400<\/div>\n\n\n\n<p>Indeed, if we would have started by painting the 8 white objects, followed by nine objects red, the formula would have been:<\/p>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\( {20 \\choose 8}\\cdot{{20-8}\\choose9}\\cdot{{20-8-9}\\choose3}={20 \\choose 8}\\cdot{{12}\\choose9}\\cdot{{3}\\choose3} \\)<\/div>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\( =\\frac{20!}{12!\\cdot8!}\\cdot\\frac{12!}{3!\\cdot9!}\\cdot\\frac{3!}{3!}=\\frac{20!}{8!\\cdot9!\\cdot3!} \\)=27713400<\/div>\n\n\n\n<p>So, it doesn\u2019t make any difference where you start. We have shown that: <\/p>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\( {20 \\choose 9}\\cdot{{11}\\choose8}\\cdot{{3}\\choose3}={20 \\choose 3}\\cdot{{17}\\choose8}\\cdot{{9}\\choose9}={20 \\choose 8}\\cdot{{12}\\choose9}\\cdot{{3}\\choose3}\\)<\/div>\n\n\n\n<p>3. This is very similar to the example given in the main text. The probability is:<\/p>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\( Probability=\\frac{{4\\choose 1 }\\cdot{1\\choose 1}\\cdot{3 \\choose 2}\\cdot{44\\choose1}}{52\\choose 5} \\)<\/div>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\( Probability=\\frac{\\frac{4!}{3!\\cdot1!}\\cdot1\\cdot\\frac{3!}{1!\\cdot2!}\\cdot\\frac{44!}{43!\\cdot1!}}{\\frac{52!}{47!\\cdot5!}}=\\frac{4!\\cdot3!\\cdot44!\\cdot47!\\cdot5!}{3!\\cdot2!\\cdot43!\\cdot52!} \\)<\/div>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\(Probability=\\frac{47!\\cdot44!\\cdot5!\\cdot4!\\cdot3!}{52!\\cdot43!\\cdot3!\\cdot2!}=\\frac{44\\cdot5\\cdot4\\cdot3\\cdot4\\cdot3\\cdot2}{52\\cdot51\\cdot50\\cdot49\\cdot48}=\\frac{63360}{311875200} \\) <\/div>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\( Probability \\) ~ 0.000203<\/div>\n\n\n\n<p>4.<\/p>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\( Probability=\\frac{{4\\choose 1 }\\cdot{1\\choose 1}\\cdot{3 \\choose 0}\\cdot{44\\choose3}}{52\\choose 5} \\)<\/div>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\(Probability =\\frac{\\frac{4!}{3!\\cdot1!}\\cdot1\\cdot1\\cdot\\frac{44!}{41!\\cdot3!}}{\\frac{52!}{47!\\cdot5!}}=\\frac{4!\\cdot44!\\cdot47!\\cdot5!}{3!\\cdot1!\\cdot41!\\cdot3!\\cdot52!} \\)<\/div>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\(Probability = \\frac{44\\cdot43\\cdot42\\cdot5\\cdot4\\cdot4}{52\\cdot51\\cdot50\\cdot49\\cdot48}=\\frac{6357120}{311875200} \\) ~ 0.0204 ~ 2%<\/div>\n\n\n\n<p>5. This probability can be calculated by adding the probability of 1 king plus the probability on two kings plus the probability on three kings plus the probability of four kings:<\/p>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\( Probability=\\frac{{4\\choose 1 }\\cdot{48\\choose 4}}{52\\choose 5}+\\frac{{4\\choose 2 }\\cdot{48\\choose 3}}{52\\choose 5}+\\frac{{4\\choose 3 }\\cdot{48\\choose 2}}{52\\choose 5}+ \\frac{{4\\choose 4 }\\cdot{48\\choose 1}}{52\\choose 5} \\)<\/div>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\(Probability =\\frac{\\frac{4!}{3!\\cdot1!}\\cdot\\frac{48!}{44!\\cdot4!}}{\\frac{52!}{47!\\cdot5!}}+\\frac{\\frac{4!}{2!\\cdot2!}\\cdot\\frac{48!