{"id":932,"date":"2015-08-13T18:06:09","date_gmt":"2015-08-13T17:06:09","guid":{"rendered":"http:\/\/pcool.dyndns.org:8080\/statsbook\/?page_id=932"},"modified":"2025-06-27T08:15:53","modified_gmt":"2025-06-27T07:15:53","slug":"probability-and-chance","status":"publish","type":"page","link":"https:\/\/pcool.dyndns.org\/index.php\/probability-and-chance\/","title":{"rendered":"Probability and Chance"},"content":{"rendered":"\n<p>When flipping a coin, it seems obvious that the chance (or probability) of head is \u00bd and the chance of tail is also \u00bd. However, what is the probability of two times head in succession? One might think that the probability of this happening is \u00bd + \u00bd = 1. However, if the probability is 1 (i.e. 100%), we mean it is a certainty. We all know it is quite possible to have tails three times in succession, so the probability cannot be 1 and should be lower than that.<\/p>\n\n\n\n<p>So, what is the probability of heads twice in succession?<\/p>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\( P = \\frac{1}{2} \\cdot \\frac {1}{2} = \\frac{1}{4}\\)<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><strong>Probabilities are multiplied rather than added<\/strong><\/p>\n\n\n\n<p>The example of the coin above is rather unique, in that the probability of heads is equal to the probability of tails ( \u00bd ). Consequently, the probability of heads twice is equal to the probability of tails twice and indeed the probability of heads once and tails once in succession.<\/p>\n\n\n\n<p>So what if we look at a slightly more complicated example; the die? When rolling a die, the probability of throwing \u201c1\u201d = 1\/6. This is equal to the probability of throwing \u201c2\u201d, \u201c3\u201d, \u201c4\u201d, \u201c5\u201d and \u201c6\u201d.<\/p>\n\n\n\n<p>As explained above, the probability of twice \u201c1\u201d in succession is:<\/p>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\(P = \\frac{1}{6} \\cdot \\frac{1}{6} = \\frac{1}{36}\\)<\/div>\n\n\n\n<p>Similarly, the probability of throwing six times \u201c6\u201d is:<\/p>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\(P = \\frac{1}{6} \\cdot \\frac{1}{6}  \\cdot \\frac{1}{6} \\cdot \\frac{1}{6} \\cdot \\frac{1}{6} \\cdot \\frac{1}{6}= \\frac{1}{46656} = 0.00002143\\)<\/div>\n\n\n\n<p>What is the probability of <em>once<\/em> \u201c1\u201d, when throwing the die twice? The probability of this happening is the probability of:<\/p>\n\n\n\n<p>First time \u201c1\u201d, second time not \u201c1\u201d : <\/p>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\(P = \\frac{1}{6} \\cdot \\frac{5}{6}\\)<\/div>\n\n\n\n<p>First time not \u201c1\u201d, second time \u201c1\u201d : <\/p>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\(P = \\frac{5}{6} \\cdot \\frac{1}{6}\\)<\/div>\n\n\n\n<p>So the probability is:<\/p>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\( P = \\frac{1}{6} \\cdot \\frac{5}{6} + \\frac{5}{6} \\cdot \\frac{1}{6} = \\frac{5}{36} + \\frac{5}{36} = \\frac{10}{36} \\approx 0.278 \\approx 28\\% \\)<\/div>\n\n\n\n<p>So why are these probabilities suddenly added together (rather than multiplied)? The probability of the first time \u201c1\u201d depends on the second time not being \u201c1\u201d. These probabilities are therefore dependent and should be multiplied.<\/p>\n\n\n\n<p><strong>Probabilities that are dependent on each other are multiplied, but probabilities that are not dependent on each other are added<\/strong>.