{"id":2102,"date":"2021-08-03T17:45:23","date_gmt":"2021-08-03T21:45:23","guid":{"rendered":"https:\/\/openbooks.macewan.ca\/rcommander\/?post_type=chapter&#038;p=2102"},"modified":"2024-06-06T17:20:38","modified_gmt":"2024-06-06T21:20:38","slug":"4-2-probability-distribution-of-a-discrete-variable","status":"publish","type":"chapter","link":"https:\/\/openbooks.macewan.ca\/introstats\/chapter\/4-2-probability-distribution-of-a-discrete-variable\/","title":{"raw":"4.2 Probability Distribution of a Discrete Variable","rendered":"4.2 Probability Distribution of a Discrete Variable"},"content":{"raw":"The <strong>probability distribution<\/strong> of a discrete random variable [latex]X[\/latex] lists all possible values and their corresponding probabilities. In general, the probability distribution of a discrete variable is given in a table with two rows or two columns: one row (column) for the possible values [latex]x[\/latex] and the other row (column) for the corresponding probability of taking each value [latex]P(X = x)[\/latex]. A probability distribution has two important properties:\r\n<div class=\"textbox textbox--key-takeaways\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Key Fact: Two Important Properties of Probability Distribution<\/p>\r\n\r\n<\/header>\r\n<ul>\r\n \t<li>[latex] 0 \\leq P(X=x) \\leq 1[\/latex]<\/li>\r\n \t<li>[latex]\\sum_{\\text{all possible } x} P(X=x) = 1[\/latex]<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"textbox textbox--examples\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Example: Probability Distribution of a Discrete Variable<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n<ol>\r\n \t<li>Consider the chance experiment of flipping a balanced coin twice; the sample space is <strong><em>S <\/em><\/strong>= {HH, HT, TT, TH}. Let the random variable X = # of tails. Determine the probability distribution of X.\r\n<ul>\r\n \t<li>First, determine the possible values of [latex]X[\/latex]. If we flip a coin twice, we might observe zero tail (HH), one tail (HT, TH), and two tails (TT). Therefore, possible values are [latex]x = 0, 1, 2[\/latex].<\/li>\r\n \t<li>Next, determine the probabilities [latex]P(X=x), x = 0,1, 2[\/latex]. Since the coin is balanced, it follows that [latex]P(H)=P(T)=0.5[\/latex]. Moreover, the two flips are independent, so the special multiplication rule applies. Thus,\r\n<ul>\r\n \t<li>[latex]P(X=0)=P(HH)=P(H) \\times P(H)=0.5 \\times 0.5=0.25[\/latex].<\/li>\r\n \t<li>[latex]P(X=1)= P(HT \\text{ or }TH)=P(HT)+P(TH)=0.25+0.25=0.5[\/latex].<\/li>\r\n \t<li>[latex]P(X=2)=P(TT)=0.25[\/latex].<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ul>\r\n<p style=\"padding-left: 40px;\">Therefore, the probability distribution of X is<\/p>\r\n\r\n<div align=\"center\">\r\n<p style=\"text-align: center;\"><strong>Table 4.1<\/strong>: Probability Distribution of X=# of Heads<\/p>\r\n\r\n<table class=\"aligncenter first-col-border\" style=\"width: 80%; height: 30px;\" border=\"0\">\r\n<tbody>\r\n<tr style=\"height: 15px;\">\r\n<th style=\"width: 25%; height: 15px; text-align: center;\" scope=\"col\">[latex]x[\/latex]<\/th>\r\n<th style=\"width: 25%; height: 15px; text-align: center;\" scope=\"col\">0<\/th>\r\n<th style=\"width: 25%; height: 15px; text-align: center;\" scope=\"col\">1<\/th>\r\n<th style=\"width: 25%; height: 15px; text-align: center;\" scope=\"col\">2<\/th>\r\n<\/tr>\r\n<tr class=\"border-top\" style=\"height: 15px;\">\r\n<td style=\"width: 25%; height: 15px; text-align: center;\"><strong>[latex] P(X=x)[\/latex]<\/strong><\/td>\r\n<td style=\"width: 25%; height: 15px; text-align: center;\">0.25<\/td>\r\n<td style=\"width: 25%; height: 15px; text-align: center;\">0.5<\/td>\r\n<td style=\"width: 25%; height: 15px; text-align: center;\">0.25<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\n<p style=\"padding-left: 40px;\">Note that the sum of the probabilities is one. That is<\/p>\r\n<p align=\"center\">[latex]\\begin{align*} \\sum P(X=x) &amp;= P(X=0) + P(X=1) + P(X=2) \\\\\r\n&amp;= 0.25 + 0.5 + 0.25 = 1. \\end{align*}[\/latex]<\/p>\r\n<\/li>\r\n \t<li>A population consists of five students: Mark has no siblings, John has one sibling, both Rebecca and Sarah have two siblings, and Mary has three. Randomly pick one student and let [latex]X[\/latex] be the number of siblings the student has. Determine the probability distribution of [latex]X[\/latex].