Theoretical Probability vs Experimental Probability

Error

There was an error loading the page.

Contributors

Feedback

From users who are members of Transition to university :

Christian Lawson-Perfect	said	Ready to use	7 years, 9 months ago
Elliott Fletcher	said	Needs to be tested	7 years, 9 months ago
Chris Graham	said	Has some problems	7 years, 9 months ago

From users who are not members of Transition to university :

Anna Cartlidge

said

Ready to use

2 years, 11 months ago

History

○

Anna Cartlidge 2 years, 11 months ago

Gave some feedback: Ready to use

○

Christian Lawson-Perfect 7 years, 9 months ago

Saved a checkpoint:

Looks good!

Restore this checkpoint

○

Christian Lawson-Perfect 7 years, 9 months ago

Gave some feedback: Ready to use

○

Elliott Fletcher 7 years, 9 months ago

Published this.

○

Elliott Fletcher 7 years, 9 months ago

Gave some feedback: Needs to be tested

○

Chris Graham 7 years, 9 months ago

Gave some feedback: Has some problems

○

Chris Graham commented 7 years, 9 months ago

I'm not keen on d). It depends on your definition of getting closer - it is staying at the same value whilst the experimental is moving towards it. That's closer in my book. I think I would lose this part altogether, as I'm not sure it adds any value.

In b), I don't think that you need the bullet points under "In this case you have..." if you are later summarising the outcomes in table 2.

You advice in c) ends up being very confusing. The distribution of outcomes in the table are way off the expected, given the probabilities in table 3. I was told that the experimental probability of rolling a 3 is 0.09 and is very close to the theoretical, but it is not, it's way off.

RE your question below, yes, displaying the simplified fraction would be better.

○

Elliott Fletcher commented 7 years, 9 months ago

I think part d) will need checking, i'm not 100% sure that it's correct.

○

Elliott Fletcher 7 years, 9 months ago

Gave some feedback: Needs to be tested

○

Elliott Fletcher commented 7 years, 9 months ago

Thank you for the feedback Christian, i agree that it isn't really important that fractions are simplified, but do you think i should give the answer in the advice as a simplified fraction or do you think it doesn't really matter?

And i agree that it doesn't matter if the dice are thrown one after the other or at the same time so i will reword the statement accordingly.

○

Christian Lawson-Perfect 7 years, 9 months ago

Gave some feedback: Has some problems

○

Christian Lawson-Perfect 7 years, 9 months ago

Saved a checkpoint:

The explanatory text that was in the statement should be at the top of the advice - I've moved it.

The description of the experiment in the prompt for part a should be in the statement - it applies to every part.

I've changed "find the experimental probability of rolling a {number}" to "find the experimental probability of rolling a total of {number}".

I put the total number of trials in the statement - it's rare you'd do an experiment and not know how many times you did it, and adding up 11 numbers is a chore.

Lots of sentences in the advice start "therefore, ..." - try leaving it out, and if the sentence still makes sense leave it that way.

Is it important that the fractions are simplified? Do you need to explain how to do that in this question? I've taken away the "must be reduced" restriction, and I think the explanation of how to reduce a fraction in the advice can go.

"When two unbiased 6-sided dice are rolled together, the total shown on the faces has to be between 2 and 12. There are 36 possible outcomes when adding the faces of two dice together" is incorrect. There are 11 outcomes after adding. There are 36 outcomes when rolling two dice, one after the other. When two indistinguishable dice are thrown at the same time, you could make a case that there are 18 outcomes. This will trip students up, so it's important to be very clear.

Can you convince yourself that it doesn't matter if the dice are thrown one after the other or at the same time, if you're only looking at the total score? Can you rephrase the statement about the number of outcomes so that it's obvious?

I haven't got time to look at the last piece of advice, but it looks like there isn't any for part d. The table for part c should make it clear that this is just one sample, not the results you'd get every time.

Restore this checkpoint

○

Elliott Fletcher 7 years, 9 months ago

Gave some feedback: Needs to be tested

○

Chris Graham commented 7 years, 9 months ago

As discussed, would be better as two questions with separate parts. In part (b) the student will answer the second two gaps based on their interpretation of the coin's bias, and offering feedback after the first bit would then set them down the right track.

The drop down question should work with the opposing terms either way round. This would require a marking matrix (it would need to be its own part, but I've suggested you do that anyway), or this could be replaced with multiple choice statements.