<\/p>\n\n\n\n<p>Similarly, we can calculate the probability of the die being <em>at least once<\/em> \u201c1\u201d in two throws. This probability is:<\/p>\n\n\n\n<p>First time \u201c1\u201d, second time not \u201c1\u201d : <\/p>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\(P = \\frac{1}{6} \\cdot \\frac{5}{6}\\)<\/div>\n\n\n\n<p>First time not \u201c1\u201d, second time \u201c1\u201d : <\/p>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\(P = \\frac{5}{6} \\cdot \\frac{1}{6}\\)<\/div>\n\n\n\n<p>First time \u201d1\u201d and second time \u201c1\u201d : <\/p>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\(P = \\frac{1}{6} \\cdot \\frac{1}{6}\\)<\/div>\n\n\n\n<p>So the probability is:<\/p>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\( P = \\frac{1}{6} \\cdot \\frac{5}{6} + \\frac{5}{6} \\cdot \\frac{1}{6} + \\frac{1}{6} \\cdot \\frac{1}{6} = \\frac{5}{36} + \\frac{5}{36} + \\frac{1}{36}= \\frac{11}{36} \\approx 0.3056 \\approx 31\\% \\)<\/div>\n\n\n\n<p>Let us look at a slightly more complex example. What is the probability on twice \u201c1\u201d on throwing a die six times? One might think this probability would be:<\/p>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\( P = \\frac{1}{6} \\cdot \\frac{1}{6} \\cdot \\frac{5}{6} \\cdot \\frac{5}{6} \\cdot \\frac{5}{6} \\cdot \\frac{5}{6} \\)<\/div>\n\n\n\n<p>However, this is the probability of throwing \u201c1\u201d in the first and second throw followed by four throws when the die is not \u201c1\u201d. But if we throw a \u201c1\u201d the first and third time, we also have thrown \u201c1\u201d twice out of six:<\/p>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\( P = \\frac{1}{6} \\cdot \\frac{5}{6} \\cdot \\frac{1}{6} \\cdot \\frac{5}{6} \\cdot \\frac{5}{6} \\cdot \\frac{5}{6} \\)<\/div>\n\n\n\n<p>And if we throw a \u201c1\u201d the first and fourth throw:<\/p>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\( P = \\frac{1}{6} \\cdot \\frac{5}{6} \\cdot \\frac{5}{6} \\cdot \\frac{1}{6} \\cdot \\frac{5}{6} \\cdot \\frac{5}{6} \\)<\/div>\n\n\n\n<p>Obviously, there are many combinations possible. If \u201c1\u201d depicts the die being \u201c1\u201d and \u201cx\u201d the die not being \u201c1\u201d, the possible combinations are:<\/p>\n\n\n\n<p>11xxxx&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; x11xxx&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; xx11xx&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; xxx11x&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; xxxx11<\/p>\n\n\n\n<p>1x1xxx&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; x1x1xx&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; xx1x1x&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; xxx1x1<\/p>\n\n\n\n<p>1xx1xx&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; x1xx1x&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; xx1xx1<\/p>\n\n\n\n<p>1xxx1x&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; x1xxx1<\/p>\n\n\n\n<p>1xxxx1<\/p>\n\n\n\n<p>So there are in total 15 possible combinations. And the probability of throwing \u201c1\u201d twice out of six throws is:<\/p>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\( P = \\frac{1}{6} \\cdot \\frac{1}{6} \\cdot \\frac{5}{6} \\cdot \\frac{5}{6} \\cdot \\frac{5}{6} \\cdot \\frac{5}{6} +  \\)\n\\( \\frac{1}{6} \\cdot \\frac{5}{6} \\cdot \\frac{1}{6} \\cdot \\frac{5}{6} \\cdot \\frac{5}{6} \\cdot \\frac{5}{6} +  \\)\n\\( \\frac{1}{6} \\cdot \\frac{5}{6} \\cdot \\frac{5}{6} \\cdot \\frac{1}{6} \\cdot \\frac{5}{6} \\cdot \\frac{5}{6} \\cdots +  \\)\n\\( \\frac{5}{6} \\cdot \\frac{5}{6} \\cdot \\frac{5}{6} \\cdot \\frac{5}{6} \\cdot \\frac{1}{6} \\cdot \\frac{1}{6}\n \\)<\/div>\n\n\n\n<p>In other words the number of combinations times the