An insurance company finds that .005 percent of the population die from a certain kind of accident each year. What is the probability that the company must pay off on more than 3 of 10,000 insured risks against such accidents in a given year?<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n We are given that the probability of an individual dying from a certain kind of accident in a year is 0.005%<\/strong>, or:<\/p>\n\n\n\n [ For a total of 10,000 insured individuals<\/strong>, we define ( X ) as the number of deaths in a given year. ( X ) follows a Binomial distribution<\/strong>:<\/p>\n\n\n\n [ Since ( n ) is large and ( p ) is small, we approximate the Binomial distribution<\/strong> using a Poisson distribution<\/strong> with:<\/p>\n\n\n\n [ Thus, we approximate ( X ) as:<\/p>\n\n\n\n [ We need to find the probability that more than 3<\/strong> deaths occur:<\/p>\n\n\n\n [ Using the Poisson probability mass function:<\/p>\n\n\n\n [ we compute:<\/p>\n\n\n\n [ [ [ [ Summing these:<\/p>\n\n\n\n [ [ The probability that the insurance company must pay off on more than 3 insured risks in a given year is 0.0018<\/strong>, or 0.18%<\/strong>.<\/p>\n","protected":false},"excerpt":{"rendered":" An insurance company finds that .005 percent of the population die from a certain kind of accident each year. What is the probability that the company must pay off on more than 3 of 10,000 insured risks against such accidents in a given year? The Correct Answer and Explanation is : We are given 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-190764","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/190764","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=190764"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/190764\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=190764"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=190764"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=190764"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
p = \\frac{0.005}{100} = 0.00005
]<\/p>\n\n\n\n
X \\sim \\text{Binomial}(n=10,000, p=0.00005)
]<\/p>\n\n\n\nStep 1: Check if Poisson Approximation is Valid<\/h3>\n\n\n\n
\\lambda = np = 10,000 \\times 0.00005 = 0.5
]<\/p>\n\n\n\n
X \\sim \\text{Poisson}(\\lambda = 0.5)
]<\/p>\n\n\n\n
P(X > 3) = 1 – P(X \\leq 3)
]<\/p>\n\n\n\nStep 2: Compute Cumulative Probability ( P(X \\leq 3) )<\/h3>\n\n\n\n
P(X = k) = \\frac{e^{-\\lambda} \\lambda^k}{k!}
]<\/p>\n\n\n\n
P(X = 0) = \\frac{e^{-0.5} (0.5)^0}{0!} = e^{-0.5} \\approx 0.6065
]<\/p>\n\n\n\n
P(X = 1) = \\frac{e^{-0.5} (0.5)^1}{1!} = 0.6065 \\times 0.5 = 0.3033
]<\/p>\n\n\n\n
P(X = 2) = \\frac{e^{-0.5} (0.5)^2}{2!} = 0.6065 \\times 0.25 = 0.0758
]<\/p>\n\n\n\n
P(X = 3) = \\frac{e^{-0.5} (0.5)^3}{3!} = 0.6065 \\times \\frac{0.125}{6} = 0.0126
]<\/p>\n\n\n\n
P(X \\leq 3) = 0.6065 + 0.3033 + 0.0758 + 0.0126 = 0.9982
]<\/p>\n\n\n\nStep 3: Compute ( P(X > 3) )<\/h3>\n\n\n\n
P(X > 3) = 1 – P(X \\leq 3) = 1 – 0.9982 = 0.0018
]<\/p>\n\n\n\nConclusion:<\/h3>\n\n\n\n