{"id":842015,"date":"2024-09-11T13:44:15","date_gmt":"2024-09-11T08:14:15","guid":{"rendered":"https:\/\/leverageedu.com\/discover\/?p=842015"},"modified":"2024-09-11T13:44:15","modified_gmt":"2024-09-11T08:14:15","slug":"exam-prep-binomial-distribution","status":"publish","type":"post","link":"https:\/\/leverageedu.com\/discover\/indian-exams\/exam-prep-binomial-distribution\/","title":{"rendered":"Binomial Distribution: Definition, Properties and Solved Examples"},"content":{"rendered":"\n<p>The binomial distribution is a discrete probability distribution that models the number of successes in a fixed number of independent trials, each with the same probability of success. It is defined by two parameters: the number of trials (n) and the probability of success in each trial (p). The formula to calculate the probability of exactly k successes is <strong>P(X=k) = (<sup>n<\/sup><sub>k<\/sub>)p<sup>k<\/sup>(1\u2212p)<sup>n-k<\/sup><\/strong>. Key properties include mean (np) and variance (npq, where q = 1 &#8211; p). This distribution is fundamental for analyzing binary outcomes and is frequently tested in competitive exams like <strong><a href=\"https:\/\/leverageedu.com\/discover\/indian-exams\/how-to-prepare-for-jee-main\/\">IIT-JEE<\/a><\/strong>, <strong>GATE<\/strong>, <strong><a href=\"https:\/\/leverageedu.com\/tests\/gre\/\">GRE<\/a><\/strong>, and <strong><a href=\"https:\/\/leverageedu.com\/discover\/indian-exams\/management-cat-exam\/\">CAT<\/a><\/strong>, where solving related problems helps understand real-life applications like quality control and decision-making under uncertainty.<\/p>\n\n\n\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-what-is-binomial-distribution\">What is Binomial Distribution?<\/h2>\n\n\n\n<p>The binomial distribution is a discrete probability distribution that models the number of successes in a fixed number of independent and identical trials, where each trial has only two possible outcomes: success or failure. The probability of success remains constant across trials, and each trial is independent of others. The distribution is used to calculate the likelihood of achieving a specific number of successes in a given set of trials. Important terms related to binomial distribution are:<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Trial<\/strong>: A single event with two possible outcomes (success\/failure).<\/li>\n\n\n\n<li><strong>Success (p)<\/strong>: The probability of the desired outcome in a single trial.<\/li>\n\n\n\n<li><strong>Failure (q)<\/strong>: The probability of the undesired outcome, where q=1\u2212p.<\/li>\n\n\n\n<li><strong>Number of Trials (n)<\/strong>: The total number of independent trials or experiments conducted.<\/li>\n\n\n\n<li><strong>Number of Successes (k)<\/strong>: The number of successful outcomes in the trials.<\/li>\n\n\n\n<li><strong>Binomial Coefficient<\/strong> <strong>(<\/strong><strong><sup>n<\/sup><\/strong><strong><sub>k<\/sub><\/strong><strong>)<\/strong>: A combination that calculates the number of ways to choose k successes from n trials, represented as <strong>n!\/k!(n\u2212k)!<\/strong>\u200b.<\/li>\n\n\n\n<li><strong>Probability Mass Function (PMF)<\/strong>: The formula used to calculate the probability of exactly k successes: <strong>P(X=k) = (<\/strong><strong><sup>n<\/sup><\/strong><strong><sub>k<\/sub><\/strong><strong>)p<\/strong><strong><sup>k<\/sup><\/strong><strong>(1\u2212p)<\/strong><strong><sup>n-k<\/sup><\/strong>.<\/li>\n\n\n\n<li><strong>Mean (np)<\/strong>: The expected number of successes in n trials.<\/li>\n\n\n\n<li><strong>Variance (npq)<\/strong>: A measure of the spread or variability in the number of successes.