probability of one of these combinations:<\/p>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\( P = 15 \\cdot {\\frac{1}{6} \\cdot \\frac{1}{6} \\cdot \\frac{5}{6} \\cdot \\frac{5}{6} \\cdot \\frac{5}{6} \\cdot \\frac{5}{6} } \\approx 0.201\\)<\/div>\n\n\n\n<p>So how do we calculate the number of possible combinations? The number of possible combinations of two out of six is:<\/p>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\( {6 \\choose 2} \\)<\/div>\n\n\n\n<p class=\"is-style-text-annotation is-style-text-annotation--1\">&#8216;six choose two&#8217;<\/p>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\( {6 \\choose 2} = \\frac{6!}{4!\\cdot2!} = \\frac{6\\cdot5\\cdot4\\cdot3\\cdot2\\cdot1}{(4\\cdot3\\cdot2\\cdot1)\\cdot(2\\cdot1)} = \\frac{6\\cdot5}{2\\cdot1} = \\frac{30}{2} = 15 \\)<\/div>\n\n\n\n<p>In general, the total number of combinations N from n (number of occurrences) out of t (total number of occurrences):<\/p>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\( N = {t \\choose n} = \\frac{t!}{(t-n)! \\cdot (n!)} \\)<\/div>\n\n\n\n<p>If c is the probability on a single occurrence; the probability of this not happening is 1 &#8211; c:<\/p>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\(P = {t \\choose n} \\cdot c^n\\cdot(1-c)^{t-n} \\)<\/div>\n\n\n\n<p>So in our example of the die, what is the probability of throwing twice \u201c1\u201d out of six throws?<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>t = 6<\/li>\n\n\n\n<li>n = 2 <\/li>\n\n\n\n<li>c = 1\/6<\/li>\n<\/ul>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\(P = {6 \\choose 2} \\cdot (\\frac{1}{6})^2\\cdot(\\frac{5}{6})^{6-2} \\)\n\\( = \\frac{6!}{(6-2)!\\cdot(2)!}\\cdot(\\frac{1}{6})^2 \\cdot (\\frac{5}{6})^4 \\)\n<\/div>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\(P = \\frac{6\\cdot5}{2\\cdot1}\\cdot(\\frac{1}{6})^2 \\cdot (\\frac{5}{6})^4 \\)<\/div>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\(P = 15 \\cdot (\\frac{625}{46656}) = \\frac{9375}{46656} \\approx 0.201\\)<\/div>\n\n\n\n<p>What is the probability of throwing <em>at least <\/em>four \u201c6\u201d out of six throws? This is the probability of throwing 4 times \u201c6\u201d plus the probability of throwing 5 times \u201c6\u201d plus the probability of throwing six times \u201c6\u201d:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>t = 6<\/li>\n\n\n\n<li>n1 = 4<\/li>\n\n\n\n<li>n2 = 5<\/li>\n\n\n\n<li>n3 = 6<\/li>\n\n\n\n<li>c&nbsp;&nbsp; = 1\/6<\/li>\n<\/ul>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\(P = {6 \\choose 4} \\cdot (\\frac{1}{6})^4\\cdot(\\frac{5}{6})^{6-4} + {6 \\choose 5} \\cdot (\\frac{1}{6})^5\\cdot(\\frac{5}{6})^{6-5} + {6 \\choose 6} \\cdot (\\frac{1}{6})^6\\cdot(\\frac{5}{6})^{6-6}  \\)<\/div>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\(P= \\frac{6!}{2!\\cdot4!}\\cdot(\\frac{1}{6})^4 \\cdot (\\frac{5}{6})^2  + \\frac{6!}{1!\\cdot5!}\\cdot(\\frac{1}{6})^5 \\cdot (\\frac{5}{6})^1 + \\frac{6!}{0!\\cdot6!}\\cdot(\\frac{1}{6})^6 \\cdot (\\frac{5}{6})^0\\)<\/div>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\(P= \\frac{6\\cdot5}{2}\\cdot(\\frac{1}{6})^4 \\cdot(\\frac{5}{6})^2  +  \\frac{6}{1}\\cdot(\\frac{1}{6})^5 \\cdot(\\frac{5}{6})^1  + 1\\cdot(\\frac{1}{6})^6 \\cdot(\\frac{5}{6})^0 \\)<\/div>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\(P= 15\\cdot(\\frac{1^4\\cdot5^2}{6^6}) + 6\\cdot(\\frac{1^5\\cdot5^1}{6^6}) + 1\\cdot(\\frac{1^6}{6^6}) \\)<\/div>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\(P= \\frac{15\\cdot5^2 + 6\\cdot5 + 1}{6^6} = \\frac{406}{46656} \\approx 0.0087 \\approx 