{"id":44205,"date":"2025-06-30T14:15:53","date_gmt":"2025-06-30T14:15:53","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=44205"},"modified":"2025-06-30T14:15:55","modified_gmt":"2025-06-30T14:15:55","slug":"each-entry-in-a-table-of-random-digits-like-table-b-has-probability-0-1-of-being-a-0-and-digits-are-independent-of-each-other","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/each-entry-in-a-table-of-random-digits-like-table-b-has-probability-0-1-of-being-a-0-and-digits-are-independent-of-each-other\/","title":{"rendered":"Each entry in a table of random digits like Table B has probability 0.1 of being a 0, and digits are independent of each other."},"content":{"rendered":"\n
    \n
  1. Each entry in a table of random digits like Table B has probability 0.1 of being a 0, and digits are independent of each other. (5pts) a) Find the probability that a group of five random digits will contain at least one 0. b) Find the probability that a group of five random digits will contain exactly three 0s. c) Suppose we have 40 random digits. What is the mean number of 0s in these 40 digits?<\/li>\n<\/ol>\n\n\n\n

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

    Part a) Find the probability that a group of five random digits will contain at least one 0.<\/h3>\n\n\n\n

    Each digit in the table has a probability of 0.1 of being a 0, and 0.9 of being any other digit (1-9). We can use the complement rule to find the probability of having at least one 0 in five digits. First, we calculate the probability of having no 0s at all, and then subtract that from 1 to find the probability of having at least one 0.<\/p>\n\n\n\n

    The probability that a single digit is not<\/strong> a 0 is 0.9. Therefore, the probability that none<\/strong> of the five digits are 0s is:P(no 0s)=(0.9)5=0.59049P(\\text{no 0s}) = (0.9)^5 = 0.59049P(no 0s)=(0.9)5=0.59049<\/p>\n\n\n\n

    Thus, the probability of having at least one 0 is:P(at least one 0)=1\u2212P(no 0s)=1\u22120.59049=0.40951P(\\text{at least one 0}) = 1 – P(\\text{no 0s}) = 1 – 0.59049 = 0.40951P(at least one 0)=1\u2212P(no 0s)=1\u22120.59049=0.40951<\/p>\n\n\n\n

    Part b) Find the probability that a group of five random digits will contain exactly three 0s.<\/h3>\n\n\n\n

    This is a binomial probability problem where the number of trials is 5 (since there are 5 digits), and the probability of success (getting a 0) on each trial is 0.1. The formula for the binomial probability is:P(k successes)=(nk)pk(1\u2212p)n\u2212kP(k \\text{ successes}) = \\binom{n}{k} p^k (1-p)^{n-k}P(k successes)=(kn\u200b)pk(1\u2212p)n\u2212k<\/p>\n\n\n\n

    where:<\/p>\n\n\n\n

      \n
    • n=5n = 5n=5 is the number of trials,<\/li>\n\n\n\n
    • k=3k = 3k=3 is the number of successes (0s),<\/li>\n\n\n\n
    • p=0.1p = 0.1p=0.1 is the probability of success.<\/li>\n<\/ul>\n\n\n\n

      The binomial coefficient (nk)\\binom{n}{k}(kn\u200b) is calculated as:(53)=5!3!(5\u22123)!=5\u00d742\u00d71=10\\binom{5}{3} = \\frac{5!}{3!(5-3)!} = \\frac{5 \\times 4}{2 \\times 1} = 10(35\u200b)=3!(5\u22123)!5!\u200b=2\u00d715\u00d74\u200b=10<\/p>\n\n\n\n

      Now, calculate the probability:P(exactly 3 0s)=(53)(0.1)3(0.9)2=10\u00d7(0.1)3\u00d7(0.9)2P(\\text{exactly 3 0s}) = \\binom{5}{3} (0.1)^3 (0.9)^2 = 10 \\times (0.1)^3 \\times (0.9)^2P(exactly 3 0s)=(35\u200b)(0.1)3(0.9)2=10\u00d7(0.1)3\u00d7(0.9)2P(exactly 3 0s)=10\u00d70.001\u00d70.81=0.0081P(\\text{exactly 3 0s}) = 10 \\times 0.001 \\times 0.81 = 0.0081P(exactly 3 0s)=10\u00d70.001\u00d70.81=0.0081<\/p>\n\n\n\n

      Part c) Suppose we have 40 random digits. What is the mean number of 0s in these 40 digits?<\/h3>\n\n\n\n

      The number of 0s in 40 digits follows a binomial distribution. The mean of a binomial distribution is given by:\u03bc=n\u00d7p\\mu = n \\times p\u03bc=n\u00d7p<\/p>\n\n\n\n

      where:<\/p>\n\n\n\n

        \n
      • n=40n = 40n=40 is the number of trials,<\/li>\n\n\n\n
      • p=0.1p = 0.1p=0.1 is the probability of getting a 0.<\/li>\n<\/ul>\n\n\n\n

        Thus, the mean number of 0s in 40 digits is:\u03bc=40\u00d70.1=4\\mu = 40 \\times 0.1 = 4\u03bc=40\u00d70.1=4<\/p>\n\n\n\n

        So, on average, you can expect 4 zeros in a group of 40 random digits.<\/p>\n\n\n\n

        Summary of Results:<\/h3>\n\n\n\n

        a) The probability that a group of five random digits will contain at least one 0 is 0.40951.
        b) The probability that a group of five random digits will contain exactly three 0s is 0.0081.
        c) The mean number of 0s in a group of 40 random digits is 4.<\/p>\n\n\n\n

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

        The Correct Answer and Explanation is: Part a) Find the probability that a group of five random digits will contain at least one 0. Each digit in the table has a probability of 0.1 of being a 0, and 0.9 of being any other digit (1-9). We can use the complement rule to find the […]<\/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-44205","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/44205","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=44205"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/44205\/revisions"}],"predecessor-version":[{"id":44207,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/44205\/revisions\/44207"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=44205"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=44205"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=44205"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}