<\/li>\n<\/ol>\n\n\n\n<p class=\"has-text-align-center has-electric-grass-gradient-background has-background has-medium-font-size\"><strong>Must Read: <a href=\"https:\/\/leverageedu.com\/discover\/indian-exams\/exam-prep-commercial-maths\/\">Commercial Maths<\/a><\/strong><\/p>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-properties-of-binomial-distribution\">Properties of Binomial Distribution<\/h2>\n\n\n\n<p>The <strong>binomial distribution<\/strong> has several key properties that describe its behavior and characteristics. These properties are crucial in understanding the distribution&#8217;s shape, central tendency, and variability<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Two Possible Outcomes<\/strong>: Each trial has only two possible outcomes: success (with probability p) and failure (with probability q=1\u2212p).<\/li>\n\n\n\n<li><strong>Fixed Number of Trials (n)<\/strong>: The number of trials, n, is fixed and predetermined. Each trial is independent of others, meaning the outcome of one trial does not affect the others.<\/li>\n\n\n\n<li><strong>Constant Probability (p)<\/strong>: The probability of success, p, remains the same for all trials.<\/li>\n\n\n\n<li><strong>Discrete Distribution<\/strong>: The binomial distribution is discrete, meaning it takes only integer values (the number of successes) ranging from 0 to n.<\/li>\n\n\n\n<li><strong>Mean (Expected Value)<\/strong>: The mean or expected value of the binomial distribution is given by:<br><strong>\u03bc = np<\/strong><br>This represents the average number of successes expected in n trials.<\/li>\n\n\n\n<li><strong>Variance<\/strong>: The variance, which measures the spread of the distribution, is:<br><strong>\u03c3<sup>2 <\/sup>= npq<\/strong><br>where q = 1\u2212p is the probability of failure. This shows how much the number of successes is expected to deviate from the mean.<\/li>\n\n\n\n<li><strong>Standard Deviation<\/strong>: The standard deviation is the square root of the variance: <strong>\u03c3 = \u221anpq<\/strong><\/li>\n\n\n\n<li><strong>Symmetry and Skewness<\/strong>: The binomial distribution becomes more symmetric as n increases and when p is close to 0.5. If p is much greater than 0.5, the distribution is skewed to the left (negatively skewed), while if p is much less than 0.5, it is skewed to the right (positively skewed).<\/li>\n\n\n\n<li><strong>Maximum Probability<\/strong>: The most probable value (the mode) of the binomial distribution is around k = np (the mean), but it may vary slightly depending on whether p is closer to 0 or 1.<\/li>\n\n\n\n<li><strong>Relationship with Normal Distribution<\/strong>: For large n, the binomial distribution approaches a normal distribution, especially when p is close to 0.5. This is known as the <strong>normal approximation<\/strong> to the binomial distribution, often applied when np \u2265 5 and nq \u2265 5.<\/li>\n<\/ol>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-formula-of-binomial-distribution\">Formula of Binomial Distribution<\/h2>\n\n\n\n<figure class=\"wp-block-embed is-type-video is-provider-youtube wp-block-embed-youtube wp-embed-aspect-16-9 wp-has-aspect-ratio\"><div class=\"wp-block-embed__wrapper\">\n<iframe loading=\"lazy\" title=\"Binomial distribution | Probability and Statistics | Khan Academy\" width=\"1200\" height=\"675\" src=\"https:\/\/www.youtube.com\/embed\/WWv0RUxDfbs?feature=oembed\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture; web-share\" referrerpolicy=\"strict-origin-when-cross-origin\" allowfullscreen><\/iframe>\n<\/div><figcaption class=\"wp-element-caption\">Source: Khan Academy<\/figcaption><\/figure>\n\n\n\n<p>The formula of the binomial distribution gives the probability of obtaining exactly <strong>k<\/strong> successes in <strong>n<\/strong> independent trials, where each trial has two possible outcomes (success or failure) and the probability of success is <strong>p<\/strong>.