// Numbas version: exam_results_page_options {"custom_part_types": [], "contributors": [{"name": "Robert Aykroyd", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1944/"}, {"name": "Maria Aneiros", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3388/"}], "extensions": ["random_person"], "timing": {"timeout": {"action": "none", "message": ""}, "timedwarning": {"action": "none", "message": ""}, "allowPause": true}, "showQuestionGroupNames": false, "percentPass": 0, "question_groups": [{"pickQuestions": 1, "pickingStrategy": "all-ordered", "name": "Group", "questions": [{"name": "Probability - Notation and Conversion between Percentages, Decimals and Fractions", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"preventleave": false, "allowregen": true, "showfrontpage": false}, "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}, {"name": "Elliott Fletcher", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1591/"}], "variable_groups": [], "statement": "

Probabilities can be expressed as fractions, decimals or percentages.

\n

Convert each of the following probabilities into each of the two alternative numerical forms.

", "parts": [{"prompt": "

The probability that it rains today is $\\var{percentage}\\%$.

\n

i)

\n

Convert this probability to a decimal.

\n

$\\mathrm{P}(\\text{Rain}) =$ []

\n

ii)

\n

Convert this probability to a fraction.

\n

$\\mathrm{P}(\\text{Rain}) =$ []

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The probability that your bus is late today is $\\var{decimal}$.

\n

i)

\n

Convert this probability to a percentage (if necessary, round your answer to the nearest percent).

\n

$\\mathrm{P}(\\text{Late}) =$ []$\\%$.

\n

ii)

\n

Convert this probability to a fraction.

\n

$\\mathrm{P}(\\text{Late}) =$ []

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The probability that a football match between Newcastle United and Manchester United ends in a draw is $\\displaystyle\\frac{1}{\\var{denominator}}$.

\n

i)

\n

Convert this probability to a percentage (if necessary, round your answer to the nearest percent).

\n

$\\mathrm{P}(\\text{Draw}) =$ []$\\%$.

\n

ii)

\n

Convert this probability to a decimal.

\n

$\\mathrm{P}(\\text{Draw}) =$ []

Part a.

", "definition": "random(10..90 #10)", "templateType": "anything"}, "denominator": {"group": "Ungrouped variables", "name": "denominator", "description": "

Part c.

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Part b.

", "definition": "random(0.10..0.90 #0.10 except percentage/100 )", "templateType": "anything"}}, "tags": ["conversion", "Decimals", "decimals", "fractions", "Fractions", "percentages", "Probability", "probability", "taxonomy"], "variablesTest": {"maxRuns": 100, "condition": ""}, "metadata": {"licence": "Creative Commons Attribution 4.0 International", "description": "

Represent a given probability to a decimal, fraction or percentage.

#### a)

\n

We are told that $\\mathrm{P}(\\text{Rain}) = \\var{percentage}\\%$.

\n

i)

\n

To convert this to a decimal, we divide $\\var{percentage}$ by $100$.

\n

\$\\frac{\\var{percentage}}{100} = \\simplify{{{percentage}/100}}.\$

\n

Then, $\\mathrm{P}(\\text{Rain}) = \\simplify{{{percentage}/100}}$.

\n

ii)

\n

Similary, to convert this to a fraction, we divide $\\var{percentage}$ by $100$, but leave the fraction in its simplified form.

\n

\$\\frac{\\var{percentage}}{100} = \\simplify{{percentage}/100}.\$

\n

Then, $\\mathrm{P}(\\text{Rain}) = \\displaystyle\\simplify{{percentage}/100}$.

\n

#### b)

\n

We are told that $\\mathrm{P}(\\text{Late}) = \\var{decimal}$.

\n

i)

\n

To convert this to a percentage, we multiply $\\var{decimal}$ by $100$.

\n

\$\\var{decimal} \\times 100 = \\simplify{{decimal}*100}.\$

\n

Then, $\\mathrm{P}(\\text{Late}) = \\simplify{{decimal}*100}\\%$.

\n

ii)

\n

To convert this to a fraction, we multiply $\\var{decimal}$ by $\\displaystyle\\frac{100}{100}$.

\n

\\\begin{align} \\var{decimal} \\times \\frac{100}{100} &= \\frac{\\simplify{{decimal}*100}}{100} \\\\[0.5em] &= \\simplify{({decimal*100})/100}. \\end{align} \

\n

Then, $\\mathrm{P}(\\text{Late}) =\\displaystyle\\simplify{({decimal}*100)/100}$.

\n

#### c)

\n

We are told that $\\mathrm{P}(\\text{Draw}) = \\displaystyle\\frac{1}{\\var{denominator}}$.

\n

i)

\n

To convert this to a percentage, we multiply $\\displaystyle\\frac{1}{\\var{denominator}}$ by $100$.

\n

\\\begin{align} \\frac{1}{\\var{denominator}} \\times 100 &= \\var{100/{denominator}} \\\\ &= \\var{dpformat(100/{denominator},0)} & (\\text{rounded to the nearest integer}). \\end{align} \

\n

So, $\\mathrm{P}(\\text{Draw}) = \\var{dpformat(100/{denominator},0)}\\%$.

\n

ii)

\n

To convert this to a decimal, we can use a calculator to calculate $1 \\div \\var{denominator}$.

\n

\$\\frac{1}{\\var{denominator}} = \\simplify{{1/{denominator}}}.\$

\n

So, $\\mathrm{P}(\\text{Draw}) = \\var{dpformat(1/{denominator},2)}$ (rounded to two decimal places).

", "functions": {}, "preamble": {"css": "", "js": ""}}, {"name": "Probabilities of certain and impossible events", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"preventleave": false, "allowregen": true, "showfrontpage": false}, "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}, {"name": "Elliott Fletcher", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1591/"}], "variable_groups": [], "statement": "

Probability is the chance that an event will occur.

\n

A certain event is an event that will definitely happen and therefore has a probability of $1$.

\n

An impossible event is an event that will never happen and therefore has a probability of $0$.

", "parts": [{"prompt": "

A fair $6$-sided die is rolled. The probability that the die lands on a $\\var{dice}$ is []. Therefore, we say that this event is [].

\n

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1

", "

1/2

", "

1/6

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0

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impossible

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certain

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unlikely

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A teacher chooses a student at random from a class of $\\var{students}$ girls. The probability that the student chosen is a girl is []. Therefore, we say that this event is [].

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1

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{students-1}/{students}

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1/{students}

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0

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certain

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impossible

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unlikely

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Determine whether outcomes are impossible or certain to occur.

#### a)

\n

We are told that a fair, $6$-sided die is thrown.

