// Numbas version: finer_feedback_settings {"name": "Maria's copy of Probability of scoring at basketball given probability of not scoring", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"metadata": {"description": "
Given the probability that a basketball shot misses the hoop, find the probability that it's on target - use the law of total probability.
", "licence": "Creative Commons Attribution 4.0 International"}, "parts": [{"variableReplacements": [], "mustBeReducedPC": 0, "type": "numberentry", "showCorrectAnswer": true, "marks": 1, "showFeedbackIcon": true, "allowFractions": true, "mustBeReduced": false, "variableReplacementStrategy": "originalfirst", "correctAnswerFraction": true, "prompt": "What is the probability that the ball goes through the hoop?
\nInput your answer as a fraction.
", "minValue": "1-1/{Hoop}", "maxValue": "1-1/{Hoop}", "correctAnswerStyle": "plain", "scripts": {}, "notationStyles": ["plain", "en", "si-en"]}], "preamble": {"css": "", "js": ""}, "advice": "All probability situations can be reduced to two possible outcomes: success or failure.
\nWhen we express the outcomes in this way we say that they are complementary.
\nThe sum of the probability of an event and its complement is always $1$.
\nIf $\\mathrm{P}(\\mathrm{E})$ is the probability of an event $\\mathrm{E}$ happening and $\\mathrm{P}(\\bar{\\mathrm{E}})$ is the probability of that event not happening then
\n\\[\\mathrm{P}(\\mathrm{E}) +\\mathrm{P}(\\bar{\\mathrm{E}}) = 1.\\]
\nRearranging this equation gives:
\n\\[\\mathrm{P}(\\bar{\\mathrm{E}}) = 1 - \\mathrm{P}(\\mathrm{E})\\]
\nWhen we throw the ball we can say that there are two possible outcomes: either the ball goes through the hoop or the ball does not go through the hoop (the ball misses the hoop).
\nLet $\\mathrm{H}$ be the event that the ball goes through the hoop. Then
\n\\[\\mathrm{P}(\\mathrm{H}) + \\mathrm{P}(\\bar{\\mathrm{H}}) = 1.\\]
\nBut we are given that $\\mathrm{P}(\\bar{\\mathrm{H}}) = \\displaystyle\\frac{1}{\\var{Hoop}}$.
\nRearranging the above equation to obtain $\\mathrm{P}(\\mathrm{H})$.
\n\\begin{align}
\\mathrm{P}(\\mathrm{H}) &= 1 - \\mathrm{P}(\\bar{\\mathrm{H}}) \\\\[0.5em]
&= 1 - \\displaystyle\\frac{1}{\\var{Hoop}}\\\\[0.5em]
&= \\simplify[fractionNumbers]{{1-1/{Hoop}}}.
\\end{align}
Denominator of the fraction for the probability that the ball misses the hoop.
", "name": "Hoop", "definition": "random(5..9)", "group": "Ungrouped variables", "templateType": "anything"}}, "rulesets": {}, "tags": ["taxonomy"], "variable_groups": [], "statement": "You are playing a game of basketball against your friend. You have the ball but your friend is blocking you from moving forwards so you throw the ball and hope that it goes through the hoop.
\nThe probability that the ball misses the hoop is $\\displaystyle \\frac{1}{\\var{Hoop}}$.
", "name": "Maria's copy of Probability of scoring at basketball given probability of not scoring", "ungrouped_variables": ["Hoop"], "type": "question", "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}, {"name": "Maria Aneiros", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3388/"}]}]}], "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}, {"name": "Maria Aneiros", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3388/"}]}