<\/p>\n\n\n\n<p><strong>Binomial Distribution Formula:<\/strong><\/p>\n\n\n\n<p><strong>P(X=k) = (<\/strong><strong><sup>n<\/sup><\/strong><strong><sub>k<\/sub><\/strong><strong>)p<\/strong><strong><sup>k<\/sup><\/strong><strong>(1\u2212p)<\/strong><strong><sup>n-k<\/sup><\/strong><\/p>\n\n\n\n<p>Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>P(X=k): The probability of getting exactly <strong>k<\/strong> successes in <strong>n<\/strong> trials.<\/li>\n\n\n\n<li>n: The total number of trials.<\/li>\n\n\n\n<li>k: The number of successes.<\/li>\n\n\n\n<li>p: The probability of success in a single trial.<\/li>\n\n\n\n<li>1\u2212p or q: The probability of failure in a single trial.<\/li>\n\n\n\n<li>(<strong><sup>n<\/sup><\/strong><strong><sub>k<\/sub><\/strong>): The binomial coefficient, representing the number of ways to choose <strong>k<\/strong> successes from <strong>n<\/strong> trials, is calculated as:&nbsp;<\/li>\n<\/ul>\n\n\n\n<p><strong>(<\/strong><strong><sup>n<\/sup><\/strong><strong><sub>k<\/sub><\/strong><strong>) = n!\/k!(n\u2212k)!\u200b&nbsp;<\/strong><\/p>\n\n\n\n<p>where n! is the factorial of n.<\/p>\n\n\n\n<p><strong>Explanation:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The term p<sup>k<\/sup> represents the probability of getting k successes.<\/li>\n\n\n\n<li>The term (1\u2212p)<sup>n-k<\/sup> represents the probability of getting n &#8211; k failures.<\/li>\n\n\n\n<li>The binomial coefficient (<sup>n<\/sup><sub>k<\/sub>) accounts for the different combinations in which the k successes can occur in n trials.<\/li>\n<\/ul>\n\n\n\n<p class=\"has-text-align-center has-electric-grass-gradient-background has-background has-medium-font-size\"><strong>Must Read:<\/strong> <strong><a href=\"https:\/\/leverageedu.com\/discover\/indian-exams\/exam-prep-relations-and-functions\/\">Relations and Functions<\/a><\/strong><\/p>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-solved-examples-of-binomial-distribution\">Solved Examples of Binomial Distribution<\/h2>\n\n\n\n<p>Here are 5 solved examples of binomial distribution to illustrate its application:<\/p>\n\n\n\n<p><strong>Example 1: Tossing a Coin<\/strong><\/p>\n\n\n\n<p>You toss a fair coin 5 times. What is the probability of getting exactly 3 heads?<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Solution<\/strong>:\n<ul class=\"wp-block-list\">\n<li>Number of trials n = 5<\/li>\n\n\n\n<li>Probability of success (head) p=0.5<\/li>\n\n\n\n<li>Probability of failure (tail) q = 1\u2212p = 0.5<\/li>\n\n\n\n<li>Number of successes k = 3<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n\n\n\n<p>Using the binomial distribution formula:<\/p>\n\n\n\n<p>P(X=3) = (<sup>5<\/sup><sub>3<\/sub>)(0.5)<sup>3<\/sup>(0.5)<sup>5-3<\/sup>&nbsp;<\/p>\n\n\n\n<p>P(X=3) = {5!\/3!(5\u22123)!}(0.5)<sup>5 <\/sup>= 10\u00d70.03125 = 0.3125<\/p>\n\n\n\n<p>So, the probability of getting exactly 3 heads is <strong>0.3125<\/strong>.<\/p>\n\n\n\n<p><strong>Example 2: Quality Control<\/strong><\/p>\n\n\n\n<p>A factory produces light bulbs, and 95% of them are good. If you randomly select 10 bulbs, what is the probability that exactly 9 of them are good?<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Solution<\/strong>:\n<ul class=\"wp-block-list\">\n<li>Number of trials n = 10<\/li>\n\n\n\n<li>Probability of success (good bulb) p = 0.95<\/li>\n\n\n\n<li>Probability of failure (bad bulb) q = 0.05<\/li>\n\n\n\n<li>Number of successes k = 9<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n\n\n\n<p>Using the formula:<\/p>\n\n\n\n<p>P(X=9) = (<sup>10<\/sup><sub>9<\/sub>)(0.95)<sup>9<\/sup>(0.05)<sup>1<\/sup>&nbsp;<\/p>\n\n\n\n<p>P(X=9) = 10\u00d7(0.95)9\u00d7(0.05)<\/p>\n\n\n\n<p>P(X=9) \u2248 10\u00d70.6302\u00d70.05 = 0.3151<\/p>\n\n\n\n<p>So, the probability of exactly 9 good bulbs is <strong>0.3151<\/strong>.