\n

The sides on a fair, $6$-sided die are numbered $1$ to $6$. There is no $7$ on such a die.

\n

Therefore, there is no chance of the die landing on a $7$, which means that the probability of the die landing on a $7$ must be $0$.

\n

Hence, this is an impossible event.

\n

#### b)

\n

We are told that a teacher chooses a student at random from a class of $20$ girls.

\n

The key piece of information to notice here is that the teacher's class is comprised of $20$ girls. Everybody in the class is a girl.

\n

Therefore, the student that the teacher randomly chooses from the class must be a girl, which means that the probability that the student chosen is a girl is $1$.

\n

Hence, this is a certain event.

", "functions": {}, "preamble": {"css": "", "js": ""}}, {"name": "Ways of rolling two dice ", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"preventleave": false, "allowregen": true, "showfrontpage": false}, "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}, {"name": "Elliott Fletcher", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1591/"}], "variable_groups": [], "statement": "

A game asks for two dice scores to be added together, but doesn't specify how. You want to work out the probability of each possible outcome occurring.

\n

You come up with three different methods for obtaining two dice scores:

\n

Method $1$: Rolling one die twice.

\n

Method $2$: Rolling two dice consecutively.

\n

Method $3$: Rolling two identical dice simultaneously.

", "parts": [{"prompt": "

How many different outcomes are there in each of the three methods, before adding up the scores on the dice?

\n

Number of outcomes for Method $1 =$ []

\n

Number of outcomes for Method $2 =$ []

\n

Number of outcomes for Method $3 =$ []

In which of these methods are the outcomes not all equally likely to occur?

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Method 1

", "

Method 2

", "

Method 3

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Calculate outcomes for different configurations of rolling two dice.

This question asks you to think about a very subtle property of probability. Don't worry if it seems complicated - it is!

\n

#### a)

\n

You're asked to consider what can happen when you roll dice following each of the given methods, before adding up the scores on the dice. So you need to think about how many different outcomes you can observe.

\n
##### Method 1 - Rolling one die twice
\n

There are a total of $36$ different outcomes when we roll one die twice; these are shown in the table below. for each outcome, the value of the first throw is shown before the outcome of the second throw.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
123456
1(1,1)(1,2)(1,3)(1,4)(1,5)(1,6)
2(2,1)(2,2)(2,3)(2,4)(2,5)(2,6)
3(3,1)(3,2)(3,3)(3,4)(3,5)(3,6)
4(4,1)(4,2)(4,3)(4,4)(4,5)(4,6)
5(5,1)(5,2)(5,3)(5,4)(5,5)(5,6)
6(6,1)(6,2)(6,3)(6,4)(6,5)(6,6)
\n

When we roll one die twice, we know the order in which the rolls happened.

\n

This means that we can differentiate between rolling a 1 followed by a 2 - that's written (1,2) in the table above - and rolling a 2 followed by a 1 - that's written (2,1).

\n

Although the sum of the numbers in these outcomes is the same, these two outcomes are different because we are able to distinguish between the two rolls of the die.

\n
##### Method 2 - Rolling two dice consecutively
\n

Similarly to Method 1, there are a total of $36$ different outcomes when we roll two dice consecutively (one after the other); these are the same outcomes as in the table for Method 1.

\n

As in Method 1, we know that the dice were rolled in a certain order so we can distinguish between them.

\n
##### Method 3 - Rolling two identical dice simultaneously
\n

If you roll two indistinguishable dice simultaneously (at the same time), you can't tell the difference between (a 1 and a 2) or (a 2 and a 1), because there is nothing different about the dice and they weren't rolled in any order. From your point of view, they are the same outcome with probability $\\displaystyle\\frac{2}{36}$. The probabilities of the underlying events (each die's score) haven't changed, but the outcomes you observe have.

\n

In this case you have $21$ outcomes:

\n
\n
• (1,1)
• \n
• (1,2) or (2,1)
• \n
• (1,3) or (3,1)
• \n
• (1,4) or (4,1)
• \n
• (1,5) or (5,1)
• \n
• (2,2)
• \n
• (2,3) or (2,3)
• \n
• (2,4) or (4,2)
• \n
• (2,5) or (5,2)
• \n
• (2,6) or (6,2)
• \n
• (3,3)
• \n
• (3,4) or (4,3)
• \n
• (3,5) or (5,3)
• \n
• (3,6) or (6,3)
• \n
• (4,4)
• \n
• (4,5) or (5,4)
• \n
• (4,6) or (6,4)
• \n
• (5,5)
• \n
• (5,6) or (6,5)
• \n
• (6,6)
• \n
\n

#### b)

\n

In methods 1 and 2 we have $36$ outcomes, which are all different. Hence any one of these $36$ outcomes is equally likely to occur and the probability of each outcome is $\\displaystyle\\frac{1}{36}$.

\n

Contrastingly in Method 3, we have $21$ outcomes but these outcomes are not all equally likely.

\n

For example, there's only one way of obtaining two 1s, while there are two ways of obtaining a 1 and a 2, corresponding to two squares in the table in part a).

\n

Therefore, the probability of obtaining two 1s would be $\\displaystyle\\frac{1}{36}$ but the probability of obtaining a 1 and a 2 would be

\n

\$\\displaystyle\\frac{2}{36} = \\displaystyle\\frac{1}{18}.\$

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You're going to roll a fair six-sided die.

", "parts": [{"prompt": "

What is the probability of rolling an even number?

\n

[]

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$1$

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$\\displaystyle\\frac{2}{3}$

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$\\displaystyle\\frac{1}{2}$

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$\\displaystyle\\frac{1}{3}$

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What is the probability of not rolling a $\\var{die1}$ or $\\var{die2}$?

\n

[]

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$0$

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$\\displaystyle\\frac{2}{3}$

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$\\displaystyle\\frac{1}{3}$

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$\\displaystyle\\frac{1}{4}$

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Not included number for a) ii)

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number of red balls in part c

", "definition": "random(15,19)", "templateType": "anything"}, "die2": {"group": "Ungrouped variables", "name": "die2", "description": "

Not included number for a) ii)

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First part asks for the probability of rolling an even number. Second part asks for the probability of not rolling either of two given numbers.

For equally likely outcomes, you can calculate the probability of a particular event occurring by using the formula

\n

$\\text{Probability of an event} = \\displaystyle\\frac{\\text{number of favourable outcomes}}{\\text{total number of outcomes}}$.

\n

Rolling a fair six-sided die has six possible outcomes, each of which is equally likely.

\n

Let's say we want to find the probability of rolling a $2$. There is only one outcome which involves a $2$ being rolled, so the number of favourable outcomes is $1$.

