// Numbas version: exam_results_page_options {"name": "Hugh's copy of The probability of an event not happening - five friends play mini golf", "extensions": ["random_person"], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"parts": [{"prompt": "

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What is $\\mathrm{P}(\\text{not } \\var{name})$?

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[[0]]

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What is $\\mathrm{P}(\\var{name})$?

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[[0]]

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All probability situations can be reduced to two possible outcomes: success or failure.

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When we express the outcomes in this way we say that they are complementary.

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The sum of the probability of an event and its complement is always $1$.

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If $\\mathrm{P}(\\mathrm{E})$ is the probability of an event $\\mathrm{E}$ happening and $\\mathrm{P}(\\bar{\\mathrm{E}})$ is the probability of that event not happening then

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\\[\\mathrm{P}(\\mathrm{E}) +\\mathrm{P}(\\bar{\\mathrm{E}}) = 1.\\]

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Rearranging this equation gives:

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\\[\\mathrm{P}(\\bar{\\mathrm{E}}) = 1 - \\mathrm{P}(\\mathrm{E})\\]

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We can think of this game as having two possible outcomes: either Dexter wins or Dexter doesn't win.

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This means that

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\\[\\mathrm{P}(\\var{name}) + \\mathrm{P}(\\text{not } \\var{name}) = 1 \\text{.}\\]

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a)

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If {name} doesn't win the game then that means that one of the other four players must win the game.

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So the probability of {name} not winning the game is the same as the probability of any of the other four players winning the game.

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Therefore

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\\begin{align}
\\mathrm{P}(\\text{not }\\var{name}) &= \\mathrm{P}(\\var{people[0]['name']})+\\mathrm{P}(\\var{people[1]['name']})+\\mathrm{P}(\\var{people[1]['name']})+\\mathrm{P}(\\var{people[1]['name']}) \\\\
&= \\var{latex(join(probs,' + '))}\\\\
&= \\var{sum(probs)}.
\\end{align}

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b)

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Rearranging the equation above gives

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\\[\\mathrm{P}(\\var{name}) = 1 - \\mathrm{P}(\\text{not } \\var{name}).\\]

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We know from a) that $\\mathrm{P}(\\text{not } \\var{name}) = \\var{sum(probs)}$.

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Therefore

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\\begin{align}
\\mathrm{P}(\\var{name}) &= 1 - \\mathrm{P}(\\text{not } \\var{name})\\\\
&= 1 - \\var{sum(probs)}\\\\
&= \\var{1-sum(probs)}.
\\end{align}

", "variables": {"name": {"name": "name", "description": "", "definition": "person['name']", "group": "Ungrouped variables", "templateType": "anything"}, "raw_probs": {"name": "raw_probs", "description": "

Uniform random values for each of the five friends. Their winning probabilities will be in proportion to this.

", "definition": "repeat(random(0..1#0),5)", "group": "Ungrouped variables", "templateType": "anything"}, "people": {"name": "people", "description": "", "definition": "random_people(5)", "group": "Ungrouped variables", "templateType": "anything"}, "person": {"name": "person", "description": "

The person whose probability is not given.

", "definition": "people[2]", "group": "Ungrouped variables", "templateType": "anything"}, "probs": {"name": "probs", "description": "

The probability of each of the first 4 friends winning the game. The missing person isn't included, so their probability can be 1 minus the sum of the rest, accumulating any rounding errors.

", "definition": "map(precround(raw_probs[j]/sum(raw_probs),2),j,0..3)", "group": "Ungrouped variables", "templateType": "anything"}}, "functions": {}, "metadata": {"description": "

Given the probabilities that each of four out of five friends will win a round of mini-golf, work out the probability that the fifth friend won't win, then use that to find the probability that they will win.

", "licence": "Creative Commons Attribution 4.0 International"}, "variable_groups": [], "tags": ["Complement", "complement", "complementary", "Probabilities sum to 1", "Probability", "probability", "taxonomy"], "ungrouped_variables": ["people", "raw_probs", "probs", "person", "name"], "name": "Hugh's copy of The probability of an event not happening - five friends play mini golf", "rulesets": {}, "statement": "

Five friends are playing a game of mini-golf.

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The probability that each person wins the game, $\\mathrm{P}(\\text{Person})$, is given in the table.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
Person{people[0]['name']}{people[1]['name']}{people[2]['name']}{people[3]['name']}{people[4]['name']}
$\\mathrm{P}(\\text{Person})$$\\var{probs[0]}$$\\var{probs[1]}$$\\var{probs[2]}$$\\var{probs[3]}$
", "type": "question", "contributors": [{"name": "Hugh O'Donnell", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2158/"}]}]}], "contributors": [{"name": "Hugh O'Donnell", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2158/"}]}