<\/p>\n\n\n\n<p><strong>Example 3: Exam Questions<\/strong><\/p>\n\n\n\n<p>In a multiple-choice exam, each question has 4 options, with only one correct answer. If you randomly guess on 6 questions, what is the probability of getting exactly 2 correct answers?<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Solution<\/strong>:\n<ul class=\"wp-block-list\">\n<li>Number of trials n = 6<\/li>\n\n\n\n<li>Probability of success (correct answer) p = \u00bc = 0.25<\/li>\n\n\n\n<li>Probability of failure (wrong answer) q = 0.75<\/li>\n\n\n\n<li>Number of successes k = 2<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n\n\n\n<p>Using the formula:<\/p>\n\n\n\n<p>P(X=2) = (<sup>6<\/sup><sub>2<\/sub>)(0.25)<sup>2<\/sup>(0.75)<sup>4<\/sup>&nbsp;<\/p>\n\n\n\n<p>P(X=2) = {6!\/2!(6\u22122)!}\u00d7(0.25)<sup>2<\/sup>\u00d7(0.75)<sup>4 <\/sup>= 15\u00d70.0625\u00d70.3164P&nbsp;<\/p>\n\n\n\n<p>P(X=2) \u2248 15\u00d70.01977 = 0.2966<\/p>\n\n\n\n<p>The probability of getting exactly 2 correct answers is <strong>0.2966<\/strong>.<\/p>\n\n\n\n<p><strong>Example 4: Defective Products<\/strong><\/p>\n\n\n\n<p>A shipment contains 12 items, and there is a 20% chance that any item is defective. If you randomly check 8 items, what is the probability that exactly 2 are defective?<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Solution<\/strong>:\n<ul class=\"wp-block-list\">\n<li>Number of trials n = 8<\/li>\n\n\n\n<li>Probability of success (defective item) p = 0.2<\/li>\n\n\n\n<li>Probability of failure (non-defective) q = 0.8<\/li>\n\n\n\n<li>Number of successes k = 2<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n\n\n\n<p>Using the formula:<\/p>\n\n\n\n<p>P(X=2) = (<sup>8<\/sup><sub>2<\/sub>)(0.2)<sup>2<\/sup>(0.8)<sup>6<\/sup>&nbsp;<\/p>\n\n\n\n<p>P(X=2) = 28\u00d7(0.2)<sup>2<\/sup>\u00d7(0.8)<sup>6 <\/sup>= 28\u00d70.04\u00d70.2621&nbsp;<\/p>\n\n\n\n<p>P(X=2) \u2248 28\u00d70.010484 = 0.2936<\/p>\n\n\n\n<p>The probability of finding exactly 2 defective items is <strong>0.2936<\/strong>.<\/p>\n\n\n\n<p><strong>Example 5: Drug Trial<\/strong><\/p>\n\n\n\n<p>In a drug trial, a new treatment has a 70% success rate. If 10 patients are treated, what is the probability that exactly 7 patients will recover?<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Solution<\/strong>:\n<ul class=\"wp-block-list\">\n<li>Number of trials n = 10<\/li>\n\n\n\n<li>Probability of success (recovery) p = 0.7<\/li>\n\n\n\n<li>Probability of failure (no recovery) q = 0.3<\/li>\n\n\n\n<li>Number of successes k = 7<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n\n\n\n<p>Using the formula:<\/p>\n\n\n\n<p>P(X=7) = (<sup>10<\/sup><sub>7<\/sub>)(0.7)<sup>7<\/sup>(0.3)<sup>3<\/sup>&nbsp;<\/p>\n\n\n\n<p>P(X=7) = {10!7!(10\u22127)!}\u00d7(0.7)<sup>7<\/sup>\u00d7(0.3)<sup>3 <\/sup>= 120\u00d70.0823543\u00d70.027&nbsp;<\/p>\n\n\n\n<p>P(X=7) \u2248 120\u00d70.002224 = 0.2669<\/p>\n\n\n\n<p>The probability of exactly 7 patients recovering is <strong>0.2669<\/strong>.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-faqs\">FAQs<\/h2>\n\n\n\n<div class=\"schema-faq wp-block-yoast-faq-block\"><div class=\"schema-faq-section\" id=\"faq-question-1726041263691\"><strong class=\"schema-faq-question\"><strong>What are the 4 factors for binomial distribution?<\/strong><\/strong> <p class=\"schema-faq-answer\">The four factors for binomial distribution are:<br\/>&#8212; <strong>Fixed number of trials (n):<\/strong> The experiment must have a predetermined number of trials.