{"id":20340,"date":"2024-04-15T05:51:35","date_gmt":"2024-04-15T05:51:35","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=20340"},"modified":"2024-04-15T05:51:35","modified_gmt":"2024-04-15T05:51:35","slug":"two-n-bit-binary-strings-s1-and-s2-are-chosen-randomly-with-uniform-probability-the-probability-that-the-hamming-distance-between-these-strings-the-number-of-bit-positions-where-the-two-strings-dif","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/two-n-bit-binary-strings-s1-and-s2-are-chosen-randomly-with-uniform-probability-the-probability-that-the-hamming-distance-between-these-strings-the-number-of-bit-positions-where-the-two-strings-dif\/","title":{"rendered":"Two n bit binary strings, S1 and S2 are chosen randomly with uniform probability. The probability that the Hamming distance between these strings (the number of bit positions where the two strings differ) is equal to d is A. $$\\frac{{{}^{\\text{n}}{{\\text{C}}_{\\text{d}}}}}{{{2^{\\text{n}}}}}$$ B. $$\\frac{{{}^{\\text{n}}{{\\text{C}}_{\\text{d}}}}}{{{2^{\\text{d}}}}}$$ C. $$\\frac{{\\text{d}}}{{{2^{\\text{n}}}}}$$ D. $$\\frac{1}{{{2^{\\text{d}}}}}$$"},"content":{"rendered":"<p>\r\n    <!-- Check if it's an AMP page -->\r\n            <!-- Non-AMP version -->\r\n        <div class=\"mcq-container\" data-quiz-id=\"quizState_6a97f9632ddcf\">\r\n                                            <div class=\"option\" data-option-key=\"option1\" data-is-correct=\"false\">\r\n                    $$\frac{{{}^{\text{n}}{{\text{C}}_{\text{d}}}}}{{{2^{\text{n}}}}}$$                <\/div>\r\n                                            <div class=\"option\" data-option-key=\"option2\" data-is-correct=\"true\">\r\n                    $$\frac{{{}^{\text{n}}{{\text{C}}_{\text{d}}}}}{{{2^{\text{d}}}}}$$                <\/div>\r\n                                            <div class=\"option\" data-option-key=\"option3\" data-is-correct=\"false\">\r\n                    $$\frac{{\text{d}}}{{{2^{\text{n}}}}}$$                <\/div>\r\n                                            <div class=\"option\" data-option-key=\"option4\" data-is-correct=\"false\">\r\n                    $$\frac{1}{{{2^{\text{d}}}}}$$                <\/div>\r\n                            \r\n            <!-- Feedback messages for non-AMP -->\r\n            <div class=\"feedback\" data-feedback=\"wrong\">Answer is Right!<\/div>\r\n            <div class=\"feedback\" data-feedback=\"right\">Answer is Wrong!<\/div>\r\n        <\/div>\r\n\r\n        <script>\r\n        document.addEventListener('DOMContentLoaded', function () {\r\n            var containers = document.querySelectorAll('.mcq-container');\r\n\r\n            containers.forEach(function(container) {\r\n                var options = container.querySelectorAll('.option');\r\n                var feedbackSelect = container.querySelector('[data-feedback=\"select\"]');\r\n                var feedbackWrong = container.querySelector('[data-feedback=\"wrong\"]');\r\n                var feedbackRight = container.querySelector('[data-feedback=\"right\"]');\r\n\r\n                options.forEach(function(option) {\r\n                    option.addEventListener('click', function() {\r\n                        var selectedOption = option.getAttribute('data-option-key');\r\n                        var isCorrect = option.getAttribute('data-is-correct') === 'true';\r\n\r\n                        \/\/ Remove previous selections\r\n                        options.forEach(function(opt) {\r\n                            opt.classList.remove('correct', 'incorrect');\r\n                        });\r\n\r\n                        \/\/ Add the correct\/incorrect class\r\n                        if (isCorrect) {\r\n                            option.classList.add('correct');\r\n                            feedbackRight.hidden = false;\r\n                            feedbackWrong.hidden = true;\r\n                        } else {\r\n                            option.classList.add('incorrect');\r\n                            feedbackRight.hidden = true;\r\n                            feedbackWrong.hidden = false;\r\n                        }\r\n\r\n                        \/\/ Hide select feedback\r\n                        feedbackSelect.hidden = true;\r\n                    });\r\n                });\r\n            });\r\n        });\r\n        <\/script>\r\n    \r\n    <!--more--><\/p>\n<p>The correct answer is $\\boxed{\\frac{{{}^{\\text{n}}{{\\text{C}}_{\\text{d}}}}}{{{2^{\\text{n}}}}}}$.