1% \\)<\/div>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">Please note 0! = 1, \\({6\\choose4}={6\\choose2}=15 \\)  and \\( {6\\choose6}={6\\choose0}=1\\)<\/div>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">In general: \\({t\\choose n}={t\\choose t-n} \\)<\/div>\n\n\n\n<p>So far, we have only been concerned with probabilities that do <strong>not<\/strong> change. Our examples of the coin and die are similar to putting ten balls (numbered 1 to 10) in a hat; taking one out and <strong>putting it back<\/strong> afterwards. In that case, the probability of getting ball number \u201c1\u201d once is 1\/10 and getting it twice in succession is 1\/10 * 1\/10 = 1\/100. The probability of getting ball number \u201c1\u201d twice out of four draws is:<\/p>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\(P = {t \\choose n}\\cdot c^n\\cdot(1-c)^{(t-n)} \\)<\/div>\n\n\n\n<ul class=\"wp-block-list\">\n<li>c = 1\/10<\/li>\n\n\n\n<li>t = 4, n=2:<\/li>\n<\/ul>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\(P = {4 \\choose 2} \\cdot (\\frac{1}{10})^2 \\cdot (\\frac{9}{10})^2  = \\frac{4!}{2!\\cdot2!}\\cdot(\\frac{1}{10})^2 \\cdot(\\frac{9}{10})^2 \\) <\/div>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\(P= \\frac{4\\cdot3}{2\\cdot1}\\cdot(\\frac{1}{10})^2 \\cdot (\\frac{9}{10})^2 = 6\\cdot \\frac{1 \\cdot 9^2}{10^4} = \\frac{486}{10000} \\approx 5% \\)<\/div>\n\n\n\n<p>However, often we have to deal with situations were the probability depends on the previous history. In other words, taking the balls out of the hat <strong>without putting them back<\/strong>! If we start with all ten balls in the hat, the probability of drawing number \u201c1\u201d is 1\/10. The probability on a successive ball (\u201c2\u201d, \u201c3\u201d, \u201c4\u201d, \u201c5\u201d, \u201c6\u201d, \u201c7\u201d, \u201c8\u201d, \u201c9\u201d 0r \u201c10\u201d) is now 1\/9. The probability that we draw number \u201c1\u201d followed by number \u201c5\u201d is 1\/10 \u00d7 1\/9. This is equal to the probability we draw number \u201c5\u201d followed by number \u201c1\u201d. So in drawing two balls, the probability we draw ball number \u201c1\u201d and ball number \u201c5\u201d in any order is:<\/p>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\(P = 2 \\cdot \\frac{1}{10} \\cdot \\frac {1}{9} \\approx 0.0022 \\approx 2\\% \\)<\/div>\n\n\n\n<p>So, how do we calculate the probability in more complicated scenarios? It has to be remembered that the probability on X happening equals the number of possible combinations with X divided by the total number of possible combinations:<\/p>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\(Probability(X) = \\frac{Number Of Combinations With X}{Total Number Of Combinations} \\)<\/div>\n\n\n\n<p>So, how do we calculate the possible number of combinations? We have already seen that the number of combinations N equals:<\/p>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\( N = {t \\choose n} = \\frac{t!}{(t-n)! \\cdot (n!)} \\)<\/div>\n\n\n\n<p>The number of combinations of getting one ball out of the hat is obviously:<\/p>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\( {1 \\choose 1} = 1 \\)<\/div>\n\n\n\n<p>So the probability of getting the ball with number \u201c1\u201d out of the hat with ten balls is:<\/p>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\(P = \\frac{1 \\choose 1}{10 \\choose 1} = \\frac{1}{\\frac{10!}{9! \\cdot 1!}} = \\frac{9!}{10!} = \\frac{1}{10}\\)<\/div>\n\n\n\n<p>Which is what we calculated previously. Similarly, getting number \u201c1\u201d and number \u201c5\u201d in any order is:<\/p>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\(P = \\frac{{1 \\choose 1} \\cdot {1 \\choose 1}} {10 \\choose 2} = \\frac{1}{\\frac{10!