{"id":26245,"date":"2025-06-19T11:32:48","date_gmt":"2025-06-19T11:32:48","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=26245"},"modified":"2025-06-19T11:32:50","modified_gmt":"2025-06-19T11:32:50","slug":"samples-of-rejuvenated-mitochondria-are-mutated-defective-with-probability-0-16","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/samples-of-rejuvenated-mitochondria-are-mutated-defective-with-probability-0-16\/","title":{"rendered":"Samples of rejuvenated mitochondria are mutated (defective) with probability 0.16."},"content":{"rendered":"\n<p>Samples of rejuvenated mitochondria are mutated (defective) with probability 0.16. Find the probability you need to examine at least 6 samples to find 2 samples containing mutations. Report answer to 3 decimal places<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-vivid-red-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>We are given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Probability of mutation (success) = <strong>p = 0.16<\/strong><\/li>\n\n\n\n<li>We are looking for the probability that <strong>at least 6 samples are needed<\/strong> to find <strong>2 mutated samples<\/strong>.<\/li>\n<\/ul>\n\n\n\n<p>This is a <strong>Negative Binomial Distribution<\/strong> problem, where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>We are looking for the probability that the <strong>2nd success<\/strong> occurs <strong>on or after the 6th trial<\/strong>.<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Understanding the Setup<\/h3>\n\n\n\n<p>Let XXX be the trial on which the <strong>2nd mutation<\/strong> (success) occurs. Then XXX follows a <strong>Negative Binomial distribution<\/strong>:P(X=k)=(k\u22121r\u22121)pr(1\u2212p)k\u2212rP(X = k) = \\binom{k-1}{r-1} p^r (1-p)^{k &#8211; r}P(X=k)=(r\u22121k\u22121\u200b)pr(1\u2212p)k\u2212r<\/p>\n\n\n\n<p>Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>r=2r = 2r=2 (we are looking for the 2nd mutated sample)<\/li>\n\n\n\n<li>p=0.16p = 0.16p=0.16<\/li>\n\n\n\n<li>k\u22652k \\geq 2k\u22652 is the trial number where the 2nd success occurs<\/li>\n<\/ul>\n\n\n\n<p>We want:P(X\u22656)=1\u2212P(X\u22645)P(X \\geq 6) = 1 &#8211; P(X \\leq 5)P(X\u22656)=1\u2212P(X\u22645)<\/p>\n\n\n\n<p>So we will calculate:P(X\u22645)=P(X=2)+P(X=3)+P(X=4)+P(X=5)P(X \\leq 5) = P(X=2) + P(X=3) + P(X=4) + P(X=5)P(X\u22645)=P(X=2)+P(X=3)+P(X=4)+P(X=5)<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Calculate Individual Probabilities<\/h3>\n\n\n\n<p>Use the negative binomial formula for each value from 2 to 5:<\/p>\n\n\n\n<h4 class=\"wp-block-heading\">P(X=2)P(X = 2)P(X=2):<\/h4>\n\n\n\n<p>(11)(0.16)2(0.84)0=1\u22c50.0256\u22c51=0.0256\\binom{1}{1} (0.16)^2 (0.84)^0 = 1 \\cdot 0.0256 \\cdot 1 = 0.0256(11\u200b)(0.16)2(0.84)0=1\u22c50.0256\u22c51=0.0256<\/p>\n\n\n\n<h4 class=\"wp-block-heading\">P(X=3)P(X = 3)P(X=3):<\/h4>\n\n\n\n<p>(21)(0.16)2(0.84)1=2\u22c50.0256\u22c50.84=0.0430\\binom{2}{1} (0.16)^2 (0.84)^1 = 2 \\cdot 0.0256 \\cdot 0.84 = 0.0430(12\u200b)(0.16)2(0.84)1=2\u22c50.0256\u22c50.84=0.0430<\/p>\n\n\n\n<h4 class=\"wp-block-heading\">P(X=4)P(X = 4)P(X=4):<\/h4>\n\n\n\n<p>(31)(0.16)2(0.84)2=3\u22c50.0256\u22c50.7056=0.0542\\binom{3}{1} (0.16)^2 (0.84)^2 = 3 \\cdot 0.0256 \\cdot 0.7056 = 0.0542(13\u200b)(0.16)2(0.84)2=3\u22c50.0256\u22c50.7056=0.0542<\/p>\n\n\n\n<h4 class=\"wp-block-heading\">P(X=5)P(X = 5)P(X=5):<\/h4>\n\n\n\n<p>(41)(0.16)2(0.84)3=4\u22c50.0256\u22c50.5927=0.0607\\binom{4}{1} (0.16)^2 (0.84)^3 = 4 \\cdot 0.0256 \\cdot 0.5927 = 0.0607(14\u200b)(0.16)2(0.84)3=4\u22c50.0256\u22c50.5927=0.0607<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 3: Add Probabilities<\/h3>\n\n\n\n<p>P(X\u22645)=0.0256+0.0430+0.0542+0.0607=0.1835P(X \\leq 5) = 0.0256 + 0.0430 + 0.0542 + 0.0607 = 0.1835P(X\u22645)=0.0256+0.0430+0.0542+0.0607=0.1835<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 4: Subtract from 1<\/h3>\n\n\n\n<p>P(X\u22656)=1\u22120.1835=0.816P(X \\geq 6) = 1 &#8211; 0.1835 = \\boxed{0.816}P(X\u22656)=1\u22120.1835=0.816\u200b<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Final Answer: <strong>0.816<\/strong><\/h3>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation <\/h3>\n\n\n\n<p>This problem is modeled using the <strong>Negative Binomial Distribution<\/strong>, which describes the number of trials required to achieve a fixed number of successes (in this case, 2 mutations). The key here is understanding that the samples are being examined one by one until the second mutated sample is found. The probability of mutation (success) is 0.16, meaning the probability that a sample is <strong>not<\/strong> mutated (failure) is 0.84.<\/p>\n\n\n\n<p>We are asked to find the probability that <strong>at least 6 samples<\/strong> must be examined to observe the second mutation. This is equivalent to saying that the second success does <strong>not<\/strong> occur in the first 5 samples. To compute this, we first calculate the probability that the second success occurs on trials 2 through 5 and subtract this from 1.<\/p>\n\n\n\n<p>The Negative Binomial formula gives us the probability of the r-th success occurring on the k-th trial. We compute these probabilities for k = 2 through 5, and sum them to get the total probability that the second mutation occurs within the first 5 trials. Subtracting this sum from 1 gives the probability that the second mutation happens on the 6th trial or later.<\/p>\n\n\n\n<p>After computing each component carefully and summing them, we find that the probability of needing <strong>at least 6 samples<\/strong> is <strong>0.816<\/strong>, or <strong>81.6%<\/strong>. This high value makes sense because with a relatively low mutation rate (16%), it usually takes a number of samples to encounter multiple mutations.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"722\" height=\"1024\" src=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner10-79.jpeg\" alt=\"\" class=\"wp-image-26246\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner10-79.jpeg 722w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner10-79-212x300.jpeg 212w\" sizes=\"auto, (max-width: 722px) 100vw, 722px\" \/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Samples of rejuvenated mitochondria are mutated (defective) with probability 0.16. Find the probability you need to examine at least 6 samples to find 2 samples containing mutations. Report answer to 3 decimal places The Correct Answer and Explanation is: We are given: This is a Negative Binomial Distribution problem, where: Step 1: Understanding the Setup [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-26245","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/26245","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=26245"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/26245\/revisions"}],"predecessor-version":[{"id":26247,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/26245\/revisions\/26247"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=26245"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=26245"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=26245"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}