<br\/><strong>&#8211;<\/strong>&#8211; I<strong>ndependent trials:<\/strong> Each trial must be independent of the others.<br\/>&#8212; <strong>Two possible outcomes:<\/strong> Each trial must have only two possible outcomes (success or failure).<br\/>&#8212; <strong>Constant probability of success (p):<\/strong> The probability of success must remain the same for each trial.<\/p> <\/div> <div class=\"schema-faq-section\" id=\"faq-question-1726041331202\"><strong class=\"schema-faq-question\">Why is it called binomial distribution?<\/strong> <p class=\"schema-faq-answer\">Binomial distribution is called so because it deals with experiments having only two possible outcomes &#8211; success or failure, and the probability of success remains constant across trials.<\/p> <\/div> <div class=\"schema-faq-section\" id=\"faq-question-1726041494682\"><strong class=\"schema-faq-question\">What is the full formula of binomial distribution?<\/strong> <p class=\"schema-faq-answer\">The full formula of the binomial distribution is: P(X=k) = (nk)pk(1\u2212p)n-k<\/p> <\/div> <\/div>\n\n\n\n<p class=\"has-text-align-center has-vivid-red-color has-text-color has-link-color has-medium-font-size wp-elements-adc4b1619be72b19c4386dda21667032\"><strong>RELATED BLOGS<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-table is-style-stripes\"><table><tbody><tr><td class=\"has-text-align-center\" data-align=\"center\"><a href=\"https:\/\/leverageedu.com\/discover\/indian-exams\/exam-prep-elementary-linear-algebra\/\"><strong>Elementary Linear Algebra<\/strong><\/a><br><\/td><td class=\"has-text-align-center\" data-align=\"center\"><a href=\"https:\/\/leverageedu.com\/discover\/indian-exams\/exam-prep-factorisation-method\/\"><strong>Factorisation Method<\/strong><\/a><\/td><\/tr><tr><td class=\"has-text-align-center\" data-align=\"center\"><a href=\"https:\/\/leverageedu.com\/discover\/indian-exams\/exam-prep-how-to-solve-fraction-equations\/\"><strong>Fraction Equations<\/strong><\/a><br><\/td><td class=\"has-text-align-center\" data-align=\"center\"><a href=\"https:\/\/leverageedu.com\/discover\/indian-exams\/exam-prep-how-to-solve-equations-with-variables-on-both-sides\/\"><strong>Equations With Variables on Both Sides<\/strong><\/a><\/td><\/tr><tr><td class=\"has-text-align-center\" data-align=\"center\"><a href=\"https:\/\/leverageedu.com\/discover\/indian-exams\/exam-prep-heights-and-distances\/\"><strong>Heights and Distances<\/strong><\/a><\/td><td class=\"has-text-align-center\" data-align=\"center\"><a href=\"https:\/\/leverageedu.com\/discover\/indian-exams\/exam-prep-additive-inverse\/\"><strong>Additive Inverse<\/strong><\/a><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p>This was all about the \u201c<strong>Binomial Distribution<\/strong>\u201d. For more such informative blogs, check out our<a href=\"https:\/\/leverageedu.com\/discover\/category\/indian-exams\/study-material\/\"> Study Material Section<\/a>, or you can learn more about us by visiting our<a href=\"https:\/\/leverageedu.com\/discover\/category\/indian-exams\/\">&nbsp; Indian exams<\/a> page.<\/p>\n","protected":false},"excerpt":{"rendered":"The binomial distribution is a discrete probability distribution that models the number of successes in a fixed number&hellip;\n","protected":false},"author":115,"featured_media":842020,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_acf_changed":false,"editor_notices":[],"footnotes":""},"categories":[369,396],"tags":[],"class_list":{"0":"post-842015","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-indian-exams","8":"category-study-material"},"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v27.3 (Yoast SEO v27.3) - 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I find peace and