}{8! \\cdot 2!}} = \\frac{8! \\cdot 2!}{10!} = \\frac{2}{10 \\cdot 9} \\approx 0.022 \\approx 2\\% \\)<\/div>\n\n\n\n<p>The number of combinations in the numerator should be multiplied, as they depend on each other. In more general terms:<\/p>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\(P(X) = \\frac{{x \\choose a} \\cdot {y \\choose b} \\cdot {z \\choose c}} {t \\choose n} \\)<\/div>\n\n\n\n<ul class=\"wp-block-list\">\n<li>t = total number of occurrences<\/li>\n<\/ul>\n\n\n\n<ul class=\"wp-block-list\">\n<li>n = number of occurrences<\/li>\n<\/ul>\n\n\n\n<ul class=\"wp-block-list\">\n<li>t = x + y  + z<\/li>\n<\/ul>\n\n\n\n<ul class=\"wp-block-list\">\n<li>n = a + b + c<\/li>\n<\/ul>\n\n\n\n<p>So let us look at playing cards. What is the probability on getting <strong>one<\/strong> king when we draw five cards? There are four kings in the set, so there are: <\/p>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\( {4 \\choose 1} \\)<\/div>\n\n\n\n<p><span class=\"MathJax_Preview\" style=\"color: inherit;\"><\/span><span id=\"MathJax-Element-45-Frame\" class=\"mjx-chtml MJXc-processed\" tabindex=\"0\" data-mathml=\"<math xmlns=&quot;http:\/\/www.w3.org\/1998\/Math\/MathML&quot;><mrow class=&quot;MJX-TeXAtom-ORD&quot;><mrow><mstyle scriptlevel=&quot;0&quot;><mrow class=&quot;MJX-TeXAtom-OPEN&quot;><mo maxsize=&quot;1.2em&quot; minsize=&quot;1.2em&quot;>(<\/mo><\/mrow><\/mstyle><mfrac linethickness=&quot;0&quot;><mn>4<\/mn><mn>1<\/mn><\/mfrac><mstyle scriptlevel=&quot;0&quot;><mrow class=&quot;MJX-TeXAtom-CLOSE&quot;><mo maxsize=&quot;1.2em&quot; minsize=&quot;1.2em&quot;>)<\/mo><\/mrow><\/mstyle><\/mrow><\/mrow><\/math>&#8221; role=&#8221;presentation&#8221; style=&#8221;font-size: 122%; position: relative;&#8221;><span id=\"MJXc-Node-1729\" class=\"mjx-math\" aria-hidden=\"true\"><span id=\"MJXc-Node-1730\" class=\"mjx-mrow\"><span id=\"MJXc-Node-1731\" class=\"mjx-texatom\"><span id=\"MJXc-Node-1732\" class=\"mjx-mrow\"><span id=\"MJXc-Node-1733\" class=\"mjx-mrow\"><span id=\"MJXc-Node-1734\" class=\"mjx-TeXmathchoice\"><span id=\"MJXc-Node-1735\" class=\"mjx-mstyle\"><span id=\"MJXc-Node-1736\" class=\"mjx-mrow\"><span id=\"MJXc-Node-1737\" class=\"mjx-texatom\"><span id=\"MJXc-Node-1738\" class=\"mjx-mrow\"><span class=\"mjx-char MJXc-TeX-size1-R\" style=\"padding-top: 0.615em; padding-bottom: 0.615em;\"><\/span><\/span><\/span><\/span><\/span><span class=\"mjx-box MJXc-stacked\" style=\"width: 0.354em;\"><span class=\"mjx-numerator\" style=\"font-size: 70.7%; width: 0.5em; top: -1.33em;\"><\/span><span class=\"mjx-numerator\" style=\"font-size: 70.7%; width: 0.5em; top: -1.33em;\"><span id=\"MJXc-Node-1741\" class=\"mjx-mn\" style=\"\"><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>possibilities. The remaining four cards should not be kings. There are 51 cards left, of which three are kings and 48 non kings. So the number of possibilities on selecting four cards that are not kings is:<\/p>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\( { 48 \\choose 4} \\).<\/div>\n\n\n\n<p>The total number of possibilities on selecting five cards out of 52 is obviously:<\/p>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\( {52 \\choose 5} \\).<\/div>\n\n\n\n<p>So the probability of getting one King is:<\/p>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\(P = \\frac{{4 \\choose 1} \\cdot {48 \\choose 4}}{52 \\choose 5} = \n\\frac{\\frac{4!}{3! \\cdot 1!} \\cdot \\frac{48!}{44! \\cdot 4!}}{\\frac{52!}{47! \\cdot 5!