{"id":20280,"date":"2024-04-15T05:50:47","date_gmt":"2024-04-15T05:50:47","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=20280"},"modified":"2024-04-15T05:50:47","modified_gmt":"2024-04-15T05:50:47","slug":"a-fair-coin-is-tossed-10-times-what-is-the-probability-that-only-the-first-two-tosses-will-yield-heads-a-left-frac12-right2-b-10textc_2left-frac12","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/a-fair-coin-is-tossed-10-times-what-is-the-probability-that-only-the-first-two-tosses-will-yield-heads-a-left-frac12-right2-b-10textc_2left-frac12\/","title":{"rendered":"A fair coin is tossed 10 times. What is the probability that ONLY the first two tosses will yield heads? A. $${\\left( {\\frac{1}{2}} \\right)^2}$$ B. $${}^{10}{{\\text{C}}_2}{\\left( {\\frac{1}{2}} \\right)^2}$$ C. $${\\left( {\\frac{1}{2}} \\right)^{10}}$$ D. $${}^{10}{{\\text{C}}_2}{\\left( {\\frac{1}{2}} \\right)^{10}}$$"},"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_6a9846e3357d8\">\r\n                                            <div class=\"option\" data-option-key=\"option1\" data-is-correct=\"true\">\r\n                    $${left( {\frac{1}{2}} \right)^2}$$                <\/div>\r\n                                            <div class=\"option\" data-option-key=\"option2\" data-is-correct=\"false\">\r\n                    $${}^{10}{{\text{C}}_2}{left( {\frac{1}{2}} \right)^2}$$                <\/div>\r\n                                            <div class=\"option\" data-option-key=\"option3\" data-is-correct=\"false\">\r\n                    $${left( {\frac{1}{2}} \right)^{10}}$$                <\/div>\r\n                                            <div class=\"option\" data-option-key=\"option4\" data-is-correct=\"false\">\r\n                    $${}^{10}{{\text{C}}_2}{left( {\frac{1}{2}} \right)^{10}}$$                <\/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 $\\boxed{{}^{10}{{\\text{C}}_2}{\\left( {\\frac{1}{2}} \\right)^2}}$.<\/p>\n<p>The probability of getting heads on any given toss of a fair coin is $\\frac{1}{2}$. The probability of getting tails is also $\\frac{1}{2}$. The probability of getting heads on the first two tosses is $\\frac{1}{2} \\cdot \\frac{1}{2} = \\frac{1}{4}$.<\/p>\n<p>To get the probability of getting only heads on the first two tosses and tails on the remaining eight tosses, we need to multiply the probability of getting heads on the first two tosses by the number of ways to choose two tosses from ten tosses, which is $10 \\choose 2$. We also need to multiply by the probability of getting tails on each of the remaining eight tosses, which is $\\frac{1}{2}$.<\/p>\n<p>The probability of getting only heads on the first <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> two tosses is therefore:<\/p>\n<p>$$\\frac{1}{4} \\cdot {}^{10}{{\\text{C}}_2} \\cdot \\left( \\frac{1}{2} \\right)^8 = {}^{10}{{\\text{C}}_2}{\\left( {\\frac{1}{2}} \\right)^2}$$<\/p>\n<p>Here is a brief explanation of each option:<\/p>\n<ul>\n<li>Option A: $\\left( {\\frac{1}{2}} \\right)^2$ is the probability of getting heads on two consecutive tosses. This is not the correct answer because it does not take into account the possibility of getting tails on the remaining eight tosses.<\/li>\n<li>Option B: $^{10}{{\\text{C}}_2}{\\left( {\\frac{1}{2}} \\right)^8}$ is the probability of getting tails on the remaining eight tosses, given that the first two tosses are heads. This is not the correct answer because it does not take into account the possibility of getting heads on the first two tosses and tails on the remaining eight tosses.<\/li>\n<li>Option C: $\\left( {\\frac{1}{2}} \\right)^{10}$ is the probability of getting tails on all ten <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            <\/span>\r\n            Subscribe on YouTube\r\n        <\/a>\r\n    <\/div> tosses. This is not the correct answer because it does not take into account the possibility of getting heads on the first two tosses.<\/li>\n<\/ul>\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>A fair coin is tossed 10 times. What is the probability that ONLY the first two tosses will yield heads? A. $${\\left( {\\frac{1}{2}} \\right)^2}$$ B. $${}^{10}{{\\text{C}}_2}{\\left( {\\frac{1}{2}} \\right)^2}$$ C. $${\\left( {\\frac{1}{2}} \\right)^{10}}$$ D. $${}^{10}{{\\text{C}}_2}{\\left( {\\frac{1}{2}} \\right)^{10}}$$<\/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\/a-fair-coin-is-tossed-10-times-what-is-the-probability-that-only-the-first-two-tosses-will-yield-heads-a-left-frac12-right2-b-10textc_2left-frac12\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"A fair coin is tossed 10 times. What is the probability that ONLY the first two tosses will yield heads? 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