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 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 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 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 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 a) The probability that a group of five random digits will contain at least one 0 is 0.40951. Each entry in a table of random digits like Table D has probability 0.1 of being aand the digits are independent of one another. If many lines of 40 random digits are selected, the mean and standard deviation of the number of 0 s will be approximately (a) meanstandard deviation. (b) meanstandard deviation(c) meanstandard deviation. […]<\/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-44208","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/44208","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=44208"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/44208\/revisions"}],"predecessor-version":[{"id":44210,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/44208\/revisions\/44210"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=44208"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=44208"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=44208"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Part b) Find the probability that a group of five random digits will contain exactly three 0s.<\/h3>\n\n\n\n
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Part c) Suppose we have 40 random digits. What is the mean number of 0s in these 40 digits?<\/h3>\n\n\n\n
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Summary of Results:<\/h3>\n\n\n\n
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![\"\"](\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-1481.jpeg\")
<\/figure>\n","protected":false},"excerpt":{"rendered":"