\r\n<ul>\r\n \t<li>First, observe that the possible values of [latex]X[\/latex] are [latex]x=0, 1, 2, 3[\/latex]<\/li>\r\n \t<li>Next, determine the probabilities [latex]P(X=x), x=0, 1, 2, 3[\/latex]. One student has no siblings, two students have two siblings, and one student has three siblings. Thus,\r\n<ul>\r\n \t<li>[latex]P(X=0) = \\frac{f}{N} = \\frac{1}{5} = 0.2[\/latex].<\/li>\r\n \t<li>[latex]P(X=1) = \\frac{f}{N} = \\frac{1}{5} = 0.2[\/latex].<\/li>\r\n \t<li>[latex]P(X=2) = \\frac{f}{N} = \\frac{2}{5} = 0.4[\/latex].<\/li>\r\n \t<li>[latex]P(X=3) = \\frac{f}{N} = \\frac{1}{5} = 0.2[\/latex].<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ul>\r\n<p style=\"padding-left: 40px;\">The probability distribution of X is<\/p>\r\n<p style=\"text-align: center;\"><strong>Table 4.2<\/strong>: Probability Distribution of X=# of Siblings<\/p>\r\n<\/li>\r\n<\/ol>\r\n<table class=\"aligncenter first-col-border\" style=\"width: 80%;\" border=\"0\">\r\n<tbody>\r\n<tr>\r\n<td style=\"width: 20%; text-align: center;\"><strong>[latex]x[\/latex]<\/strong><\/td>\r\n<td style=\"width: 20%; text-align: center;\">0<\/td>\r\n<td style=\"width: 20%; text-align: center;\">1<\/td>\r\n<td style=\"width: 20%; text-align: center;\">2<\/td>\r\n<td style=\"width: 20%; text-align: center;\">3<\/td>\r\n<\/tr>\r\n<tr class=\"border-top\">\r\n<td style=\"width: 20%; text-align: center;\"><strong>[latex] P(X=x)[\/latex]<\/strong><\/td>\r\n<td style=\"width: 20%; text-align: center;\">0.2<\/td>\r\n<td style=\"width: 20%; text-align: center;\">0.2<\/td>\r\n<td style=\"width: 20%; text-align: center;\">0.4<\/td>\r\n<td style=\"width: 20%; text-align: center;\">0.2<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p style=\"padding-left: 40px;\">Note that the sum of the probabilities is one. That is<\/p>\r\n[latex]\\begin{align*}\\sum P(X=x) &amp;= P(X=0)+P(X=1)+P(X=2)+P(X=3)\\\\\r\n&amp;= 0.2+0.2+0.4+0.2=1. \\end{align*}[\/latex]\r\n\r\nA probability distribution can also be represented as a histogram. Every possible value should have a separate bar for a discrete random variable.<a id=\"retfig4.3\"><\/a>\r\n\r\n&nbsp;\r\n\r\n[caption id=\"attachment_2755\" align=\"aligncenter\" width=\"300\"]<img class=\"wp-image-2755 size-medium\" src=\"https:\/\/openbooks.macewan.ca\/introstats\/wp-content\/uploads\/sites\/8\/2021\/08\/hist_discrete_crop-300x287.png\" alt=\"A probability histogram of number of siblings. The highest probability is 2. Image description available.\" width=\"300\" height=\"287\" \/> <strong>Figure 4.3<\/strong>: Histogram of # of Siblings [<a href=\"https:\/\/openbooks.macewan.ca\/introstats\/back-matter\/image-description\/#fig4.3\">Image Description (See Appendix D Figure 4.3)<\/a>][\/caption]<\/div>\r\n<\/div>","rendered":"<p>The <strong>probability distribution<\/strong> of a discrete random variable [latex]X[\/latex] lists all possible values and their corresponding probabilities. In general, the probability distribution of a discrete variable is given in a table with two rows or two columns: one row (column) for the possible values [latex]x[\/latex] and the other row (column) for the corresponding probability of taking each value [latex]P(X = x)[\/latex]. A probability distribution has two important properties:<\/p>\n<div class=\"textbox textbox--key-takeaways\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Key Fact: Two Important Properties of Probability Distribution<\/p>\n<\/header>\n<ul>\n<li>[latex]0 \\leq P(X=x) \\leq 1[\/latex]<\/li>\n<li>[latex]\\sum_{\\text{all possible } x} P(X=x) = 1[\/latex]<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox textbox--examples\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Example: Probability Distribution of a Discrete Variable<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<ol>\n<li>Consider the chance experiment of flipping a balanced coin twice; the sample space is <strong><em>S <\/em><\/strong>= {HH, HT, TT, TH}. Let the random variable X = # of tails. Determine the probability distribution of X.