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{"name": "Probability: Discrete Random Variable 2", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "Probability: Discrete Random Variable 2", "tags": [], "metadata": {"description": "<p>Given a discrete random variable (as a formula) calculate:<br />E(X), Var, std dev, cumulative distribution and probabilties.</p>", "licence": "None specified"}, "statement": "<p>The probability distribution of a discrete random variable $\\var{X}$ is given by:</p>\n<p>$P(\\var{X} = r) = k(r+\\var{m})(r+\\var{n})$ for $1 \\le r \\le 3$</p>\n<p>$P(\\var{X} = r) = 0)$ otherwise.</p>", "advice": "<p>a) A probability distribution is a function that defines the probability of each possible outcome. &nbsp;Since the sum of probabilities of all outcomes must be 1, we can use this fact to calculate $k$.</p>\n<p>$P(\\var{X} = 1) = k(1+\\var{m})(1+\\var{n}) = \\var{(1+m)*(1+n)}k$</p>\n<p>$P(\\var{X} = 2) = k(2+\\var{m})(2+\\var{n})=\\var{(2+m)*(2+n)}k$</p>\n<p>$P(\\var{X} = 3) = k(3+\\var{m})(3+\\var{n})=\\var{(3+m)*(3+n)}k$</p>\n<p>with the probability of any other value of $\\var{X}$ being $0$.<br/><br/>Therefore $P(\\var{X} = 1) +&nbsp;P(\\var{X} =2)&nbsp;+&nbsp;P(\\var{X}=3) = 1$</p>\n<p>$\\implies \\var{(1+m)*(1+n)}k + \\var{(2+m)*(2+n)}k + \\var{(3+m)*(3+n)}k = 1$</p>\n<p>Which we can solve to show that $k=\\frac{1}{\\var{(1+m)*(1+n)+(2+m)*(2+n)+(3+n)*(3+m)}}$</p>\n<p>b) To find $P(\\var{l} \\le \\var{X} \\le \\var{u})$ we can calculate the probability of each value in the range and sum</p>", "rulesets": {}, "extensions": [], "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true}, "constants": [], "variables": {"X": {"name": "X", "group": "Ungrouped variables", "definition": "random('X','Y','A','F','M')", "description": "", "templateType": "anything", "can_override": 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