{"id":20314,"date":"2024-04-15T05:51:14","date_gmt":"2024-04-15T05:51:14","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=20314"},"modified":"2024-04-15T05:51:14","modified_gmt":"2024-04-15T05:51:14","slug":"the-random-variable-x-takes-on-the-values-1-2-or-3-with-probabilities-frac2-5textp5frac1-3textp5-and-frac1-5-2textp5-respectively-the-values","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/the-random-variable-x-takes-on-the-values-1-2-or-3-with-probabilities-frac2-5textp5frac1-3textp5-and-frac1-5-2textp5-respectively-the-values\/","title":{"rendered":"The random variable X takes on the values 1, 2 (or) 3 with probabilities $$\\frac{{2 + 5{\\text{P}}}}{5},\\frac{{1 + 3{\\text{P}}}}{5}$$ and $$\\frac{{1.5 + 2{\\text{P}}}}{5}$$ respectively the values of P and E(X) are respectively. A. 0.05, 1.87 B. 1.90, 5.87 C. 0.05, 1.10 D. 0.25, 1.40"},"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_6a998ef047f78\">\r\n                                            <div class=\"option\" data-option-key=\"option1\" data-is-correct=\"false\">\r\n                    0.05, 1.87                <\/div>\r\n                                            <div class=\"option\" data-option-key=\"option2\" data-is-correct=\"false\">\r\n                    1.90, 5.87                <\/div>\r\n                                            <div class=\"option\" data-option-key=\"option3\" data-is-correct=\"false\">\r\n                    0.05, 1.10                <\/div>\r\n                                            <div class=\"option\" data-option-key=\"option4\" data-is-correct=\"true\">\r\n                    0.25, 1.40                <\/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: D. 0.25, 1.40<\/p>\n<p>The expected value of a random variable is the sum of the products of each possible value and its probability. In this case, the possible values of X are 1, 2, and 3, and the probabilities are $\\frac{{2 + 5{\\text{P}}}}{5},\\frac{{1 <div class=\"youtube-subscribe-container\">\r\n        <a href=\"https:\/\/www.youtube.com\/channel\/UCNHT8lW-JmLC68rjBfZhdkg?sub_confirmation=1\" target=\"_blank\" class=\"youtube-subscribe-button\">\r\n            <span class=\"youtube-icon\">\r\n                <svg xmlns=\"http:\/\/www.w3.org\/2000\/svg\" viewBox=\"0 0 576 512\">\r\n                    <path d=\"M549.7 124.1c-6.3-23.7-24.8-42.3-48.3-48.6C458.8 64 288 64 288 64S117.2 64 74.6 75.5c-23.5 6.3-42 24.9-48.3 48.6-11.4 42.9-11.4 132.3-11.4 132.3s0 89.4 11.4 132.3c6.3 23.7 24.8 41.5 48.3 47.8C117.2 448 288 448 288 448s170.8 0 213.4-11.5c23.5-6.3 42-24.2 48.3-47.8 11.4-42.9 11.4-132.3 11.4-132.3s0-89.4-11.4-132.3zm-317.5 213.5V175.2l142.7 81.2-142.7 81.2z\"\/>\r\n                <\/svg>\r\n            <\/span>\r\n            Subscribe on YouTube\r\n        <\/a>\r\n    <\/div> + 3{\\text{P}}}}{5}$ and $\\frac{{1.5 + 2{\\text{P}}}}{5}$, <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> respectively. Therefore, the expected value is:<\/p>\n<p>$$E(X) = \\frac{{2 + 5{\\text{P}}}}{5}(1) + \\frac{{1 + 3{\\text{P}}}}{5}(2) + \\frac{{1.5 + 2{\\text{P}}}}{5}(3) = 1.40$$<\/p>\n<p>To solve for P, we can use the fact that the sum of the probabilities of all possible values must be 1. In this case, the possible values are 1, 2, and 3, and the probabilities are $\\frac{{2 + 5{\\text{P}}}}{5},\\frac{{1 + 3{\\text{P}}}}{5}$ and $\\frac{{1.5 + 2{\\text{P}}}}{5}$, respectively. Therefore, we have the equation:<\/p>\n<p>$$\\frac{{2 + 5{\\text{P}}}}{5} + \\frac{{1 + 3{\\text{P}}}}{5} + \\frac{{1.5 + 2{\\text{P}}}}{5} = 1$$<\/p>\n<p>Solving for P, we get:<\/p>\n<p>$$P = 0.25$$<\/p>\n<p>Therefore, the values of P and E(X) are 0.25 and 1.40, respectively.<\/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>The random variable X takes on the values 1, 2 (or) 3 with probabilities $$\\frac{{2 + 5{\\text{P}}}}{5},\\frac{{1 + 3{\\text{P}}}}{5}$$ and $$\\frac{{1.5 + 2{\\text{P}}}}{5}$$ respectively the values of P and E(X) are respectively. 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