// Numbas version: finer_feedback_settings {"name": "Data collection", "duration": 0, "feedback": {"feedbackmessages": [], "allowrevealanswer": true, "showactualmark": true, "showanswerstate": true, "intro": "", "showtotalmark": true, "advicethreshold": 0, "enterreviewmodeimmediately": true, "showexpectedanswerswhen": "inreview", "showpartfeedbackmessageswhen": "always", "showactualmarkwhen": "always", "showtotalmarkwhen": "always", "showanswerstatewhen": "always", "showadvicewhen": "never"}, "percentPass": 0, "question_groups": [{"name": "Group", "pickingStrategy": "all-ordered", "pickQuestions": 1, "questions": [{"name": "Theoretical Probability vs Experimental Probability ", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": 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/"}], "type": "question", "statement": "

Two unbiased, 6-sided dice were rolled together and the total of the numbers shown on the faces was recorded.

The experiment was repeated $\\var{no_rolls}$ times.

This table gives the frequency of each outcome.

\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

Total	2	3	4	5	6	7	8	9	10	11	12
Frequency	$\\var{Freq[0]}$	$\\var{Freq[1]}$	$\\var{Freq[2]}$	$\\var{Freq[3]}$	$\\var{Freq[4]}$	$\\var{Freq[5]}$	$\\var{Freq[6]}$	$\\var{Freq[7]}$	$\\var{Freq[8]}$	$\\var{Freq[9]}$	$\\var{Freq[10]}$

", "variablesTest": {"condition": "", "maxRuns": 100}, "variables": {"Freq2": {"group": "Ungrouped variables", "name": "Freq2", "description": "

List of Frequencies for theoretical probability.

", "templateType": "anything", "definition": "[1,2,3,4,5,6,5,4,3,2,1]"}, "sum": {"group": "Ungrouped variables", "name": "sum", "description": "

Sums obtained from no_rolls of two dice part a)

", "templateType": "anything", "definition": "repeat(random(2..12), no_rolls)"}, "gcd2": {"group": "Ungrouped variables", "name": "gcd2", "description": "

Gcd for answer in part a) ii)

", "templateType": "anything", "definition": "gcd(Freq2[x[0]],36)"}, "gcd1": {"group": "Ungrouped variables", "name": "gcd1", "description": "

Gcd of numerator and denominator for advice for part a) i).

", "templateType": "anything", "definition": "gcd(Freq[sum[0]-2], no_rolls)"}, "die": {"group": "Ungrouped variables", "name": "die", "description": "

number the die lands on in part a)

", "templateType": "anything", "definition": "[2,3,4,5,6,7,8,9,10,11,12]"}, "Freq": {"group": "Ungrouped variables", "name": "Freq", "description": "

Frequencies of each possible sum of numbers from rolling 2 die. part a

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Number of rolls of the die in part a.

", "templateType": "anything", "definition": "random(50..100 #10)"}, "remainder": {"group": "Ungrouped variables", "name": "remainder", "description": "", "templateType": "anything", "definition": "(ceil(10000/36)+add)+(ceil(10000/18)+add)+(ceil(2500/3)+add)+(ceil(10000/9)+add)+(ceil(12500/9)+add)+(ceil(12500/9)-add)+(ceil(10000/9)-add)+(ceil(2500/3)-add)+(ceil(5000/9)-add)+(ceil(2500/9)-add)"}, "x": {"group": "Ungrouped variables", "name": "x", "description": "

Index for part a.

", "templateType": "anything", "definition": "indices(die, sum[0])"}}, "functions": {}, "tags": ["dice", "Dice", "Experimental probability", "Experimental Probability", "experimental probability", "Probability", "probability", "sum of two dice", "taxonomy", "theoretical probability", "Theoretical Probability"], "variable_groups": [], "parts": [{"scripts": {}, "variableReplacements": [], "type": "gapfill", "variableReplacementStrategy": "originalfirst", "showFeedbackIcon": true, "gaps": [{"correctAnswerFraction": true, "scripts": {}, "type": "numberentry", "variableReplacementStrategy": "originalfirst", "allowFractions": true, "maxValue": "{Freq[{sum[0]}-2]}/{no_rolls}", "showFeedbackIcon": true, "minValue": "{Freq[{sum[0]}-2]}/{no_rolls}", "correctAnswerStyle": "plain", "mustBeReducedPC": 0, "mustBeReduced": false, "notationStyles": ["plain", "en", "si-en"], "variableReplacements": [], "marks": 1, "showCorrectAnswer": true}], "showCorrectAnswer": true, "prompt": "

Find the experimental probability of rolling a total of $\\var{sum[0]}$.