<\/p>\n<p>To calculate the probability of a Hamming distance of $d$, we need to count the number of ways to choose $d$ bit positions where the two strings differ, and then divide this by the total number of possible strings. The total number of possible strings is $2^n$, since each bit can be either 0 or 1. The number of ways to choose $d$ bit positions <div class=\"youtube-subscribe-container\">\r\n        <a href=\"https:\/\/www.youtube.com\/channel\/UCNHT8lW-JmLC68rjBfZhdkg?sub_confirmation=1\" target=\"_blank\" class=\"youtube-subscribe-button\">\r\n            <span class=\"youtube-icon\">\r\n                <svg xmlns=\"http:\/\/www.w3.org\/2000\/svg\" viewBox=\"0 0 576 512\">\r\n                    <path d=\"M549.7 124.1c-6.3-23.7-24.8-42.3-48.3-48.6C458.8 64 288 64 288 64S117.2 64 74.6 75.5c-23.5 6.3-42 24.9-48.3 48.6-11.4 42.9-11.4 132.3-11.4 132.3s0 89.4 11.4 132.3c6.3 23.7 24.8 41.5 48.3 47.8C117.2 448 288 448 288 448s170.8 0 213.4-11.5c23.5-6.3 42-24.2 48.3-47.8 11.4-42.9 <div class=\"telegram-channel-container\">\r\n        <a href=\"https:\/\/t.me\/pscnotes2025\" target=\"_blank\" class=\"telegram-channel-button\">\r\n            <span class=\"telegram-icon\">\r\n                <svg xmlns=\"http:\/\/www.w3.org\/2000\/svg\" viewBox=\"0 0 496 512\">\r\n                    <path fill=\"white\" d=\"M248,8C111,8,0,119,0,256s111,248,248,248s248-111,248-248S385,8,248,8z M362,177L320,367c-3,14-10,18-20,14l-56-41l-27,26 c-3,3-5,5-10,5l4-63L323,196c5-5-1-7-8-3l-98,62l-42-13c-9-3-10-9,2-14l162-63C351,160,365,164,362,177z\"\/>\r\n                <\/svg>\r\n            <\/span>\r\n            Join Our Telegram Channel\r\n        <\/a>\r\n    <\/div> 11.4-132.3 11.4-132.3s0-89.4-11.4-132.3zm-317.5 213.5V175.2l142.7 81.2-142.7 81.2z\"\/>\r\n                <\/svg>\r\n            <\/span>\r\n            Subscribe on YouTube\r\n        <\/a>\r\n    <\/div> where the two strings differ is ${{n \\choose d}}$, which is the binomial coefficient. The binomial coefficient is the number of ways to choose $d$ objects from a set of $n$ objects.<\/p>\n<p>Therefore, the probability of a Hamming distance of $d$ is $$\\frac{{{}^{\\text{n}}{{\\text{C}}_{\\text{d}}}}}{{{2^{\\text{n}}}}}$$<\/p>\n<p>Option A is incorrect because it divides by $2^d$ instead of $2^n$. Option B is incorrect because it divides by $2^d$ instead of $2^n$ and also includes the case where $d=0$, which is not a possible Hamming distance. Option C is incorrect because it does not take into account the number of possible ways to choose the bit positions where the two strings differ. Option D is incorrect because it is the probability that the two strings are identical, which is not the same as the probability that the Hamming distance is 0.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Join Our Telegram Channel Subscribe on YouTube<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[691],"tags":[],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v22.2 (Yoast SEO v23.3) - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>Two n bit binary strings, S1 and S2 are chosen randomly with uniform probability. The probability that the Hamming distance between these strings (the number of bit positions where the two strings differ) is equal to d is A. $$\\frac{{{}^{\\text{n}}{{\\text{C}}_{\\text{d}}}}}{{{2^{\\text{n}}}}}$$ B. $$\\frac{{{}^{\\text{n}}{{\\text{C}}_{\\text{d}}}}}{{{2^{\\text{d}}}}}$$ C. $$\\frac{{\\text{d}}}{{{2^{\\text{n}}}}}$$ D. $$\\frac{1}{{{2^{\\text{d}}}}}$$<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/exam.pscnotes.com\/mcq\/two-n-bit-binary-strings-s1-and-s2-are-chosen-randomly-with-uniform-probability-the-probability-that-the-hamming-distance-between-these-strings-the-number-of-bit-positions-where-the-two-strings-dif\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Two n bit binary strings, S1 and S2 are chosen randomly with uniform probability. 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The probability that the Hamming distance between these strings (the number of bit positions where the two strings differ) is equal to d is A. $$\\frac{{{}^{\\text{n}}{{\\text{C}}_{\\text{d}}}}}{{{2^{\\text{n}}}}}$$ B. $$\\frac{{{}^{\\text{n}}{{\\text{C}}_{\\text{d}}}}}{{{2^{\\text{d}}}}}$$ C. $$\\frac{{\\text{d}}}{{{2^{\\text{n}}}}}$$ D. $$\\frac{1}{{{2^{\\text{d}}}}}$$","robots":{"index":"index","follow":"follow","max-snippet":"max-snippet:-1","max-image-preview":"max-image-preview:large","max-video-preview":"max-video-preview:-1"},"canonical":"https:\/\/exam.pscnotes.com\/mcq\/two-n-bit-binary-strings-s1-and-s2-are-chosen-randomly-with-uniform-probability-the-probability-that-the-hamming-distance-between-these-strings-the-number-of-bit-positions-where-the-two-strings-dif\/","og_locale":"en_US","og_type":"article","og_title":"Two n bit binary strings, S1 and S2 are chosen randomly with uniform probability. 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