\n<ul>\n<li>First, determine the possible values of [latex]X[\/latex]. If we flip a coin twice, we might observe zero tail (HH), one tail (HT, TH), and two tails (TT). Therefore, possible values are [latex]x = 0, 1, 2[\/latex].<\/li>\n<li>Next, determine the probabilities [latex]P(X=x), x = 0,1, 2[\/latex]. Since the coin is balanced, it follows that [latex]P(H)=P(T)=0.5[\/latex]. Moreover, the two flips are independent, so the special multiplication rule applies. Thus,\n<ul>\n<li>[latex]P(X=0)=P(HH)=P(H) \\times P(H)=0.5 \\times 0.5=0.25[\/latex].<\/li>\n<li>[latex]P(X=1)= P(HT \\text{ or }TH)=P(HT)+P(TH)=0.25+0.25=0.5[\/latex].<\/li>\n<li>[latex]P(X=2)=P(TT)=0.25[\/latex].<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<p style=\"padding-left: 40px;\">Therefore, the probability distribution of X is<\/p>\n<div style=\"margin: auto;\">\n<p style=\"text-align: center;\"><strong>Table 4.1<\/strong>: Probability Distribution of X=# of Heads<\/p>\n<table class=\"aligncenter first-col-border\" style=\"width: 80%; height: 30px;\">\n<tbody>\n<tr style=\"height: 15px;\">\n<th style=\"width: 25%; height: 15px; text-align: center;\" scope=\"col\">[latex]x[\/latex]<\/th>\n<th style=\"width: 25%; height: 15px; text-align: center;\" scope=\"col\">0<\/th>\n<th style=\"width: 25%; height: 15px; text-align: center;\" scope=\"col\">1<\/th>\n<th style=\"width: 25%; height: 15px; text-align: center;\" scope=\"col\">2<\/th>\n<\/tr>\n<tr class=\"border-top\" style=\"height: 15px;\">\n<td style=\"width: 25%; height: 15px; text-align: center;\"><strong>[latex]P(X=x)[\/latex]<\/strong><\/td>\n<td style=\"width: 25%; height: 15px; text-align: center;\">0.25<\/td>\n<td style=\"width: 25%; height: 15px; text-align: center;\">0.5<\/td>\n<td style=\"width: 25%; height: 15px; text-align: center;\">0.25<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<p style=\"padding-left: 40px;\">Note that the sum of the probabilities is one. That is<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align*} \\sum P(X=x) &= P(X=0) + P(X=1) + P(X=2) \\\\  &= 0.25 + 0.5 + 0.25 = 1. \\end{align*}[\/latex]<\/p>\n<\/li>\n<li>A population consists of five students: Mark has no siblings, John has one sibling, both Rebecca and Sarah have two siblings, and Mary has three. Randomly pick one student and let [latex]X[\/latex] be the number of siblings the student has. Determine the probability distribution of [latex]X[\/latex].\n<ul>\n<li>First, observe that the possible values of [latex]X[\/latex] are [latex]x=0, 1, 2, 3[\/latex]<\/li>\n<li>Next, determine the probabilities [latex]P(X=x), x=0, 1, 2, 3[\/latex]. One student has no siblings, two students have two siblings, and one student has three siblings. Thus,\n<ul>\n<li>[latex]P(X=0) = \\frac{f}{N} = \\frac{1}{5} = 0.2[\/latex].<\/li>\n<li>[latex]P(X=1) = \\frac{f}{N} = \\frac{1}{5} = 0.2[\/latex].<\/li>\n<li>[latex]P(X=2) = \\frac{f}{N} = \\frac{2}{5} = 0.4[\/latex].<\/li>\n<li>[latex]P(X=3) = \\frac{f}{N} = \\frac{1}{5} = 0.2[\/latex].