{"id":221453,"date":"2025-05-29T14:49:23","date_gmt":"2025-05-29T14:49:23","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=221453"},"modified":"2025-05-29T14:49:25","modified_gmt":"2025-05-29T14:49:25","slug":"consider-a-binary-code-with-5-bits-0-or-1-in-each-code-word","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/29\/consider-a-binary-code-with-5-bits-0-or-1-in-each-code-word\/","title":{"rendered":"Consider a binary code with 5 bits (0 or 1) in each code word."},"content":{"rendered":"\n

Consider a binary code with 5 bits (0 or 1) in each code word. An example of a code word is 01010. In each code word, a bit is a zero with probability 0.8, independent of any other bit.<\/p>\n\n\n\n

(a) What is the probability of the code word 00111?<\/p>\n\n\n\n

(b) What is the probability that a code word contains exactly three ones?<\/p>\n\n\n\n

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

(a) What is the probability of the code word 00111?<\/h3>\n\n\n\n

Each bit is a 0 with probability 0.8<\/strong> and a 1 with probability 0.2<\/strong>.
The code word 00111<\/strong> has two 0s and three 1s.<\/p>\n\n\n\n

So, the probability of this specific code word is: P(00111)=(0.8)2\u00d7(0.2)3=0.64\u00d70.008=0.00512P(00111) = (0.8)^2 \\times (0.2)^3 = 0.64 \\times 0.008 = 0.00512<\/p>\n\n\n\n

Answer: 0.00512<\/strong><\/p>\n\n\n\n

\n\n\n\n

(b) What is the probability that a code word contains exactly three ones?<\/h3>\n\n\n\n

To find this, we need to compute the probability that any<\/strong> 5-bit code word has exactly three 1s (and thus, two 0s). Since the bits are independent, we use the binomial distribution<\/strong>.<\/p>\n\n\n\n

The formula for the probability of exactly kk successes (1s) in nn independent trials (bits), where each success has probability pp, is: P(X=k)=(nk)pk(1\u2212p)n\u2212kP(X = k) = \\binom{n}{k} p^k (1 – p)^{n – k}<\/p>\n\n\n\n

Here:<\/p>\n\n\n\n

    \n
  • n=5n = 5 (number of bits),<\/li>\n\n\n\n
  • k=3k = 3 (we want exactly 3 ones),<\/li>\n\n\n\n
  • p=0.2p = 0.2 (probability a bit is 1),<\/li>\n\n\n\n
  • 1\u2212p=0.81 – p = 0.8 (probability a bit is 0).<\/li>\n<\/ul>\n\n\n\n

    So, P(exactly 3 ones)=(53)(0.2)3(0.8)2=10\u00d70.008\u00d70.64=10\u00d70.00512=0.0512P(\\text{exactly 3 ones}) = \\binom{5}{3} (0.2)^3 (0.8)^2 = 10 \\times 0.008 \\times 0.64 = 10 \\times 0.00512 = 0.0512<\/p>\n\n\n\n

    Answer: 0.0512<\/strong><\/p>\n\n\n\n

    \n\n\n\n

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

    In binary coding, each code word is a fixed-length sequence of bits (0s and 1s). In this problem, we’re given that each of the 5 bits in a code word independently takes the value 0 with a probability of 0.8 and 1 with a probability of 0.2.<\/p>\n\n\n\n

    Part (a): Specific Code Word Probability<\/h4>\n\n\n\n

    To find the probability of a specific code word like \u201c00111\u201d, we multiply the probabilities of each bit occurring in that exact position. Since independence is assumed, the joint probability is simply the product:<\/p>\n\n\n\n

      \n
    • First two bits are 0s: 0.8\u00d70.80.8 \\times 0.8<\/li>\n\n\n\n
    • Next three bits are 1s: 0.2\u00d70.2\u00d70.20.2 \\times 0.2 \\times 0.2
      So, P(00111)=0.82\u00d70.23=0.00512P(00111) = 0.8^2 \\times 0.2^3 = 0.00512<\/li>\n<\/ul>\n\n\n\n

      Part (b): Probability of Exactly Three Ones<\/h4>\n\n\n\n

      To calculate the probability that any 5-bit code word has exactly 3 ones (regardless of position), we use the binomial distribution. The binomial coefficient (53)=10\\binom{5}{3} = 10 tells us how many different code words have exactly 3 ones. Each of these combinations has the same probability: 0.23\u00d70.820.2^3 \\times 0.8^2 because the bits are independent.<\/p>\n\n\n\n

      So, the total probability is: 10\u00d70.008\u00d70.64=0.051210 \\times 0.008 \\times 0.64 = 0.0512<\/p>\n\n\n\n

      This problem illustrates how probability theory and binomial distribution help analyze random patterns in binary data, which has applications in error detection, data compression, and coding theory.<\/p>\n\n\n\n

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

      Consider a binary code with 5 bits (0 or 1) in each code word. An example of a code word is 01010. In each code word, a bit is a zero with probability 0.8, independent of any other bit. (a) What is the probability of the code word 00111? (b) What is the probability that […]<\/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-221453","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/221453","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=221453"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/221453\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=221453"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=221453"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=221453"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}