}}\n \\)<\/div>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\(P = \\frac{4! \\cdot 48! \\cdot 47! \\cdot 5!}{3! \\cdot 1! \\cdot 44! \\cdot 4! \\cdot 52!} =\n\\frac{48! \\cdot 47! \\cdot 5!}{3! \\cdot 44! \\cdot 52!}\n\\)<\/div>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\(P = \\frac{47 \\cdot 46 \\cdot 45 \\cdot 5 \\cdot 4}{52 \\cdot 51 \\cdot 50 \\cdot 49} =\n\\frac{1945800}{6497400} \\approx 0.299 \\approx 30\\% \\)<\/div>\n\n\n\n<p class=\"is-style-text-annotation is-style-text-annotation--2\">It should be noted that 4 + 48 = 52 and 1 + 4 = 5 (numerator).<\/p>\n\n\n\n<p>Let us calculate the probability of drawing any one king, the queen of hearts and a further queen out of a total five playing cards. We start with the number of possibilities of drawing one king: <\/p>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\( {4 \\choose 1} \\).<\/div>\n\n\n\n<p>Now the Queen of hearts:<\/p>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\( {1 \\choose 1} \\).<\/div>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\"><\/div>\n\n\n\n<p> There are three queens remaining, so the number of possibilities on drawing one further queen is:<\/p>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\( {3 \\choose 1} \\).<\/div>\n\n\n\n<p>Followed by two cards that are neither a king nor a remaining queen (there are in total 44 non kings and queens in the set): <\/p>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\( {44 \\choose 2} \\).<\/div>\n\n\n\n<p>It should be noted that 4 + 1 + 3 + 44 = 52 and 1 + 1 + 1 + 2 = 5. The total number of possible combinations is as previously:<\/p>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\( {52 \\choose 5} \\).<\/div>\n\n\n\n<p>So, the probability of drawing any one king, the queen of hearts and a further queen in five cards out of 52 is:<\/p>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\(P = \\frac{{4 \\choose 1} \\cdot {1 \\choose 1} \\cdot {3 \\choose 1} \\cdot {44 \\choose 2}}{52 \\choose 5} \\)<\/div>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\(P = \\frac{\\frac{4!}{3! \\cdot 1!} \\cdot  1 \\cdot \\frac{3!}{2! \\cdot 1!} \\cdot \\frac{44!}{42! \\cdot 2!}}{\\frac{52!}{47! \\cdot 5!}} =\n\\frac{4! \\cdot 3! \\cdot 44! \\cdot 47! \\cdot 5!}{3! \\cdot 2! \\cdot 42! \\cdot 2! \\cdot 52!}\n \\)<\/div>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\(P = \\frac{44 \\cdot 43 \\cdot 5 \\cdot 4 \\cdot 3 \\cdot 4 \\cdot 3}{52 \\cdot 51 \\cdot 50 \\cdot 49 \\cdot 48} = \\frac{1362240}{311875200} \\approx 0.0044 \\approx 0.4\\% \\)<\/div>\n\n\n\n<p><\/p>\n\n\n\n<p>&nbsp;<\/p>\n\n\n\n<p>&nbsp;<\/p>\n\n\n\n<p>&nbsp;<\/p>\n","protected":false},"excerpt":{"rendered":"<p>When flipping a coin, it seems obvious that the chance (or probability) of head is \u00bd and the chance of tail is also \u00bd. However, what is the probability of two times head in succession? One might think that the probability of this happening is \u00bd + \u00bd = 1. However, if the probability is [&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-932","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/pcool.dyndns.org\/index.php\/wp-json\/wp\/v2\/pages\/932","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=932"}],"version-history":[{"count":17,"href":"https:\/\/pcool.dyndns.org\/index.php\/wp-json\/wp\/v2\/pages\/932\/revisions"}],"predecessor-version":[{"id":4483,"href":"https:\/\/pcool.dyndns.org\/index.php\/wp-json\/wp\/v2\/pages\/932\/revisions\/4483"}],"wp:attachment":[{"href":"https:\/\/pcool.dyndns.org\/index.php\/wp-json\/wp\/v2\/media?parent=932"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}