{"id":258698,"date":"2025-07-18T08:58:57","date_gmt":"2025-07-18T08:58:57","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=258698"},"modified":"2025-07-18T08:59:00","modified_gmt":"2025-07-18T08:59:00","slug":"assume-the-random-variable-x-has-a-binomial-distribution-with-the-given-probability-of-obtaining-a-success","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/18\/assume-the-random-variable-x-has-a-binomial-distribution-with-the-given-probability-of-obtaining-a-success\/","title":{"rendered":"Assume the random variable X has a binomial distribution with the given probability of obtaining a success"},"content":{"rendered":"\n<p>Assume the random variable X has a binomial distribution with the given probability of obtaining a success. Find the following probability, given the number of trials and the probability of obtaining a success. Round your answer to four decimal places. P(X9) , n=10 , p=0.8<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-0-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>To find the probability P(X=9)P(X = 9)P(X=9) for a binomial distribution, we use the binomial probability formula:P(X=k)=(nk)pk(1\u2212p)n\u2212kP(X = k) = \\binom{n}{k} p^k (1-p)^{n-k}P(X=k)=(kn\u200b)pk(1\u2212p)n\u2212k<\/p>\n\n\n\n<p>Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>nnn is the number of trials,<\/li>\n\n\n\n<li>kkk is the number of successes,<\/li>\n\n\n\n<li>ppp is the probability of success on a single trial,<\/li>\n\n\n\n<li>(nk)\\binom{n}{k}(kn\u200b) is the binomial coefficient, also known as &#8220;n choose k&#8221;.<\/li>\n<\/ul>\n\n\n\n<p>Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>n=10n = 10n=10 (the number of trials),<\/li>\n\n\n\n<li>p=0.8p = 0.8p=0.8 (the probability of success on a single trial),<\/li>\n\n\n\n<li>k=9k = 9k=9 (the number of successes we are interested in).<\/li>\n<\/ul>\n\n\n\n<p>Let&#8217;s break this down:<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Find the binomial coefficient<\/h3>\n\n\n\n<p>The binomial coefficient (nk)\\binom{n}{k}(kn\u200b) is calculated as:(nk)=n!k!(n\u2212k)!\\binom{n}{k} = \\frac{n!}{k!(n-k)!}(kn\u200b)=k!(n\u2212k)!n!\u200b<\/p>\n\n\n\n<p>For n=10n = 10n=10 and k=9k = 9k=9:(109)=10!9!(10\u22129)!=10!9!1!=10\u00d79!9!\u00d71!=10\\binom{10}{9} = \\frac{10!}{9!(10-9)!} = \\frac{10!}{9!1!} = \\frac{10 \\times 9!}{9! \\times 1!} = 10(910\u200b)=9!(10\u22129)!10!\u200b=9!1!10!\u200b=9!\u00d71!10\u00d79!\u200b=10<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Apply the formula<\/h3>\n\n\n\n<p>Now, we can apply the binomial probability formula:P(X=9)=(109)(0.8)9(1\u22120.8)10\u22129P(X = 9) = \\binom{10}{9} (0.8)^9 (1 &#8211; 0.8)^{10 &#8211; 9}P(X=9)=(910\u200b)(0.8)9(1\u22120.8)10\u22129P(X=9)=10\u00d7(0.8)9\u00d7(0.2)1P(X = 9) = 10 \\times (0.8)^9 \\times (0.2)^1P(X=9)=10\u00d7(0.8)9\u00d7(0.2)1<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 3: Simplify the powers<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>(0.8)9\u22480.1342(0.8)^9 \\approx 0.1342(0.8)9\u22480.1342<\/li>\n\n\n\n<li>(0.2)1=0.2(0.2)^1 = 0.2(0.2)1=0.2<\/li>\n<\/ul>\n\n\n\n<p>So:P(X=9)=10\u00d70.1342\u00d70.2P(X = 9) = 10 \\times 0.1342 \\times 0.2P(X=9)=10\u00d70.1342\u00d70.2P(X=9)=10\u00d70.02684=0.2684P(X = 9) = 10 \\times 0.02684 = 0.2684P(X=9)=10\u00d70.02684=0.2684<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Final Answer:<\/h3>\n\n\n\n<p>The probability P(X=9)P(X = 9)P(X=9) is approximately <strong>0.2684<\/strong>.<\/p>\n\n\n\n<p>This means that the probability of getting exactly 9 successes in 10 trials with a success probability of 0.8 is about 0.2684, or 26.84%.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner9-216.jpeg\" alt=\"\" class=\"wp-image-258699\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Assume the random variable X has a binomial distribution with the given probability of obtaining a success. Find the following probability, given the number of trials and the probability of obtaining a success. Round your answer to four decimal places. P(X9) , n=10 , p=0.8 The Correct Answer and Explanation is: To find the probability [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-258698","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/258698","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=258698"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/258698\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=258698"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=258698"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=258698"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}