Samples of rejuvenated mitochondria are mutated (defective) with a probability 0.17. Find the probability that at most one sample is mutated in 10 samples. Report answer to 3 decimal places.<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n To solve this problem, we can use the binomial probability formula<\/strong> because we are dealing with a fixed number of independent trials, where each trial has two possible outcomes (mutated or not mutated).<\/p>\n\n\n\n The binomial probability formula is given by:<\/p>\n\n\n\n [ Where:<\/p>\n\n\n\n We need to find the probability that at most one sample is mutated, which means ( P(X \\leq 1) ). This can be computed as:<\/p>\n\n\n\n [ [ [ [ The probability that at most one sample is mutated is 0.591<\/strong> (to three decimal places).<\/p>\n\n\n\n The problem involves calculating the probability that at most one of 10 samples is defective, given that each sample is defective with a probability of 0.17. This is a classic example of a binomial distribution, where each sample is an independent trial, and we are counting the number of successes (mutations).<\/p>\n\n\n\n The probability of success (( p )) is 0.17, and the probability of failure (( 1-p )) is 0.83. Using the binomial formula, we computed the probabilities for exactly 0 and 1 mutations, which are ( P(X = 0) ) and ( P(X = 1) ), respectively. These represent scenarios where no samples are defective or only one sample is defective.<\/p>\n\n\n\n To find the total probability of at most one mutation, we summed the probabilities for these two scenarios, ( P(X \\leq 1) = P(X = 0) + P(X = 1) ). The calculations used the binomial coefficient, which determines the number of ways to select ( k ) successes from ( n ) trials.<\/p>\n\n\n\n This method ensures an exact calculation, avoiding approximations or unnecessary assumptions. The final result, 0.591, indicates that in 59.1% of cases, 10 randomly selected samples will have 1 or fewer mutations. This result is rounded to three decimal places for precision, aligning with the problem’s requirements.<\/p>\n\n\n\n <\/p>\n","protected":false},"excerpt":{"rendered":" Samples of rejuvenated mitochondria are mutated (defective) with a probability 0.17. Find the probability that at most one sample is mutated in 10 samples. Report answer to 3 decimal places. The Correct Answer and Explanation is : To solve this problem, we can use the binomial probability formula because we are dealing with a fixed […]<\/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-183817","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/183817","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=183817"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/183817\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=183817"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=183817"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=183817"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
P(X = k) = \\binom{n}{k} p^k (1-p)^{n-k}
]<\/p>\n\n\n\n\n
P(X \\leq 1) = P(X = 0) + P(X = 1)
]<\/p>\n\n\n\nStep 1: Calculate ( P(X = 0) )<\/h3>\n\n\n\n
P(X = 0) = \\binom{10}{0} (0.17)^0 (1 – 0.17)^{10} = 1 \\cdot 1 \\cdot (0.83)^{10} = 0.1938
]<\/p>\n\n\n\nStep 2: Calculate ( P(X = 1) )<\/h3>\n\n\n\n
P(X = 1) = \\binom{10}{1} (0.17)^1 (0.83)^9 = 10 \\cdot 0.17 \\cdot (0.83)^9 = 0.3976
]<\/p>\n\n\n\nStep 3: Add ( P(X = 0) ) and ( P(X = 1) )<\/h3>\n\n\n\n
P(X \\leq 1) = 0.1938 + 0.3976 = 0.5914
]<\/p>\n\n\n\nFinal Answer:<\/h3>\n\n\n\n
\n\n\n\nExplanation (300 words):<\/h3>\n\n\n\n