I don't think "Is this coin biased?" is a fair question here: statistically it may or may not be, so as suggested you could replace this with two experiments, with a distractor set of results for small N, and then ask the student which coin is more likely to be biased. This is assessing their understanding that the experimental approaches the theoretical probability.

○

Elliott Fletcher 7 years, 9 months ago

Created this.

Name	Status	Author	Last Modified
Theoretical Probability vs Experimental Probability	Ready to use	Elliott Fletcher	08/05/2022 17:12
Hugh's copy of Theoretical Probability vs Experimental Probability	draft	Hugh O'Donnell	16/01/2018 13:46
Theoretical Probability vs Experimental Probability	draft	Xiaodan Leng	11/07/2019 01:59

There are 3 other versions that do you not have access to.

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.

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

Give any introductory information the student needs.

			Name	Type	Generated Value

die	list	List of 11 items
no_rolls	integer	100
sum	list	List of 100 items
Freq	list	List of 11 items
Freq2	list	List of 11 items
x	list	[ 2 ]
gcd1	number	2
gcd2	number	3
add	integer	2
remainder	integer	8338

Automatically regenerate variables?

Name

Data type

Value

xxxxxxxxxx
 
[2,3,4,5,6,7,8,9,10,11,12]

JME function reference

Description

number the die lands on in part a)

Describe what this variable represents, and list any assumptions made about its value.

Can an exam override the value of this variable?

Generated value: `list`

[ 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 ]

This variable doesn't seem to be used anywhere.

Parts

Part a) Gap-fill
1. Gap Number entry
  Add
Part b) Number entry
Part c) Gap-fill
1. Gap Choose one from a list
  Add

Gap-fill

Ask the student a question, and give any hints about how they should answer this part.

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

Enter your answer as a fraction.

Unnamed gap

Advice

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}$

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 $1$ .

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 $1$ .

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:

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.

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

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.

Give a worked solution to the whole question.

Use this tab to check that this question works as expected.

There was an error which means the tests can't run:

Part	Test	Passed?
Gap-fill
Gap-fill		Hasn't run yet
Number entry
Number entry		Hasn't run yet
Number entry
Number entry		Hasn't run yet
Gap-fill
Gap-fill		Hasn't run yet
Choose one from a list
Choose one from a list		Hasn't run yet

This question is used in the following exams:

Loading...

Error

Theoretical Probability vs Experimental Probability

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

Anna Cartlidge 2 years, 11 months ago

Christian Lawson-Perfect 7 years, 9 months ago

Christian Lawson-Perfect 7 years, 9 months ago

Elliott Fletcher 7 years, 9 months ago

Elliott Fletcher 7 years, 9 months ago

Chris Graham 7 years, 9 months ago

Chris Graham commented 7 years, 9 months ago

Elliott Fletcher commented 7 years, 9 months ago

Elliott Fletcher 7 years, 9 months ago

Elliott Fletcher commented 7 years, 9 months ago

Christian Lawson-Perfect 7 years, 9 months ago

Christian Lawson-Perfect 7 years, 9 months ago

Elliott Fletcher 7 years, 9 months ago

Chris Graham commented 7 years, 9 months ago

Elliott Fletcher 7 years, 9 months ago

Error in variable testing condition

Error

Warning

Generated value: `list`

Parts

Gap-fill

a)

b)

c)

Loading...

Error

Theoretical Probability vs Experimental Probability

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

Anna Cartlidge 2 years, 11 months ago

Christian Lawson-Perfect 7 years, 9 months ago

Christian Lawson-Perfect 7 years, 9 months ago

Elliott Fletcher 7 years, 9 months ago

Elliott Fletcher 7 years, 9 months ago

Chris Graham 7 years, 9 months ago

Chris Graham commented 7 years, 9 months ago

Elliott Fletcher commented 7 years, 9 months ago

Elliott Fletcher 7 years, 9 months ago

Elliott Fletcher commented 7 years, 9 months ago

Christian Lawson-Perfect 7 years, 9 months ago

Christian Lawson-Perfect 7 years, 9 months ago

Elliott Fletcher 7 years, 9 months ago

Chris Graham commented 7 years, 9 months ago

Elliott Fletcher 7 years, 9 months ago

Error in variable testing condition

Error

Warning

Generated value: list

Parts

Gap-fill

a)

b)

c)

Generated value: `list`