<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<p style=\"padding-left: 40px;\">The probability distribution of X is<\/p>\n<p style=\"text-align: center;\"><strong>Table 4.2<\/strong>: Probability Distribution of X=# of Siblings<\/p>\n<\/li>\n<\/ol>\n<table class=\"aligncenter first-col-border\" style=\"width: 80%;\">\n<tbody>\n<tr>\n<td style=\"width: 20%; text-align: center;\"><strong>[latex]x[\/latex]<\/strong><\/td>\n<td style=\"width: 20%; text-align: center;\">0<\/td>\n<td style=\"width: 20%; text-align: center;\">1<\/td>\n<td style=\"width: 20%; text-align: center;\">2<\/td>\n<td style=\"width: 20%; text-align: center;\">3<\/td>\n<\/tr>\n<tr class=\"border-top\">\n<td style=\"width: 20%; text-align: center;\"><strong>[latex]P(X=x)[\/latex]<\/strong><\/td>\n<td style=\"width: 20%; text-align: center;\">0.2<\/td>\n<td style=\"width: 20%; text-align: center;\">0.2<\/td>\n<td style=\"width: 20%; text-align: center;\">0.4<\/td>\n<td style=\"width: 20%; text-align: center;\">0.2<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p style=\"padding-left: 40px;\">Note that the sum of the probabilities is one. That is<\/p>\n<p>[latex]\\begin{align*}\\sum P(X=x) &= P(X=0)+P(X=1)+P(X=2)+P(X=3)\\\\  &= 0.2+0.2+0.4+0.2=1. \\end{align*}[\/latex]<\/p>\n<p>A probability distribution can also be represented as a histogram. Every possible value should have a separate bar for a discrete random variable.<a id=\"retfig4.3\"><\/a><\/p>\n<p>&nbsp;<\/p>\n<figure id=\"attachment_2755\" aria-describedby=\"caption-attachment-2755\" style=\"width: 300px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-2755 size-medium\" src=\"https:\/\/openbooks.macewan.ca\/introstats\/wp-content\/uploads\/sites\/8\/2021\/08\/hist_discrete_crop-300x287.png\" alt=\"A probability histogram of number of siblings. The highest probability is 2. Image description available.\" width=\"300\" height=\"287\" srcset=\"https:\/\/openbooks.macewan.ca\/introstats\/wp-content\/uploads\/sites\/8\/2021\/08\/hist_discrete_crop-300x287.png 300w, https:\/\/openbooks.macewan.ca\/introstats\/wp-content\/uploads\/sites\/8\/2021\/08\/hist_discrete_crop-1024x978.png 1024w, https:\/\/openbooks.macewan.ca\/introstats\/wp-content\/uploads\/sites\/8\/2021\/08\/hist_discrete_crop-768x734.png 768w, https:\/\/openbooks.macewan.ca\/introstats\/wp-content\/uploads\/sites\/8\/2021\/08\/hist_discrete_crop-1536x1468.png 1536w, https:\/\/openbooks.macewan.ca\/introstats\/wp-content\/uploads\/sites\/8\/2021\/08\/hist_discrete_crop-2048x1957.png 2048w, https:\/\/openbooks.macewan.ca\/introstats\/wp-content\/uploads\/sites\/8\/2021\/08\/hist_discrete_crop-65x62.png 65w, https:\/\/openbooks.macewan.ca\/introstats\/wp-content\/uploads\/sites\/8\/2021\/08\/hist_discrete_crop-225x215.png 225w, https:\/\/openbooks.macewan.ca\/introstats\/wp-content\/uploads\/sites\/8\/2021\/08\/hist_discrete_crop-350x334.png 350w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><figcaption id=\"caption-attachment-2755\" class=\"wp-caption-text\"><strong>Figure 4.3<\/strong>: Histogram of # of Siblings [<a href=\"https:\/\/openbooks.macewan.ca\/introstats\/back-matter\/image-description\/#fig4.3\">Image Description (See Appendix D Figure 4.3)<\/a>]<\/figcaption><\/figure>\n<\/div>\n<\/div>\n","protected":false},"author":19,"menu_order":2,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-2102","chapter","type-chapter","status-publish","hentry"],"part":535,"_links":{"self":[{"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/pressbooks\/v2\/chapters\/2102","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/wp\/v2\/users\/19"}],"version-history":[{"count":34,"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/pressbooks\/v2\/chapters\/2102\/revisions"}],"predecessor-version":[{"id":5379,"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/pressbooks\/v2\/chapters\/2102\/revisions\/5379"}],"part":[{"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/pressbooks\/v2\/parts\/535"}],"metadata":[{"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/pressbooks\/v2\/chapters\/2102\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/wp\/v2\/media?parent=2102"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/pressbooks\/v2\/chapter-type?post=2102"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/wp\/v2\/contributor?post=2102"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/wp\/v2\/license?post=2102"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}