}{45!\\cdot3!}}{\\frac{52!}{47!\\cdot5!}} +\\frac{\\frac{4!}{1!\\cdot3!}\\cdot\\frac{48!}{46!\\cdot2!}}{\\frac{52!}{47!\\cdot5!}}+\\frac{1\\cdot\\frac{48!}{47!\\cdot1!}}{\\frac{52!}{47!\\cdot5!}} \\)<\/div>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\( Probability=\\frac{\\frac{48\\cdot47\\cdot46\\cdot45}{3\\cdot2}+\\frac{48\\cdot47\\cdot46}{1}+\\frac{48\\cdot47\\cdot2}{1}+\\frac{48}{1}}{\\frac{52\\cdot51\\cdot50\\cdot49\\cdot48}{5\\cdot4\\cdot3\\cdot2}} \\)<\/div>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\(Probability =\\frac{886656}{2598960} \\) ~ 0.341158 <\/div>\n\n\n\n<p>It would have been easier to subtract the probability of no kings from one:<\/p>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\( Probability=1-\\frac{{4\\choose 0 }\\cdot{48\\choose 5}}{52\\choose 5}=1-(\\frac{48!}{43!\\cdot5!}\\cdot\\frac{47!\\cdot5!}{52!}) \\)<\/div>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\(Probability =1-(\\frac{48!\\cdot47!\\cdot5!}{52!\\cdot43!\\cdot5!})=1-(\\frac{48!\\cdot47!}{52!\\cdot43!}) \\)<\/div>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\( Probabiliyt =1-(\\frac{47\\cdot46\\cdot45\\cdot44}{52\\cdot51\\cdot50\\cdot49}) \\)\n<\/div>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\( Probability=1-\\frac{4280760}{6497400} \\) ~ 1-0.659 ~ 0.341158<\/div>\n\n\n\n<p>6.<\/p>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\(Probability =\\frac{{6\\choose 6 }\\cdot{43\\choose 0}}{49\\choose 6}=\\frac{1}{\\frac{49!}{43!\\cdot6!}}=\\frac{1}{\\frac{49\\cdot48\\cdot47\\cdot46\\cdot45\\cdot44}{6\\cdot5\\cdot4\\cdot3\\cdot2}} \\) ~ 0.0000000715 <\/div>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\"><\/div>\n\n\n\n<p><\/p>\n","protected":false},"excerpt":{"rendered":"<p>t = 10, n = 3 and c = 1\/10. Therefore 2. The total number of combinations is: We start by painting 9 of the 20 injects red. There are possible combinations of doing that. Of the remaining 11 objects, we paint 8 white. That can be done in possible combinations. The final three objects [&hellip;]<\/p>\n","protected":false},"author":2,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"inline_featured_image":false,"footnotes":""},"class_list":["post-999","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/pcool.dyndns.org\/index.php\/wp-json\/wp\/v2\/pages\/999","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/pcool.dyndns.org\/index.php\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/pcool.dyndns.org\/index.php\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/pcool.dyndns.org\/index.php\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/pcool.dyndns.org\/index.php\/wp-json\/wp\/v2\/comments?post=999"}],"version-history":[{"count":9,"href":"https:\/\/pcool.dyndns.org\/index.php\/wp-json\/wp\/v2\/pages\/999\/revisions"}],"predecessor-version":[{"id":4368,"href":"https:\/\/pcool.dyndns.org\/index.php\/wp-json\/wp\/v2\/pages\/999\/revisions\/4368"}],"wp:attachment":[{"href":"https:\/\/pcool.dyndns.org\/index.php\/wp-json\/wp\/v2\/media?parent=999"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}