{"id":20341,"date":"2024-04-15T05:51:36","date_gmt":"2024-04-15T05:51:36","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=20341"},"modified":"2024-04-15T05:51:36","modified_gmt":"2024-04-15T05:51:36","slug":"a-box-contains-4-red-balls-and-6-black-balls-three-balls-are-selected-randomly-from-the-box-one-after-another-without-replacement-the-probability-that-the-selected-set-contains-one-red-ball-and-two","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/a-box-contains-4-red-balls-and-6-black-balls-three-balls-are-selected-randomly-from-the-box-one-after-another-without-replacement-the-probability-that-the-selected-set-contains-one-red-ball-and-two\/","title":{"rendered":"A box contains 4 red balls and 6 black balls. Three balls are selected randomly from the box one after another, without replacement. The probability that the selected set contains one red ball and two black balls is A. $$\\frac{1}{{20}}$$ B. $$\\frac{1}{{12}}$$ C. $$\\frac{3}{{10}}$$ D. $$\\frac{1}{2}$$"},"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_6a9817fab0bf8\">\r\n                                            <div class=\"option\" data-option-key=\"option1\" data-is-correct=\"false\">\r\n                    $$\frac{1}{{20}}$$                <\/div>\r\n                                            <div class=\"option\" data-option-key=\"option2\" data-is-correct=\"false\">\r\n                    $$\frac{1}{{12}}$$                <\/div>\r\n                                            <div class=\"option\" data-option-key=\"option3\" data-is-correct=\"true\">\r\n                    $$\frac{3}{{10}}$$                <\/div>\r\n                                            <div class=\"option\" data-option-key=\"option4\" data-is-correct=\"false\">\r\n                    $$\frac{1}{2}$$                <\/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 $\\boxed{\\frac{3}{10}}$.<\/p>\n<p>The probability of event A happening, given that event B has already happened, is called the conditional probability of A given B, and is denoted by $P(A|B)$. It can be calculated using the following formula:<\/p>\n<p>$$P(A|B) = \\frac{P(A \\cap B)}{P(B)}$$<\/p>\n<p>In this case, event A is &#8220;the selected set contains one red ball and two black balls&#8221; and event B is &#8220;the first ball selected is red&#8221;. We are asked to find the probability of event A happening, given that event B has already happened.<\/p>\n<p>The probability of event A happening is the number of ways to select one red ball and two black balls from a total of 10 balls, divided by the total number of ways to select 3 balls from 10 balls. <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> This is:<\/p>\n<p>$$P(A) = \\frac{\\binom{4}{1} \\binom{6}{2}}{\\binom{10}{3}} = \\frac{4 \\times 15}{120} = \\frac{3}{20}$$<\/p>\n<p>The probability of event B happening is the number of ways to select one red ball from a total of 4 red balls, divided by the total number of ways to select 1 ball from 10 balls. This is:<\/p>\n<p>$$P(B) = \\frac{\\binom{4}{1}}{\\binom{10}{1}} = \\frac{4}{10} = \\frac{2}{5}$$<\/p>\n<p>Therefore, the conditional probability of event A happening, given <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> that event B has already happened, is:<\/p>\n<p>$$P(A|B) = \\frac{P(A \\cap B)}{P(B)} = \\frac{\\frac{4 \\times 15}{120}}{\\frac{2}{5}} = \\frac{3}{10}$$<\/p>\n","protected":false},"excerpt":{"rendered":"<p>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 box contains 4 red balls and 6 black balls. 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