{"id":26238,"date":"2025-06-19T11:26:52","date_gmt":"2025-06-19T11:26:52","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=26238"},"modified":"2025-06-19T11:26:55","modified_gmt":"2025-06-19T11:26:55","slug":"samples-of-rejuvenated-mitochondria-are-mutated-defective-3-of-cases","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/samples-of-rejuvenated-mitochondria-are-mutated-defective-3-of-cases\/","title":{"rendered":"Samples of rejuvenated mitochondria are mutated (defective) 3% of cases"},"content":{"rendered":"\n

Samples of rejuvenated mitochondria are mutated (defective) 3% of cases. Suppose 12 samples are studied, and they can be considered to be independent for mutation. Determine the following probabilities. (a) No samples are mutated. (b) At most one sample is mutated c) More than half the samples are mutated: Round your answers to two decimal places (e.g: 98.76). (a) The probability is (b) The probability is (c) The probability is<\/p>\n\n\n\n

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

We are dealing with a binomial probability<\/strong> scenario.<\/p>\n\n\n\n

Problem Setup:<\/h3>\n\n\n\n
    \n
  • Probability of a sample being mutated: p=0.03p = 0.03p=0.03<\/li>\n\n\n\n
  • Number of samples: n=12n = 12n=12<\/li>\n\n\n\n
  • Probability of not being mutated: q=1\u2212p=0.97q = 1 – p = 0.97q=1\u2212p=0.97<\/li>\n\n\n\n
  • Let XXX be the number of mutated samples in 12 trials:
    X\u223cBinomial(n=12,p=0.03)X \\sim \\text{Binomial}(n = 12, p = 0.03)X\u223cBinomial(n=12,p=0.03)<\/li>\n<\/ul>\n\n\n\n

    We will calculate:<\/p>\n\n\n\n

    \n\n\n\n

    (a) P(X = 0)<\/strong>: No samples are mutated<\/h3>\n\n\n\n

    Using the binomial probability formula:P(X=k)=(nk)pk(1\u2212p)n\u2212kP(X = k) = \\binom{n}{k} p^k (1 – p)^{n – k}P(X=k)=(kn\u200b)pk(1\u2212p)n\u2212kP(X=0)=(120)(0.03)0(0.97)12=1\u22c51\u22c5(0.97)12\u22480.694P(X = 0) = \\binom{12}{0} (0.03)^0 (0.97)^{12} = 1 \\cdot 1 \\cdot (0.97)^{12} \\approx 0.694P(X=0)=(012\u200b)(0.03)0(0.97)12=1\u22c51\u22c5(0.97)12\u22480.694<\/p>\n\n\n\n

    Answer (a): 0.69<\/strong><\/p>\n\n\n\n

    \n\n\n\n

    (b) P(X \u2264 1)<\/strong>: At most one sample is mutated<\/h3>\n\n\n\n

    We add probabilities of 0 and 1:P(X=1)=(121)(0.03)1(0.97)11=12\u22c50.03\u22c5(0.97)11\u224812\u22c50.03\u22c50.737=0.265P(X = 1) = \\binom{12}{1}(0.03)^1(0.97)^{11} = 12 \\cdot 0.03 \\cdot (0.97)^{11} \\approx 12 \\cdot 0.03 \\cdot 0.737 = 0.265P(X=1)=(112\u200b)(0.03)1(0.97)11=12\u22c50.03\u22c5(0.97)11\u224812\u22c50.03\u22c50.737=0.265P(X\u22641)=P(X=0)+P(X=1)\u22480.694+0.265=0.959P(X \\leq 1) = P(X = 0) + P(X = 1) \\approx 0.694 + 0.265 = 0.959P(X\u22641)=P(X=0)+P(X=1)\u22480.694+0.265=0.959<\/p>\n\n\n\n

    Answer (b): 0.96<\/strong><\/p>\n\n\n\n

    \n\n\n\n

    (c) P(X > 6)<\/strong>: More than half the samples are mutated<\/h3>\n\n\n\n

    We find the cumulative probability up to 6 and subtract from 1:P(X>6)=1\u2212P(X\u22646)P(X > 6) = 1 – P(X \\leq 6)P(X>6)=1\u2212P(X\u22646)<\/p>\n\n\n\n

    Because p=0.03p = 0.03p=0.03 is very small, the probability of more than 6 mutations is extremely low. We can safely compute it using binomial tables or software.<\/p>\n\n\n\n

    Using software or a calculator:P(X>6)\u22480.000P(X > 6) \\approx 0.000P(X>6)\u22480.000<\/p>\n\n\n\n

    Answer (c): 0.00<\/strong><\/p>\n\n\n\n

    \n\n\n\n

    Final Answers:<\/h3>\n\n\n\n
      \n
    • (a) 0.69<\/li>\n\n\n\n
    • (b) 0.96<\/li>\n\n\n\n
    • (c) 0.00<\/li>\n<\/ul>\n\n\n\n

      Explanation:<\/h3>\n\n\n\n

      This is a classic binomial probability problem, where each sample has a small independent chance of being defective. Since the mutation rate is only 3%, the chances of observing many mutations among 12 samples are very low. The most probable outcomes are 0 or 1 mutation. This kind of analysis is common in quality control and bio-statistics, where events are rare but independent.<\/p>\n\n\n\n

      \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

      Samples of rejuvenated mitochondria are mutated (defective) 3% of cases. Suppose 12 samples are studied, and they can be considered to be independent for mutation. Determine the following probabilities. (a) No samples are mutated. (b) At most one sample is mutated c) More than half the samples are mutated: Round your answers to two decimal […]<\/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-26238","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/26238","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=26238"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/26238\/revisions"}],"predecessor-version":[{"id":26240,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/26238\/revisions\/26240"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=26238"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=26238"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=26238"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}