\n

Hence using the above formula,

\n

\\begin{align}
P(\\text{rolling a $2$}) &= \\displaystyle\\frac{\\text{number of favourable outcomes}}{\\text{total number of outcomes}}\\\\
&= \\displaystyle\\frac{1}{6}
\\end{align}

\n

#### a)

\n

There are three possible outcomes where we roll an even number on the die:

\n
\n
• we roll a $2$;
• \n
• we roll a $4$;
• \n
• we roll a $6$.
• \n
\n

Using the formula for probability for equally likely outcomes, this means that

\n

\$P(\\text{rolling an even number}) = \\frac{\\text{number of favourable outcomes}}{\\text{total number of outcomes}}= \\frac{3}{6} = \\frac{1}{2} \$

\n

#### b)

\n

To find the probability of not rolling a $\\var{die1}$ or a $\\var{die2}$, we use the same formula again.

\n

The total number of outcomes is still $6$.

\n

Here, we have four possible outcomes which don't involve rolling a $\\var{die1}$ or a $\\var{die2}$, i.e. when we roll any of the other numbers on the die.

\n

Using the formula,

\n

\$P(\\text{not rolling a \\var{die1} or a \\var{die2}}) = \\frac{\\text{number of favourable outcomes}}{\\text{total number of outcomes}} = \\frac{4}{6} = \\frac{2}{3} \$

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An unbiased coin is flipped. What is the probability of getting a tails?

\n

[]

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$1$

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$\\displaystyle\\frac{2}{3}$

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$\\displaystyle\\frac{1}{2}$

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$\\displaystyle\\frac{1}{3}$

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Choose the probability of getting a tails, from four options.

\n

When we flip an unbiased coin there are $2$ possible outcomes: heads or tails. Both outcomes are equally likely.

\n

\\\begin{align} P(\\text{tails}) &= \\displaystyle\\frac{\\text{number of favourable outcomes}}{\\text{total number of outcomes}}\\\\ &= \\displaystyle\\frac{1}{2}. \\end{align} \

\n", "functions": {}, "preamble": {"css": "", "js": ""}}, {"name": "Probability of picking a particular colour ball from a bag", "extensions": [], "custom_part_types": [], "resources": [["question-resources/dice.svg", "/srv/numbas/media/question-resources/dice.svg"]], "navigation": {"preventleave": false, "allowregen": true, "showfrontpage": false}, "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}], "variable_groups": [], "statement": "", "parts": [{"marks": 0, "scripts": {}, "gaps": [{"variableReplacementStrategy": "originalfirst", "distractors": ["This is the probability of not picking a blue ball.", "Divide by the total number of outcomes, not the number of unfavourable outcomes.", "", "Divide the number of favourable outcomes by the total number of outcomes.", "There's more than one blue ball."], "showFeedbackIcon": true, "type": "1_n_2", "marks": 0, "minMarks": 0, "displayColumns": 0, "displayType": "radiogroup", "scripts": {}, "shuffleChoices": false, "showCorrectAnswer": true, "choices": ["

$\\displaystyle\\frac{\\var{red+green}}{\\var{total}}$

", "

$\\displaystyle\\frac{\\var{blue}}{\\var{green+red}}$

", "

$\\displaystyle\\frac{\\var{blue}}{\\var{total}}$

", "

$\\displaystyle\\frac{1}{\\var{blue}}$

", "

$\\displaystyle\\frac{1}{\\var{total}}$

"], "variableReplacements": [], "maxMarks": 0, "matrix": [0, 0, "1", 0, 0]}], "showCorrectAnswer": true, "showFeedbackIcon": true, "prompt": "

A bag contains $\\var{red}$ red balls, $\\var{blue}$ blue balls and $\\var{green}$ green balls. One ball is removed from the bag at random. What is the probability that the chosen ball will be blue? Remember to reduce any fractions into their simplest form.

\n

[]

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number of red balls in part c

", "definition": "random(15,19)", "templateType": "anything"}, "blue": {"group": "Ungrouped variables", "name": "blue", "description": "

number of blue balls in part c

", "definition": "random(6,7,11)", "templateType": "anything"}, "green": {"group": "Ungrouped variables", "name": "green", "description": "

number of green balls in part c.

", "definition": "random(4,8,10)", "templateType": "anything"}, "total": {"group": "Ungrouped variables", "name": "total", "description": "

total number of balls in part c

", "definition": "red+blue+green", "templateType": "anything"}}, "tags": ["taxonomy"], "variablesTest": {"maxRuns": "100", "condition": ""}, "metadata": {"licence": "Creative Commons Attribution 4.0 International", "description": "

A bag contains balls of three different colours. You're told how many there are of each, and asked the probability of picking a ball of a particular colour.

For equally likely outcomes, you can calculate the probability of a particular event occurring by using the formula

\n

$\\text{Probability of an event} = \\displaystyle\\frac{\\text{number of favourable outcomes}}{\\text{total number of outcomes}}$.

\n

\n

We are told that the bag contains $\\var{red}$ red balls, $\\var{blue}$ blue balls and $\\var{green}$ green balls and that one ball is removed from the bag at random.

\n

The total number of balls in the bag before the chosen ball is removed is

\n

\$\\var{red}+\\var{blue}+\\var{green} = \\var{total}.\$

\n

As the ball is being removed randomly from the bag, there is an equal probability of selecting any one of the $\\var{total}$ balls.

\n

Therefore, the probability of the chosen ball being blue is

\n

\$P(\\text{blue}) = \\displaystyle\\frac{\\text{number of favourable outcomes}}{\\text{total number of outcomes}} = \\displaystyle\\frac{\\var{blue}}{\\var{total}} \$

", "functions": {}, "preamble": {"css": "", "js": ""}}, {"name": "Which coin is more likely to be biased?", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"preventleave": false, "allowregen": true, "showfrontpage": false}, "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}, {"name": "Elliott Fletcher", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1591/"}], "variable_groups": [], "statement": "

Two different coins are flipped a different number of times. It is known that one of the coins is biased.

\n

The results for both coins are given in the table below.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
OutcomeCoin 1Coin 2
Heads$\\var{h1}$$\\var{h2} Tails\\var{t1}$$\\var{t2}$
", "parts": [{"prompt": "

Calculate the experimental probability of tossing heads, $P(\\text{heads})$, for each coin. Enter your answers as fractions.