Enter your answer as a fraction.

[[0]]

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Now calculate the theoretical probability that the sum of the scores of the two dice is $\\var{sum[0]}$.

Enter your answer as a fraction.

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gets closer to the theoretical probability

", "

gets further away from the theoretical probability

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As the number of rolls of the two dice increases, the experimental probability [[0]]

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Compute the experimental probability of a particular score on a die given a sample of throws, and compare it with the theoretical probability.

The last part asks what you expect to happen to the experimental probability as the sample size increases.

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There are two ways of assigning probability:

Experimental Probability is where you do an experiment, observe the outcome and record the relative frequency results.
Theoretical Probability is where the probability is calculated based on fairness and symmetrical properties of the results.

a)

To calculate the experimental probability (relative frequency) of an outcome we divide the frequency of the outcome in the experiment by the number of trials.

We are given that the experiment was repeated $\\var{no_rolls}$ times.

We then need the number of times that the sum of the faces of the dice was equal to $\\var{sum[0]}$. From the frequency table, we can see that the frequency of rolling a $\\var{sum[0]}$ in the experiment was $\\var{Freq[sum[0]-2]}$.

Therefore, the experimental probability of rolling a $\\var{sum[0]}$ is

\\[ \\begin{align} P(\\text{total}=\\var{sum[0]}) &= \\displaystyle\\frac{\\text{number of times a total of $\\var{sum[0]}$ was rolled}}{\\text{total number of rolls}}\\\\&= \\displaystyle\\frac{\\var{Freq[sum[0]-2]}}{\\var{no_rolls}}.\\end{align}\\]

\\[\\begin{align} P(\\text{total}=\\var{sum[0]}) &= \\displaystyle\\frac{\\text{number of times a total of $\\var{sum[0]}$ was rolled}}{\\text{total number of rolls}}\\\\&= \\displaystyle\\frac{\\var{Freq[sum[0]-2]}}{\\var{no_rolls}}\\\\&= \\displaystyle\\var[fractionNumbers, simplifyFractions]{{Freq[sum[0]-2]/no_rolls}}.\\end{align}\\]

b)

When two unbiased 6-sided dice are rolled and their scores are added, there are $11$ possible outcomes: the total must be between $2$ and $12$ inclusive.

To work out the probabilities of each outcome occurring, we must be very careful about how the experiment is performed.

There are a total of $36$ different outcomes when rolling two dice one after the other; these are shown in Table $1$.

\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

Table 1
	1	2	3	4	5	6
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)

Note that these outcomes are all different. For example, the outcomes (2,1) and (1,2) are not the same because we know the order in which the dice were thrown so we can distinguish between these two outcomes; when the first die lands on $2$ and the second die lands on $1$ the outcome is (2,1), however when the first die lands on $1$ and the second die lands on $2$ the outcome is (1,2).

The sum of the numbers in these outcomes is the same, we just count them as different outcomes.

Any one of these $36$ outcomes is equally likely to occur, so the probability of each of these outcomes is $\\displaystyle\\frac{1}{36}$.

However, if you roll two indistinguishable dice at the same time, you can't differentiate (a 1 and a 2) from (a 2 and a 1). From your point of view, they are the same outcome. The probabilities of the underlying events (each die's score) haven't changed, but the outcomes you observe have.

In this case you have a total of $21$ outcomes; these outcomes are not all equally likely. 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 Table $1$. Hence the probability of obtaining two 1s would be $\\displaystyle\\frac{1}{36}$, whereas the probability of obtaining a 1 and a 2 would be

\\[\\displaystyle\\frac{2}{36}=\\displaystyle\\frac{1}{18}.\\]

You can do the experiment this way, but it's easier when you can tell the dice apart.

Table $2$ shows the corresponding total of the faces of the dice for each of the outcomes in Table $1$.

