What is the probability that a witness will refuses to serve alcoholic beverages to only two minor if she randomly checked 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 :<\/strong><\/mark><\/p>\n\n\n\n To solve this problem, we need to find the probability that a witness will refuse to serve alcoholic beverages to exactly two minors<\/strong> when randomly checking the IDs of five students selected from a group of nine students<\/strong>, where four of them are minors (not of legal age)<\/strong>.<\/p>\n\n\n\n The total number of ways to select 5 students from 9 students is calculated using the combination formula: To ensure that exactly 2 minors are selected, we need to choose 2 minors from the 4 minors, and the remaining 3 students must be selected from the 5 legal-age students. We use combinations again for this:<\/p>\n\n\n\n The total number of favorable outcomes (selecting exactly 2 minors and 3 legal-age students) is the product of the two combinations: Finally, the probability is the ratio of favorable outcomes to the total number of outcomes. The probability is: The probability that the witness will refuse to serve alcoholic beverages to exactly two minors is 10\/21<\/strong> or approximately 0.476<\/strong>.<\/p>\n\n\n\n In this problem, we used the concept of combinations because the order of selection does not matter. We first calculated the total possible ways to select 5 students from 9, and then we calculated the number of ways to select exactly 2 minors and 3 legal-age students. The ratio of these two values gives the probability that exactly 2 minors will be selected when 5 students are randomly chosen. This is a classic example of probability with combinations, where we need to calculate outcomes with restrictions (i.e., selecting exactly 2 minors out of 4).<\/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 checked of five students from among nine students of which four are not of legal age ? The correct answer and explanation is : To solve this problem, we need to find the probability that […]<\/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-199864","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/199864","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=199864"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/199864\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=199864"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=199864"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=199864"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Key Information:<\/h3>\n\n\n\n
\n
Step 1: Total number of ways to choose 5 students from 9<\/h3>\n\n\n\n
[
\\binom{n}{k} = \\frac{n!}{k!(n-k)!}
]
where ( n = 9 ) and ( k = 5 ). So, the total number of ways to choose 5 students from 9 is:
[
\\binom{9}{5} = \\frac{9!}{5!(9-5)!} = \\frac{9 \\times 8 \\times 7 \\times 6}{4 \\times 3 \\times 2 \\times 1} = 126
]<\/p>\n\n\n\nStep 2: Number of favorable outcomes (selecting exactly 2 minors)<\/h3>\n\n\n\n
\n
[
\\binom{4}{2} = \\frac{4 \\times 3}{2 \\times 1} = 6
]<\/li>\n\n\n\n
[
\\binom{5}{3} = \\frac{5 \\times 4 \\times 3}{3 \\times 2 \\times 1} = 10
]<\/li>\n<\/ul>\n\n\n\nStep 3: Total favorable outcomes<\/h3>\n\n\n\n
[
\\binom{4}{2} \\times \\binom{5}{3} = 6 \\times 10 = 60
]<\/p>\n\n\n\nStep 4: Probability<\/h3>\n\n\n\n
[
P(\\text{exactly 2 minors}) = \\frac{60}{126} = \\frac{10}{21}
]<\/p>\n\n\n\nFinal Answer:<\/h3>\n\n\n\n
Explanation:<\/h3>\n\n\n\n