{"id":20309,"date":"2024-04-15T05:51:10","date_gmt":"2024-04-15T05:51:10","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=20309"},"modified":"2024-04-15T05:51:10","modified_gmt":"2024-04-15T05:51:10","slug":"there-are-25-calculators-in-a-box-two-of-them-are-defective-suppose-5-calculators-are-randomly-picked-for-inspection-i-e-each-has-the-same-chance-of-being-selected-what-is-the-probability-that","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/there-are-25-calculators-in-a-box-two-of-them-are-defective-suppose-5-calculators-are-randomly-picked-for-inspection-i-e-each-has-the-same-chance-of-being-selected-what-is-the-probability-that\/","title":{"rendered":"There are 25 calculators in a box. Two of them are defective. Suppose 5 calculators are randomly picked for inspection (i.e., each has the same chance of being selected), what is the probability that only one of the defective calculators will be included in the inspection? A. $$\\frac{1}{2}$$ B. $$\\frac{1}{3}$$ C. $$\\frac{1}{4}$$ D. $$\\frac{1}{5}$$"},"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_6a99bf2238630\">\r\n                                            <div class=\"option\" data-option-key=\"option1\" data-is-correct=\"false\">\r\n                    $$\frac{1}{2}$$                <\/div>\r\n                                            <div class=\"option\" data-option-key=\"option2\" data-is-correct=\"false\">\r\n                    $$\frac{1}{3}$$                <\/div>\r\n                                            <div class=\"option\" data-option-key=\"option3\" data-is-correct=\"true\">\r\n                    $$\frac{1}{4}$$                <\/div>\r\n                                            <div class=\"option\" data-option-key=\"option4\" data-is-correct=\"false\">\r\n                    $$\frac{1}{5}$$                <\/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 $\\boxed{\\frac{1}{20}}$.<\/p>\n<p>There are two ways to choose one defective calculator and four good calculators:<\/p>\n<ol>\n<li>Choose one defective calculator from the two defective calculators <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 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    <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>         <\/span>\r\n            Subscribe on YouTube\r\n        <\/a>\r\n    <\/div> and then choose four good calculators from the 23 good calculators. This can be done in $\\binom{2}{1}\\binom{23}{4} = \\frac{25 \\times 24 \\times 23 \\times 22}{4 \\times 3 \\times 2 \\times 1} = 2300$ ways.<\/li>\n<li>Choose four good calculators from the 23 good calculators and then choose one defective calculator from the two defective calculators. This can be done in $\\binom{23}{4}\\binom{2}{1} = \\frac{23 \\times 22 \\times 21 \\times 20}{4 \\times 3 \\times 2 \\times 1} = 1380$ ways.<\/li>\n<\/ol>\n<p>Therefore, the probability of choosing one defective calculator and four good calculators is $\\frac{2300 + 1380}{25 \\times 24} = \\frac{3680}{600} = \\boxed{\\frac{1}{20}}$.<\/p>\n<p>Option A is incorrect because it is the probability of choosing two defective calculators. Option B is incorrect because it is the probability of choosing three defective calculators. Option C is incorrect because it is the probability of choosing four defective calculators. Option D is incorrect because it is the probability of choosing none of the defective calculators.<\/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>There are 25 calculators in a box. Two of them are defective. Suppose 5 calculators are randomly picked for inspection (i.e., each has the same chance of being selected), what is the probability that only one of the defective calculators will be included in the inspection? A. $$\\frac{1}{2}$$ B. $$\\frac{1}{3}$$ C. $$\\frac{1}{4}$$ D. $$\\frac{1}{5}$$<\/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\/there-are-25-calculators-in-a-box-two-of-them-are-defective-suppose-5-calculators-are-randomly-picked-for-inspection-i-e-each-has-the-same-chance-of-being-selected-what-is-the-probability-that\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"There are 25 calculators in a box. Two of them are defective. Suppose 5 calculators are randomly picked for inspection (i.e., each has the same chance of being selected), what is the probability that only one of the defective calculators will be included in the inspection? 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Two of them are defective. Suppose 5 calculators are randomly picked for inspection (i.e., each has the same chance of being selected), what is the probability that only one of the defective calculators will be included in the inspection? A. $$\\frac{1}{2}$$ B. $$\\frac{1}{3}$$ C. $$\\frac{1}{4}$$ D. $$\\frac{1}{5}$$","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\/there-are-25-calculators-in-a-box-two-of-them-are-defective-suppose-5-calculators-are-randomly-picked-for-inspection-i-e-each-has-the-same-chance-of-being-selected-what-is-the-probability-that\/","og_locale":"en_US","og_type":"article","og_title":"There are 25 calculators in a box. Two of them are defective. Suppose 5 calculators are randomly picked for inspection (i.e., each has the same chance of being selected), what is the probability that only one of the defective calculators will be included in the inspection? 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