\n

$P(\\text{heads})\\; \\text{for Coin$1$} =$ []

\n

$P(\\text{heads})\\; \\text{for Coin$2$} =$ []

\n

", "scripts": {}, "gaps": [{"mustBeReducedPC": 0, "marks": 1, "notationStyles": ["plain", "en", "si-en"], "mustBeReduced": false, "correctAnswerStyle": "plain", "showFeedbackIcon": true, "minValue": "{h1}/10000", "type": "numberentry", "variableReplacementStrategy": "originalfirst", "variableReplacements": [], "correctAnswerFraction": true, "scripts": {}, "showCorrectAnswer": true, "maxValue": "{h1}/10000", "allowFractions": true}, {"mustBeReducedPC": 0, "marks": 1, "notationStyles": ["plain", "en", "si-en"], "mustBeReduced": false, "correctAnswerStyle": "plain", "showFeedbackIcon": true, "minValue": "{h2}/15", "type": "numberentry", "variableReplacementStrategy": "originalfirst", "variableReplacements": [], "correctAnswerFraction": true, "scripts": {}, "showCorrectAnswer": true, "maxValue": "{h2}/15", "allowFractions": true}], "showCorrectAnswer": true, "showFeedbackIcon": true, "type": "gapfill", "variableReplacementStrategy": "originalfirst", "marks": 0, "variableReplacements": []}, {"prompt": "

Based on the available results and the experimental probabilities calculated in part a), for which coin is there more evidence of bias?

", "steps": [{"prompt": "

We say that an event is biased if one outcome of the event is more likely than the other outcomes.

", "scripts": {}, "showCorrectAnswer": true, "showFeedbackIcon": true, "type": "information", "variableReplacementStrategy": "originalfirst", "marks": 0, "variableReplacements": []}], "maxMarks": 0, "type": "1_n_2", "showFeedbackIcon": true, "marks": 0, "variableReplacementStrategy": "originalfirst", "minMarks": 0, "displayColumns": 0, "displayType": "radiogroup", "scripts": {}, "showCorrectAnswer": true, "shuffleChoices": false, "choices": ["

Coin 1

", "

Coin 2

"], "variableReplacements": [], "stepsPenalty": 0, "distractors": ["", ""], "matrix": ["1", 0]}, {"prompt": "

You decide to take part in a bet with your friend Alex.

\n

Alex says that if you toss either one of the above coins and the coin lands on tails then you win $£\\var{win}$. However, if the coin lands on heads then Alex wins $£\\var{win}$.

\n

Which coin should you choose in order to have a more reliable chance of winning the bet?

", "maxMarks": 0, "type": "1_n_2", "showFeedbackIcon": true, "marks": 0, "variableReplacementStrategy": "originalfirst", "variableReplacements": [], "displayColumns": 0, "displayType": "radiogroup", "scripts": {}, "showCorrectAnswer": true, "shuffleChoices": false, "choices": ["

Coin 1

", "

Coin 2

"], "minMarks": 0, "distractors": ["", ""], "matrix": [0, "1"]}], "ungrouped_variables": ["h1", "t1", "h2", "t2", "win"], "rulesets": {}, "type": "question", "variables": {"h2": {"definition": "15 - t2", "name": "h2", "description": "

Number of heads for coin 2.

", "group": "Ungrouped variables", "templateType": "anything"}, "h1": {"definition": "random(6000..7000 #250)", "name": "h1", "description": "

number of heads for coin 1.

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Money won in part b.

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Number of tails for coin 2.

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Number of tails for coin 1

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This question aims to assess the student's understanding of the difference between biased and unbiased events and also to assess the student's understanding of the fact that the experimental probability tends towards the theoretical probability as the number of trials increases.

The results from the tosses of both coins are given in the table below.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
 Outcome Coin 1 Coin 2 Heads $\\var{h1}$ $\\var{h2}$ Tails $\\var{t1}$ $\\var{t2}$
\n

\n

\n

\n

\n

\n

#### a)

\n

We can calculate the number of tosses of each coin by adding together the number of heads and tails obtained for each coin.

\n

Coin $1$ was tossed $\\var{h1+t1}$ times.

\n

Coin $2$ was tossed $\\var{h2+t2}$ times.

\n

This means that for Coin $1$, the experimental probability of tossing heads is

\n

\\\begin{align} \\displaystyle\\frac{\\var{h1}}{10000} &= \\var{{h1}/10000}\\\\ &= \\var{100*({h1}/10000)}\\%. \\end{align} \

\n

Whereas for Coin $2$, the experimental probability of tossing heads is

\n

\\\begin{align}\\displaystyle\\frac{\\var{h2}}{15} &= \\var{dpformat({h2}/15,2)} \\; (\\text{rounded to two decimal places})\\\\&= \\var{dpformat(100*({h2}/15),0)}\\%. \\end{align} \

\n

\\\begin{align}\\displaystyle\\frac{\\var{h2}}{15} &= \\var{{h2}/15}\\\\&= \\var{100*({h2}/15)}\\%. \\end{align} \

\n

#### b)

\n

We can see from part a) that the number of tosses of Coin $1$ is much larger than the number of tosses of Coin $2$.

\n

It is important to know that as the number of trials in an experiment gets very large the experimental probability tends towards the theoretical probability.

\n

For an unbiased coin, the theoretical probability of tossing heads is $50\\%$ so as the number of tosses gets very large we would expect the experimental probability of tossing heads to get closer to $50\\%$.

\n

Therefore, as the number of tosses of Coin $1$ is very large and the experimental probability of tossing heads with this coin is significantly different from $50\\%$, then it is quite likely that Coin $1$ could be biased.

\n

On the other hand, as the number of tosses for Coin $2$ is small, we cannot give an accurate opinion of whether this coin is biased or unbiased.

\n

So, there is more evidence of bias for Coin $1$ than for Coin $2$.

\n

#### c)

\n

As discussed in part b), there is more evidence of bias for Coin $1$  than for Coin $2$.

\n

Furthermore, we saw that the experimental probability of tossing heads with Coin $1$ is significantly different from $50\\%$, which means that the coin could be biased in favour of heads.

\n

So, as we want the coin to land tails then we want to choose the coin that has the most reliable chance of landing tails-up.

\n

Since there is less evidence of bias for Coin $2$ than for Coin $1$, we should choose to use Coin $2$ in the bet as there is a more reliable chance of tossing tails with this coin.

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An unbiased coin is flipped $\\var{no_flips}$ times. Given that the coin landed on heads each time, what is the probability of the coin landing on heads the next time it is flipped?

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Number of flips of the coin

", "group": "Ungrouped variables", "templateType": "anything"}}, "tags": ["taxonomy"], "variablesTest": {"maxRuns": 100, "condition": ""}, "metadata": {"licence": "Creative Commons Attribution 4.0 International", "description": "

Previous throws don't affect the probability distribution of subsequent throws. Believing otherwise is the gambler's fallacy.