\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

Table 2
+	1	2	3	4	5	6
1	2	3	4	5	6	7
2	3	4	5	6	7	8
3	4	5	6	7	8	9
4	5	6	7	8	9	10
5	6	7	8	9	10	11
6	7	8	9	10	11	12

For equally likely outcomes you can calculate the probability using the formula

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

Table $3$ gives the theoretical probabilities of each of the possible totals of the two dice occurring:

\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

Table 3
Total	2	3	4	5	6	7	8	9	10	11	12
Probability	$\\displaystyle\\frac{1}{36}$	$\\displaystyle\\frac{2}{36}$	$\\displaystyle\\frac{3}{36}$	$\\displaystyle\\frac{4}{36}$	$\\displaystyle\\frac{5}{36}$	$\\displaystyle\\frac{6}{36}$	$\\displaystyle\\frac{5}{36}$	$\\displaystyle\\frac{4}{36}$	$\\displaystyle\\frac{3}{36}$	$\\displaystyle\\frac{2}{36}$	$\\displaystyle\\frac{1}{36}$

\\[P(\\text{total}=\\var{sum[0]}) = \\displaystyle\\frac{\\var{Freq2[x[0]]}}{36}.\\]

\\[\\begin{align} P(\\text{total}=\\var{sum[0]}) &= \\displaystyle\\frac{\\var{Freq2[x[0]]}}{36}\\\\ &= \\displaystyle\\var[fractionNumbers,simplifyFractions]{Freq2[x[0]]/36}. \\end{align}\\]

c)

For any experiment, as we make the number of trials very large the experimental probability tends towards the theoretical probability.

For this experiment, for example, the theoretical probability of the sum of the faces of the two dice being $3$ is

\\[\\displaystyle\\frac{2}{36} = \\var{sigformat(1/18,3)} \\text{ (rounded to $3$ significant figures)}.\\]

If we were to roll the two dice a very large number of times, for example $10000$ times, the frequencies of each outcome would be different.

The table below was produced by the recording the frequency of each outcome after rolling the two dice 10,000 times.

\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

Total	2	3	4	5	6	7	8	9	10	11	12
\n Frequency \n	$\\var{ceil(10000/36)+{add}}$	$\\var{ceil(10000/18)+{add}}$	$\\var{ceil(2500/3)+{add}}$	$\\var{ceil(10000/9)+{add}}$	$\\var{ceil(12500/9)+{add}}$	$\\var{10000-{remainder}}$	$\\var{ceil(12500/9)-{add}}$	$\\var{ceil(10000/9)-{add}}$	$\\var{ceil(2500/3)-{add}}$	$\\var{ceil(5000/9)-{add}}$	$\\var{ceil(2500/9)-{add}}$

This means that the new experimental probability of the sum of the faces of the two dice being $3$ is

\\[\\displaystyle\\frac{\\var{ceil(10000/18) + {add}}}{10000} = \\var{sigformat((ceil(10000/18)+{add})/10000,3)} \\text{ (rounded to $3$ significant figures)}.\\]

This is very close to the theoretical probability, so we can see how the experimental probability changes as we increases the number of trials.

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Returns list of values of units of numbers in list for the unit part of the stem and leaf diagram.

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Construct a stem and leaf plot for these data. The \"stem\" is given; you should enter the \"leaves\" in the boxes to the right, separating neighbouring values with a comma.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

1	[[0]]
2	[[1]]
3	[[2]]
4	[[3]]
5	\n [[4]] \n

What are the stem and leaf units?

Stem unit = [[5]]

Leaf unit = [[6]]

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The following data are points scored by rugby team Newcastle Sparrows in their last $10$ matches.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

$\\var{d[0]}$

$\\var{d[1]}$

$\\var{d[2]}$

$\\var{d[3]}$

$\\var{d[4]}$

$\\var{d[5]}$

$\\var{d[6]}$

$\\var{d[7]}$

$\\var{d[8]}$

$\\var{d[9]}$

", "tags": [], "advice": "\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

Stem	Leaves
1	{d_unit_value[0]},{d_unit_value[1]},{d_unit_value[2]},{d_unit_value[3]},{d_unit_value[4]}	Corresponds to {dsort[0]},{dsort[1]},{dsort[2]},{dsort[3]} and {dsort[4]}.
2	{d_unit_value[5]},{d_unit_value[6]}	Corresponds to {dsort[5]} and {dsort[6]}
3	{d_unit_value[7]}	...and so on.
4	{d_unit_value[8]}
5	{d_unit_value[9]}

The stem unit is $10$. The leaf unit is $1$.

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Questions on the subtleties and pitfalls of methods of data collection.

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