{"id":217165,"date":"2025-05-21T06:21:22","date_gmt":"2025-05-21T06:21:22","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=217165"},"modified":"2025-05-21T06:21:25","modified_gmt":"2025-05-21T06:21:25","slug":"what-is-the-probability-that-a-witness-will-refuses-to-serve-alcoholic-beverages-to-only-two-minor-if-she-randomly-check-id-of-five-students-from-among-nine-students-of-which-four-are-not-of-legal-10","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/21\/what-is-the-probability-that-a-witness-will-refuses-to-serve-alcoholic-beverages-to-only-two-minor-if-she-randomly-check-id-of-five-students-from-among-nine-students-of-which-four-are-not-of-legal-10\/","title":{"rendered":"What is the probability that a witness will refuses to serve alcoholic beverages to only two minor if she randomly check Id of five students from among nine students of which four are not of legal age"},"content":{"rendered":"\n

What is the probability that a witness will refuses to serve alcoholic beverages to only two minor if she randomly check Id of five students from among nine students of which four are not of legal age ?<\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

To find the probability that a witness will refuse to serve alcoholic beverages to only two minors<\/strong> if she randomly checks the IDs of 5 students<\/strong> from a group of 9 students (4 minors and 5 legal-age students)<\/strong>, we can model this using hypergeometric probability<\/strong>.<\/p>\n\n\n\n

\n\n\n\n

Given:<\/strong><\/h3>\n\n\n\n
    \n
  • Total students = 9<\/li>\n\n\n\n
  • Number of minors = 4<\/li>\n\n\n\n
  • Number of legal-age students = 5<\/li>\n\n\n\n
  • Number of students whose IDs are checked = 5<\/li>\n\n\n\n
  • Desired outcome: Exactly 2 minors<\/strong> in the checked group of 5 students.<\/li>\n<\/ul>\n\n\n\n
    \n\n\n\n

    Step-by-Step Calculation:<\/strong><\/h3>\n\n\n\n

    The hypergeometric probability formula<\/strong> is: P(X=k)=(Kk)\u22c5(N\u2212Kn\u2212k)(Nn)P(X = k) = \\frac{{\\binom{K}{k} \\cdot \\binom{N-K}{n-k}}}{\\binom{N}{n}}<\/p>\n\n\n\n

    Where:<\/p>\n\n\n\n

      \n
    • N=9N = 9: total students<\/li>\n\n\n\n
    • K=4K = 4: total minors<\/li>\n\n\n\n
    • n=5n = 5: number of IDs checked<\/li>\n\n\n\n
    • k=2k = 2: desired number of minors among checked IDs<\/li>\n<\/ul>\n\n\n\n

      P(X=2)=(42)\u22c5(53)(95)P(X = 2) = \\frac{{\\binom{4}{2} \\cdot \\binom{5}{3}}}{\\binom{9}{5}}<\/p>\n\n\n\n

      First, calculate the binomial coefficients:<\/p>\n\n\n\n

        \n
      • (42)=6\\binom{4}{2} = 6<\/li>\n\n\n\n
      • (53)=10\\binom{5}{3} = 10<\/li>\n\n\n\n
      • (95)=126\\binom{9}{5} = 126<\/li>\n<\/ul>\n\n\n\n

        P(X=2)=6\u00d710126=60126=1021P(X = 2) = \\frac{6 \\times 10}{126} = \\frac{60}{126} = \\frac{10}{21}<\/p>\n\n\n\n

        \n\n\n\n

        \u2705 Correct Answer: 1021\\boxed{\\frac{10}{21}}<\/strong><\/h3>\n\n\n\n
        \n\n\n\n

        300-Word Explanation:<\/strong><\/h3>\n\n\n\n

        This problem is a classic case of hypergeometric probability<\/strong>, which applies when we randomly sample from a finite population without replacement. The key idea is to determine the probability of selecting exactly 2 minors<\/strong> (students not of legal drinking age) when 5 out of 9 total students are randomly selected.<\/p>\n\n\n\n

        Among the 9 students, 4 are minors and 5 are legal-age. When the witness checks 5 students’ IDs, she could potentially select any combination of minors and legal-age individuals. We’re specifically interested in the scenario where only 2 of the checked students are minors<\/strong>.<\/p>\n\n\n\n

        The hypergeometric formula allows us to compute the probability of getting exactly 2 minors (from the 4 available) and 3 legal-age students (from the 5 available), in a selection of 5 students. To do this, we calculate:<\/p>\n\n\n\n

          \n
        • The number of ways to choose 2 minors out of 4: (42)=6\\binom{4}{2} = 6<\/li>\n\n\n\n
        • The number of ways to choose 3 legal-age students out of 5: (53)=10\\binom{5}{3} = 10<\/li>\n\n\n\n
        • The number of total ways to choose any 5 students out of 9: (95)=126\\binom{9}{5} = 126<\/li>\n<\/ul>\n\n\n\n

          Multiplying the favorable combinations (6 \u00d7 10 = 60) and dividing by the total possible combinations (126), gives us: 60126=1021\\frac{60}{126} = \\frac{10}{21}<\/p>\n\n\n\n

          This result means there is a 10 in 21 chance<\/strong>, or roughly 47.6% probability<\/strong>, that the witness will check exactly two minors’ IDs out of five students, thereby refusing service to only those two.<\/p>\n","protected":false},"excerpt":{"rendered":"

          What is the probability that a witness will refuses to serve alcoholic beverages to only two minor if she randomly check Id of five students from among nine students of which four are not of legal age ? The Correct Answer and Explanation is: To find the probability that a witness will refuse to serve […]<\/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-217165","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/217165","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=217165"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/217165\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=217165"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=217165"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=217165"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}