When we flip an unbiased coin there are two possible events that we could measure: the coin lands on heads or the coin lands on tails.

\n

Each toss of the coin is independent; if we flip a coin once and it lands on heads then the next time we flip the coin it is still equally likely to land on either heads or tails.

\n

It doesn't matter what the coin landed on previously as this outcome does not affect the outcome of the next flip of the coin.

\n

Even when we flip an unbiased coin $\\var{no_flips}$ times and it lands on heads each time; the next time we flip the coin, it is still equally likely to land on either heads or tails.

\n

So the probability that the coin lands on heads the next time that the coin is flipped is still $\\displaystyle\\frac{1}{2}$.

\n", "functions": {}, "preamble": {"css": "", "js": ""}}, {"name": "Calculating Expected Values given a table of probabilities", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"preventleave": false, "allowregen": true, "showfrontpage": false}, "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}, {"name": "Elliott Fletcher", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1591/"}], "variable_groups": [], "statement": "

There are four films being shown in a cinema on a particular day.

\n

The probability that a person buys a ticket to see each film, denoted $P(\\text{Film})$, is given in the table below.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
 Film $P(\\text{Film})$ Genre Forgotten Game $\\var{Avatar}$ Sci-Fi The Diamond Valley $\\var{SW}$ Sci-Fi School of Return $\\var{NYSM}$ Thriller The Silk's Nobody $\\var{TIJ}$ Crime
\n

$\\var{no_people}$ people each buy a ticket at the cinema to see a film of their own choosing during the day.

", "parts": [{"mustBeReducedPC": 0, "prompt": "

How many of these people would you expect to have bought tickets to see Forgotten Game?

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How many of these people would you expect to have bought tickets to see a Sci-Fi film?

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Number of people who see a movie.

", "definition": "random(100..180 #20)", "templateType": "anything"}, "TIJ": {"group": "Ungrouped variables", "name": "TIJ", "description": "

Probability someone goes to see the Italian Job

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Probability someone goes to see Star Wars

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Probability someone sees Avatar

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Probability someone goes to see Now you see me

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This question assesses the students ability to find the expected number of times an event occurs given the probability of the event occurring for a single trial and the total number of trials.

If we are given the probability of an event occurring in a single trial then we can calculate the expected number of times that this event would occur in a larger number of trials.

\n

To do this, we multiply the probability of the event occurring in a single trial by the total number of trials:

\n

\$\\text{Expected number of times an event occurs} = \\text{Probability of event} \\times \\text{Number of trials}.\$

\n

We are given the probabilities that someone buys a ticket to see each film in the table below.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
 Film $P(\\text{Film})$ Genre Forgotten Game $\\var{Avatar}$ Sci-Fi The Diamond Valley $\\var{SW}$ Sci-Fi School of Return $\\var{NYSM}$ Thriller The Silk's Nobody $\\var{TIJ}$ Crime
\n

We are also told that $\\var{no_people}$ people each buy a ticket at the cinema to see a film of their own choosing during this day.

\n

#### a)

\n

To calculate the expected number of people who bought tickets to see one of these films we multiply the probability that a person buys a ticket for that film by how many people bought tickets for a film at the cinema.

\n

So the expected number of people who bought tickets to see Forgotten Game is

\n

\$\\var{Avatar} \\times \\var{no_people} = \\var{{Avatar}*{no_people}}. \$

\n

#### b)

\n

We are now asked to calculate the expected number of people who bought tickets to see a Sci-Fi film.

\n

From the table above we can see that there are two films which belong to the Sci-Fi genre: Forgotten Game and The Diamond Valley.

\n

Firstly, we need to calculate the probability that a person buys a ticket to see a Sci-Fi film, which we will denote $P(\\text{Sci-Fi})$.

\n

Since the probability that a person buys a ticket to see each film is different, it would be incorrect to say that the probability that a person buys a ticket to see a Sci-Fi film is

\n

\$\\displaystyle\\frac{2}{4} = \\displaystyle\\frac{1}{2}.\$

\n

Instead we must recognise that the probability that a person buys a ticket to see a Sci-Fi film is the probability that a person buys a ticket to see either Forgotten or The Diamond Valley.

\n

Therefore to calculate this probability, we add the probabilities of a person buying a ticket to see each of these films:

\n

\\\begin{align} P(\\text{Sci-Fi}) &= P(\\text{Forgotten Game})+P(\\text{The Diamond Valley})\\\\ &= \\var{Avatar}+\\var{SW}\\\\ &= \\var{Avatar+SW}. \\end{align} \

\n

Then the expected number of people who bought tickets to see a Sci-Fi film is

\n

\$\\var{Avatar+SW} \\times \\var{no_people} = \\var{({Avatar+SW})*{no_people}}. \$

\n

", "functions": {}, "preamble": {"css": "", "js": ""}}, {"name": "Calculating expected values using theoretical probability and experimental probability", "extensions": ["random_person"], "custom_part_types": [], "resources": [], "navigation": {"preventleave": false, "allowregen": true, "showfrontpage": false}, "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}, {"name": "Elliott Fletcher", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1591/"}], "variable_groups": [], "statement": "

{name} rolls an unbiased six-sided die $\\var{no_rolls}$ times.

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Based on the theoretical probability of rolling a $\\var{num1}$ or a $\\var{num2}$, how many times would you expect {pronouns['them']} to roll either one of these numbers?

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After performing the experiment, {name} reports that {pronouns['they']} rolled either a $\\var{num1}$ or a $\\var{num2}$ on $\\var{Obtained}$ occasions.

\n

Calculate the relative frequency of rolling either a $\\var{num1}$ or a $\\var{num2}$.

\n

\n

$\\text{Relative Frequency} =$ []

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If {name} rolled the same die $\\var{more_rolls}$ more times, how many times could {pronouns['they']} expect to roll either a $\\var{num1}$ or a $\\var{num2}$?

\n

Based on the experimental data: []

\n

Based on the theoretical probability: []

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Number of rolls of the die.

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Second number.

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First number.

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Number of extra rolls of the die

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Number of times the event is obtained in the experiment.