purpose in crafting verses that dance between the lines of poetry. With my pen as my wand, I weave intricate tales and heartfelt musings, breathing life into the blank canvas of each page. Blogging is my window to the world way of sharing thoughts, emotions, and a perspective uniquely my own. Every word I write is a brushstroke in the ever-evolving painting of my literary journey.","sameAs":["https:\/\/www.instagram.com\/xx_a.m.strings_xiv\/","https:\/\/www.linkedin.com\/in\/mohit-rajak-a9a5a2162\/"],"url":"https:\/\/leverageedu.com\/discover\/author\/mohit\/"},{"@type":"Question","@id":"https:\/\/leverageedu.com\/discover\/indian-exams\/exam-prep-binomial-distribution\/#faq-question-1726041263691","position":1,"url":"https:\/\/leverageedu.com\/discover\/indian-exams\/exam-prep-binomial-distribution\/#faq-question-1726041263691","name":"What are the 4 factors for binomial distribution?","answerCount":1,"acceptedAnswer":{"@type":"Answer","text":"The four factors for binomial distribution are:<br\/>-- <strong>Fixed number of trials (n):<\/strong> The experiment must have a predetermined number of trials.<br\/><strong>-<\/strong>- I<strong>ndependent trials:<\/strong> Each trial must be independent of the others.<br\/>-- <strong>Two possible outcomes:<\/strong> Each trial must have only two possible outcomes (success or failure).<br\/>-- <strong>Constant probability of success (p):<\/strong> The probability of success must remain the same for each trial.","inLanguage":"en-US"},"inLanguage":"en-US"},{"@type":"Question","@id":"https:\/\/leverageedu.com\/discover\/indian-exams\/exam-prep-binomial-distribution\/#faq-question-1726041331202","position":2,"url":"https:\/\/leverageedu.com\/discover\/indian-exams\/exam-prep-binomial-distribution\/#faq-question-1726041331202","name":"Why is it called binomial distribution?","answerCount":1,"acceptedAnswer":{"@type":"Answer","text":"Binomial distribution is called so because it deals with experiments having only two possible outcomes - success or failure, and the probability of success remains constant across trials.","inLanguage":"en-US"},"inLanguage":"en-US"},{"@type":"Question","@id":"https:\/\/leverageedu.com\/discover\/indian-exams\/exam-prep-binomial-distribution\/#faq-question-1726041494682","position":3,"url":"https:\/\/leverageedu.com\/discover\/indian-exams\/exam-prep-binomial-distribution\/#faq-question-1726041494682","name":"What is the full formula of binomial distribution?","answerCount":1,"acceptedAnswer":{"@type":"Answer","text":"The full formula of the binomial distribution is: P(X=k) = (nk)pk(1\u2212p)n-k","inLanguage":"en-US"},"inLanguage":"en-US"}]}},"acf":[],"_links":{"self":[{"href":"https:\/\/leverageedu.com\/discover\/wp-json\/wp\/v2\/posts\/842015","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/leverageedu.com\/discover\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/leverageedu.com\/discover\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/leverageedu.com\/discover\/wp-json\/wp\/v2\/users\/115"}],"replies":[{"embeddable":true,"href":"https:\/\/leverageedu.com\/discover\/wp-json\/wp\/v2\/comments?post=842015"}],"version-history":[{"count":0,"href":"https:\/\/leverageedu.com\/discover\/wp-json\/wp\/v2\/posts\/842015\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/leverageedu.com\/discover\/wp-json\/wp\/v2\/media\/842020"}],"wp:attachment":[{"href":"https:\/\/leverageedu.com\/discover\/wp-json\/wp\/v2\/media?parent=842015"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/leverageedu.com\/discover\/wp-json\/wp\/v2\/categories?post=842015"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/leverageedu.com\/discover\/wp-json\/wp\/v2\/tags?post=842015"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}