", "definition": "(no_rolls*multiplier)/10", "templateType": "anything"}, "person": {"group": "Ungrouped variables", "name": "person", "description": "", "definition": "random_person()", "templateType": "anything"}, "multiplier": {"group": "Ungrouped variables", "name": "multiplier", "description": "

multiplier for the value of Obtained variable

", "definition": "random(5,6,7)", "templateType": "anything"}}, "tags": ["Dice", "dice", "Expected values", "Expected Values", "Experimental Probability", "Experimental probability", "experimental probability", "probability", "Probability", "relative frequency", "Relative Frequency", "taxonomy", "theoretical probability", "Theoretical Probability"], "variablesTest": {"maxRuns": 100, "condition": ""}, "metadata": {"licence": "Creative Commons Attribution 4.0 International", "description": "

This question assesses

\n
\n
• the students ability to apply both theoretical and experimental probability to calculate expected values
• \n
• the students understanding of how to calculate the relative frequency of an outcome
• \n
\n

The question also helps to show students how using experimental probability and theoretical probability results in different expected values of an outcome.

#### a)

\n

Firstly, we must calculate the theoretical probability of rolling either a $\\var{num1}$ or a $\\var{num2}$.

\n

Both $\\var{num1}$ and $\\var{num2}$ only appear once on an unbiased six-sided die, so there are only $2$ possible outcomes where we roll either a $\\var{num1}$ or a $\\var{num2}$.

\n

There are $6$ possible outcomes when we roll an unbiased six-sided die.

\n

Therefore, the theoretical probability of rolling either a $\\var{num1}$ or a $\\var{num2}$ is

\n

\$\\displaystyle\\frac{2}{6} = \\displaystyle\\frac{1}{3}.\$

\n

Then the expected number of times that {name} rolls either a $\\var{num1}$ or a $\\var{num2}$ is

\n

\$\\var{no_rolls} \\times \\displaystyle\\frac{1}{3} = \\var{{no_rolls}/3}.\$

\n

#### b)

\n

We are told that in {pronouns['their']} experiment, {name} obtained either a $\\var{num1}$ or a $\\var{num2}$ on $\\var{Obtained}$ occasions.

\n

Recall the formula for the relative frequency of an outcome.

\n

\$\\text{Relative Frequency} = \\displaystyle\\frac{\\text{Frequency of an outcome}}{\\text{Number of trials}}.\$

\n

The Number of trials in the experiment is $\\var{no_rolls}$ and the frequency of the desired outcome is $\\var{Obtained}$.

\n

So the relative frequency of rolling either a $\\var{num1}$ or a $\\var{num2}$ is $\\displaystyle\\frac{\\var{Obtained}}{\\var{no_rolls}}$.

\n

#### c)

\n

The same die is now thrown $\\var{more_rolls}$ times.

\n

We know from b) that the relative frequency of rolling either a $\\var{num1}$ or a $\\var{num2}$ with this die was $\\displaystyle\\simplify{{Obtained}/{no_rolls}}$.

\n

Therefore using the experimental data, the number of times we would expect {name} to roll either a $\\var{num1}$ or a $\\var{num2}$ in $\\var{more_rolls}$ throws of the die is

\n

\$\\var{more_rolls} \\times \\displaystyle\\simplify{{Obtained}/{no_rolls}} = \\var{{more_rolls}*{Obtained}/{no_rolls}}.\$

\n

On the other hand, we know from a) that the theoretical probability of rolling either a $\\var{num1}$ or a $\\var{num2}$ with this die is $\\displaystyle\\frac{1}{3}$.

\n

Using the theoretical probability, the number of times we would expect {name} to roll either a $\\var{num1}$ or a $\\var{num2}$ in $\\var{more_rolls}$ throws of the die is

\n

\$\\var{more_rolls} \\times \\displaystyle\\frac{1}{3} = \\var{{more_rolls}/3}.\$

", "functions": {}, "preamble": {"css": "fraction {\n display: inline-block;\n vertical-align: middle;\n}\nfraction > numerator, fraction > denominator {\n float: left;\n width: 100%;\n text-align: center;\n line-height: 2.5em;\n}\nfraction > numerator {\n border-bottom: 1px solid;\n padding-bottom: 5px;\n}\nfraction > denominator {\n padding-top: 5px;\n}\nfraction input {\n line-height: 1em;\n}\n\nfraction .part {\n margin: 0;\n}\n\n.table-responsive, .fractiontable {\n display:inline-block;\n}\n.fractiontable {\n padding: 0; \n border: 0;\n}\n\n.fractiontable .tddenom \n{\n text-align: center;\n}\n\n.fractiontable .tdnum \n{\n border-bottom: 1px solid black; \n text-align: center;\n}\n\n\n.fractiontable tr {\n height: 3em;\n}\n", "js": ""}}, {"name": "The probability of an event not happening - five friends play mini golf", "extensions": ["random_person"], "custom_part_types": [], "resources": [], "navigation": {"preventleave": false, "allowregen": true, "showfrontpage": false}, "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}, {"name": "Elliott Fletcher", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1591/"}], "variable_groups": [], "statement": "

Five friends are playing a game of mini-golf.

\n

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['name']} {people['name']} {people['name']} {people['name']} {people['name']} $\\mathrm{P}(\\text{Person})$ $\\var{probs}$ $\\var{probs}$ $\\var{probs}$ $\\var{probs}$
", "parts": [{"prompt": "

\n

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

\n

[]

\n

", "scripts": {}, "gaps": [{"mustBeReducedPC": 0, "marks": 1, "notationStyles": ["plain", "en", "si-en"], "allowFractions": false, "correctAnswerStyle": "plain", "showFeedbackIcon": true, "minValue": "sum(probs)", "type": "numberentry", "variableReplacementStrategy": "originalfirst", "variableReplacements": [], "correctAnswerFraction": false, "scripts": {}, "showCorrectAnswer": true, "maxValue": "sum(probs)", "mustBeReduced": false}], "showCorrectAnswer": true, "showFeedbackIcon": true, "type": "gapfill", "variableReplacementStrategy": "originalfirst", "marks": 0, "variableReplacements": []}, {"prompt": "

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

\n

[]

", "scripts": {}, "gaps": [{"mustBeReducedPC": 0, "marks": 1, "notationStyles": ["plain", "en", "si-en"], "allowFractions": false, "correctAnswerStyle": "plain", "showFeedbackIcon": true, "minValue": "1-sum(probs)", "type": "numberentry", "variableReplacementStrategy": "originalfirst", "variableReplacements": [], "correctAnswerFraction": false, "scripts": {}, "showCorrectAnswer": true, "maxValue": "1-sum(probs)", "mustBeReduced": false}], "showCorrectAnswer": true, "showFeedbackIcon": true, "type": "gapfill", "variableReplacementStrategy": "originalfirst", "marks": 0, "variableReplacements": []}], "ungrouped_variables": ["people", "raw_probs", "probs", "person", "name"], "rulesets": {}, "type": "question", "variables": {"people": {"definition": "random_people(5)", "name": "people", "description": "", "group": "Ungrouped variables", "templateType": "anything"}, "name": {"definition": "person['name']", "name": "name", "description": "", "group": "Ungrouped variables", "templateType": "anything"}, "raw_probs": {"definition": "repeat(random(0..1#0),5)", "name": "raw_probs", "description": "

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

", "group": "Ungrouped variables", "templateType": "anything"}, "person": {"definition": "people", "name": "person", "description": "

The person whose probability is not given.

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

", "group": "Ungrouped variables", "templateType": "anything"}}, "tags": ["Complement", "complement", "complementary", "Probabilities sum to 1", "Probability", "probability", "taxonomy"], "variablesTest": {"maxRuns": 100, "condition": ""}, "metadata": {"licence": "Creative Commons Attribution 4.0 International", "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.

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

\n

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

\n

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

\n

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

\n

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

\n

Rearranging this equation gives:

\n

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

\n

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

\n

This means that

\n

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

\n\n

#### a)

\n

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

\n

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.

\n

Therefore

\n

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

\n

#### b)

\n

Rearranging the equation above gives

\n

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

\n

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

\n

Therefore

\n

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

", "functions": {}, "preamble": {"css": "", "js": ""}}, {"name": "Probability of scoring at basketball given probability of not scoring", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"preventleave": false, "allowregen": true, "showfrontpage": false}, "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}], "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.

\n

The probability that the ball misses the hoop is $\\displaystyle \\frac{1}{\\var{Hoop}}$.

", "parts": [{"mustBeReducedPC": 0, "prompt": "

What is the probability that the ball goes through the hoop?

\n

", "notationStyles": ["plain", "en", "si-en"], "allowFractions": true, "correctAnswerStyle": "plain", "showFeedbackIcon": true, "minValue": "1-1/{Hoop}", "type": "numberentry", "variableReplacementStrategy": "originalfirst", "marks": 1, "variableReplacements": [], "correctAnswerFraction": true, "scripts": {}, "showCorrectAnswer": true, "maxValue": "1-1/{Hoop}", "mustBeReduced": false}], "ungrouped_variables": ["Hoop"], "rulesets": {}, "type": "question", "variables": {"Hoop": {"group": "Ungrouped variables", "name": "Hoop", "description": "

Denominator of the fraction for the probability that the ball misses the hoop.

", "definition": "random(5..9)", "templateType": "anything"}}, "tags": ["taxonomy"], "variablesTest": {"maxRuns": 100, "condition": ""}, "metadata": {"licence": "Creative Commons Attribution 4.0 International", "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.

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

\n

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

\n

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

\n

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

\n

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

\n

Rearranging this equation gives:

\n

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

\n

When 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).

\n

Let $\\mathrm{H}$ be the event that the ball goes through the hoop. Then

\n

\$\\mathrm{P}(\\mathrm{H}) + \\mathrm{P}(\\bar{\\mathrm{H}}) = 1.\$

\n

But we are given that $\\mathrm{P}(\\bar{\\mathrm{H}}) = \\displaystyle\\frac{1}{\\var{Hoop}}$.

\n

Rearranging 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} \n", "functions": {}, "preamble": {"css": "", "js": ""}}, {"name": "Mutually exclusive events MCQ", "extensions": [], "custom_part_types": [], "resources": [["question-resources/aceOfSpades.svg", "/srv/numbas/media/question-resources/aceOfSpades.svg"], ["question-resources/aceOfSpades_NnxOXmM.svg", "/srv/numbas/media/question-resources/aceOfSpades_NnxOXmM.svg"]], "navigation": {"preventleave": false, "allowregen": true, "showfrontpage": false}, "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}, {"name": "Elliott Fletcher", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1591/"}], "variable_groups": [], "statement": "", "parts": [{"minAnswers": "1", "marks": 0, "warningType": "prevent", "showFeedbackIcon": true, "prompt": " Which of the following are mutually exclusive events? ", "distractors": ["", "", "", ""], "type": "m_n_2", "variableReplacementStrategy": "originalfirst", "variableReplacements": [], "minMarks": "0", "displayColumns": "1", "displayType": "checkbox", "scripts": {}, "shuffleChoices": false, "showCorrectAnswer": true, "choices": [" Rolling a \\var{even} on a die, and rolling an odd number. ", " Picking an Ace at random from a deck of cards, and picking a Spade. ", " Walking, and scratching your head. ", " Flipping a coin and getting heads, and flipping a coin and getting tails. "], "maxAnswers": "4", "maxMarks": "2", "matrix": ["1", "-1", "-1", "1"]}], "ungrouped_variables": ["even"], "rulesets": {}, "type": "question", "variables": {"even": {"definition": "random(2,4,6)", "name": "even", "description": "", "group": "Ungrouped variables", "templateType": "anything"}}, "tags": ["Multiple Choice", "multiple choice", "Multiple choice", "Mutually exclusive events", "Probability", "probability", "taxonomy"], "variablesTest": {"maxRuns": 100, "condition": ""}, "metadata": {"licence": "Creative Commons Attribution 4.0 International", "description": " Given descriptions of some pairs of random events, pick the ones which are mutually exclusive. "}, "advice": " Mutually exclusive events are events that cannot happen at the same time. \n \n When you roll a die, it is only possible for the die to land on one number. Therefore it is impossible to roll both a \\var{even} and an odd number, which means that these events are mutually exclusive. \n It is possible to walk while scratching your head, so these events are not mutually exclusive. \n It is impossible to obtain both a heads and a tails when you flip a coin, so these events are mutually exclusive. \n In a standard deck of cards, there is a card which is both an Ace and a Spade, the Ace of Spades. \n \n Therefore it is possible to randomly select a card from such a deck which is both an Ace and a Spade, which means that these events are not mutually exclusive. ", "functions": {}, "preamble": {"css": "", "js": ""}}, {"name": "Probability of the union of two non-mutually exclusive events", "extensions": [], "custom_part_types": [], "resources": [["question-resources/breakfastven2.svg", "/srv/numbas/media/question-resources/breakfastven2.svg"], ["question-resources/breakfastvenplus2.svg", "/srv/numbas/media/question-resources/breakfastvenplus2.svg"]], "navigation": {"preventleave": false, "allowregen": true, "showfrontpage": false}, "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}, {"name": "Elliott Fletcher", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1591/"}], "variable_groups": [], "statement": " A survey asked people what they eat for breakfast. Participants had to select foods that they typically eat for breakfast from a list. \n A story in the newspaper displayed the results of the survey in this table: \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n  Food % of participants Cereal Toast Fruit Fry-up Other {proportions} {proportions} {proportions} {proportions} {proportions} \n The most popular combination was cereal and toast, with \\var{b}\\% of the participants selecting both. ", "parts": [{"prompt": " Consider the event that someone has cereal for breakfast and the event that the same person has toast for breakfast. Are these events mutually exclusive? ", "variableReplacementStrategy": "originalfirst", "distractors": ["", ""], "showFeedbackIcon": true, "type": "1_n_2", "marks": 0, "minMarks": 0, "displayColumns": 0, "displayType": "radiogroup", "scripts": {}, "shuffleChoices": false, "showCorrectAnswer": true, "choices": [" Yes ", " No "], "variableReplacements": [], "maxMarks": 0, "matrix": [0, "1"]}, {"prompt": " What is the probability that a participant selected at random typically eats cereal or toast for breakfast? ", "variableReplacementStrategy": "originalfirst", "distractors": ["", "You have double-counted the people who eat both.", "Subtract the people who take both, rather than adding them on again.", "Some people eat neither cereal nor toast, so the probability is less than 1."], "showFeedbackIcon": true, "type": "1_n_2", "marks": 0, "minMarks": 0, "displayColumns": 0, "displayType": "radiogroup", "scripts": {}, "shuffleChoices": false, "showCorrectAnswer": true, "choices": [" \\var{({c}+{t}-{b})/100} ", " \\var{({c}+{t})/100} ", " \\var{({c}+{t}+{b})/100} ", " 1 "], "variableReplacements": [], "maxMarks": 0, "matrix": ["1", 0, 0, "0"]}], "ungrouped_variables": ["c", "t", "b", "raw_proportions", "proportions", "total"], "rulesets": {}, "type": "question", "variables": {"t": {"group": "Ungrouped variables", "name": "t", "description": " Percentage of people who have toast for breakfast. ", "definition": "proportions//random(6..15)", "templateType": "anything"}, "c": {"group": "Ungrouped variables", "name": "c", "description": " Percentage of people who have cereal for breakfast. ", "definition": "proportions//random(20..40)", "templateType": "anything"}, "raw_proportions": {"group": "Ungrouped variables", "name": "raw_proportions", "description": "", "definition": "[random(4..6#0),random(2..3#0)]+repeat(random(1..4#0),3)", "templateType": "anything"}, "proportions": {"group": "Ungrouped variables", "name": "proportions", "description": " The percentage of people who ticked each option. ", "definition": "map(floor(total*x/sum(raw_proportions)),x,raw_proportions)", "templateType": "anything"}, "b": {"group": "Ungrouped variables", "name": "b", "description": " Percentage of people who have both toast and cereal for breakfast. Between an eighth and a third of the lowest of the two options. ", "definition": "let(s,min(c,t), random(round(s/8)..round(s/3)))", "templateType": "anything"}, "total": {"group": "Ungrouped variables", "name": "total", "description": " The total of the percentages for each option. \n This is greater than 100 because some people tick more than one option. ", "definition": "random(120..160)", "templateType": "anything"}}, "tags": ["Multiple Choice", "multiple choice", "Multiple choice", "Mutually exclusive events", "Not mutually exclusive events", "Probability", "probability", "taxonomy"], "variablesTest": {"maxRuns": 100, "condition": ""}, "metadata": {"licence": "Creative Commons Attribution 4.0 International", "description": " Given results from a survey about what people eat for breakfast, where some people eat one or both of cereal and toast. Student is asked to pick the probability of eating either one or the other from a list. Distractors pick out common errors. "}, "advice": " #### a) \n Mutually exclusive events are events that cannot happen at the same time. \n We know from the results of the survey that \\var{b}\\% of participants stated that they have cereal as well as toast for breakfast. \n Therefore it is possible to have both cereal and toast for breakfast, which means that the events \"cereal\" and \"toast\" are not mutually exclusive. \n #### b) \n We know from the results of the survey that some people have both cereal and toast for breakfast, so we can present the information given to us in the question in the form of a Venn diagram. \n \n \n The number of people who have cereal or toast for breakfast is: \n \n • all the people who have cereal (including the participants who have cereal as well as toast) • \n • all the people who have toast (including the participants who have toast as well as cereal) • \n \n However, this counts the people who have cereal as well as toast twice! \n To correct our answer, we subtract the extra \"and\" part: \n \n As a general formula this is: \n \\[\\mathrm{P}(\\mathrm{A} \\cup \\mathrm{B}) = \\mathrm{P}(\\mathrm{A}) + \\mathrm{P}(\\mathrm{B}) - \\mathrm{P}(\\mathrm{A} \\cap \\mathrm{B}).\

\n

Note that here we have made use of some notation that is frequently used in probability calculations:

\n
\n
• The \"Intersection\" symbol $\\cap$, used instead of \"and\".
• \n
• The \"Union\" symbol $\\cup$, used instead of \"or\".
• \n
\n

Using this equation, the probability that a participant selected at random will either have cereal or toast for breakfast is

\n

\\\begin{align} \\mathrm{P}(\\text{cereal} \\cup \\text{toast}) &= \\mathrm{P}(\\text{cereal})+\\mathrm{P}(\\text{toast}) - \\mathrm{P}(\\text{cereal} \\cap \\text{toast})\\\\ &= \\var{{c}/100}+\\var{{t}/100}-\\var{{b}/100}\\\\ &= \\var{({c}+{t}-{b})/100}. \\end{align} \

", "functions": {}, "preamble": {"css": "", "js": ""}}]}], "type": "exam", "resources": [["question-resources/dice.svg", "/srv/numbas/media/question-resources/dice.svg"], ["question-resources/aceOfSpades.svg", "/srv/numbas/media/question-resources/aceOfSpades.svg"], ["question-resources/aceOfSpades_NnxOXmM.svg", "/srv/numbas/media/question-resources/aceOfSpades_NnxOXmM.svg"], ["question-resources/breakfastven2.svg", "/srv/numbas/media/question-resources/breakfastven2.svg"], ["question-resources/breakfastvenplus2.svg", "/srv/numbas/media/question-resources/breakfastvenplus2.svg"]], "feedback": {"feedbackmessages": [], "showanswerstate": true, "showactualmark": true, "intro": "", "allowrevealanswer": true, "showtotalmark": true, "advicethreshold": 0}, "duration": 0, "name": "Maria's copy of Introduction to Probability", "showstudentname": true, "metadata": {"licence": "Creative Commons Attribution 4.0 International", "description": "

Calculations involving elementary probability, and several questions designed to draw out misconceptions to do with probability.

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