Basic Probability Questions

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Elliott Fletcher

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Christian Lawson-Perfect	said	Should not be used	7 years, 8 months ago
Elliott Fletcher	said	Needs to be tested	7 years, 8 months ago
Vicky Hall	said	Has some problems	7 years, 9 months ago

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Christian Lawson-Perfect 7 years, 8 months ago

Saved a checkpoint:

I've split each part into a separate question.

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Christian Lawson-Perfect 7 years, 8 months ago

Gave some feedback: Should not be used

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Elliott Fletcher 7 years, 8 months ago

Published this.

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Elliott Fletcher 7 years, 8 months ago

Gave some feedback: Needs to be tested

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Elliott Fletcher 7 years, 8 months ago

Gave some feedback: Has some problems

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Aiden McCall commented 7 years, 8 months ago

Question:

Needs a comma between outcomes and you. "For equally likely outcomes, you can..."

I would personally change 'happening' to 'occuring'.

I would also change 'So' to 'Therefore'.

Change 'Let's say' to 'If'.

The question with the example does seem quite wordy. I was wondering if you would be able to reduce the words by having a table for the line starting "We roll a...".

I don't think you need a comma after "fair" in "fair 6-sided die". I would then place the comma as follows, "fair 6-sided die, we are equally..."

I also don't think you need a comma after 'Hence'.

For c) I would add another answer with the same numerator but different denominator. This is because you do not have to calculate the probability but just look for the numerator with the same value as the blue balls. I understand that in the cases of simplification this may not be the case, but I think these occur less often then not.

Advice:

a) i) "Now we must think how many outcomes are there that result in an even number being rolled." I would remove are there that so it reads as follows.
"Now we must think how many outcomes result in an even number being rolled.

c) "As the ball is being selected randomly from the bag, there is an equal probability of selecting a ball which is any of the 3 colours."

I am not sure this statement is true or at least reads such that, there is an equal probability of selecting any of the 3 colours when the probabilities for the selection of the three colours are different.

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Elliott Fletcher 7 years, 9 months ago

Gave some feedback: Needs to be tested

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Chris Graham commented 7 years, 9 months ago

As discussed, it would be nice to have an example in the statement.

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Vicky Hall 7 years, 9 months ago

Gave some feedback: Has some problems

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Vicky Hall commented 7 years, 9 months ago

Good intro to basic concepts. Part c) seems a little contrived though - there's no way I would let my daughter pick a pet at random! Also, the outcomes are probably not equally likely because the child probably doesn't like each type of pet equally, and within each group there are likely to be distinctions too. I would think of another example where it makes sense to be randomly assigned something or to randomly pick something. The numbers need to be variables and you should let the student know to simplify their answer.

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Elliott Fletcher 7 years, 9 months ago

Gave some feedback: Needs to be tested

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Elliott Fletcher 7 years, 9 months ago

Created this.

Name	Status	Author	Last Modified
Basic Probability Questions	Should not be used	Elliott Fletcher	20/11/2019 14:33
Choose the probability of getting certain scores on a die	Ready to use	Christian Lawson-Perfect	20/11/2019 14:34
Probability of flipping a coin and getting a tails	Ready to use	Christian Lawson-Perfect	20/11/2019 14:41
Probability of picking a particular colour ball from a bag	Ready to use	Christian Lawson-Perfect	20/11/2019 14:41
Maria's copy of Choose the probability of getting certain scores on a die	draft	Maria Aneiros	27/05/2019 02:29
Maria's copy of Probability of picking a particular colour ball from a bag	draft	Maria Aneiros	27/05/2019 02:32
Choose the probability of getting certain scores on a die	draft	Xiaodan Leng	11/07/2019 01:57
Probability of flipping a coin and getting a tails	draft	Xiaodan Leng	11/07/2019 01:58
Probability of picking a particular colour ball from a bag	draft	Xiaodan Leng	11/07/2019 01:58
Ugur's copy of Probability of picking a particular colour ball from a bag	Ready to use	Ugur Efem	28/04/2022 11:04

There are 24 other versions that do you not have access to.

Question statement

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

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

For example, rolling a fair $6$ -sided die has $6$ possible outcomes which are given in the table.

Outcomes

We roll a

$1$

We roll a

$2$

We roll a

$3$

We roll a

$4$

We roll a

$5$

We roll a

$6$

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

When we roll a fair $6$ -sided die, we are equally likely to roll any of the numbers from $1$ to $6$ on the die as each number only appears once on the die.

Hence using the above formula,

$\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}$

Give any introductory information the student needs.

			Name	Type	Generated Value

red	integer	19
blue	integer	7
green	integer	4
total	integer	30
die1	integer	1
die2	integer	5
die3	integer	6
die4	integer	2
die5	integer	3
die6	integer	4

Automatically regenerate variables?

Name

Data type

Value

xxxxxxxxxx
 
random(15,19)

JME function reference

Description

number of red balls in part c

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

Can an exam override the value of this variable?

Generated value: `integer`

→ Used by:

total

This variable doesn't seem to be used anywhere.

Parts

Part a) Gap-fill
1. Gap Gap 0. Choose one from a list
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2. Gap Gap 1. Choose one from a list
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Part b) Gap-fill
1. Gap Choose one from a list
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Part c) Gap-fill
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Gap-fill

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

Choose the correct probability for each outcome when you roll a fair, $6$ -sided die. Remember to reduce any fractions into their simplest form.

i) $P(\text{rolling an even number})$

Gap 0.

ii) $P(\text{not rolling a $\var{die1}$ or a $\var{die2}$})$

Gap 1.

Advice

a)

For events where each outcome is equally likely to occur, we can calculate the probability using the formula

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

When we roll a fair $6$ -sided die, we are equally likely to roll any of the numbers from $1$ to $6$ on the die as each number only appears once on the die.

i)

To find the probability of rolling an even number, we use the formula above.

The total number of outcomes is $6$ (as we are using a fair $6$ -sided die).

Now we must think how many outcomes result in an even number being rolled.

There are $3$ possible outcomes where we roll an even number on the die:

we roll a $2$
we roll a $4$
we roll a $6$

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

$\begin{align} P(\text{rolling an even number}) &= \displaystyle\frac{\text{number of favourable outcomes}}{\text{total number of outcomes}}\\ &= \displaystyle\frac{3}{6}\\ &= \displaystyle\frac{1}{2}. \end{align}$

ii)

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

The total number of outcomes is still $6$ .

Here, we have $4$ possible outcomes where we don't roll a $\var{die1}$ or a $\var{die2}$ (i.e where we roll any of the other numbers on the die):

we roll a $\var{die3}$
we roll a $\var{die4}$
we roll a $\var{die5}$
we roll a $\var{die6}$

Using the formula,

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

b)

In this question, we are finding the probability of getting tails when an unbiased coin is flipped.

The probability of getting either heads or tails is equally likely, which means that we can use the same formula as we used throughout part a).

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

There is only $1$ outcome where we obtain tails.

Using the formula,

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

c)

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. Hence, the total number of balls in the bag before the chosen ball is removed is

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

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

Therefore, using the same formula as in parts a) and b), we can calculate the probability of the chosen ball being blue as

$\begin{align} P(\text{blue}) &= \displaystyle\frac{\text{number of favourable outcomes}}{\text{total number of outcomes}}\\ &= \displaystyle\frac{\var{blue}}{\var{total}}. \end{align}$

So, $P(\text{blue}) = \displaystyle\simplify{{blue}/{total}}$ .

Give a worked solution to the whole question.

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This question is used in the following exams:

Probability by Elliott Fletcher in Elliott's workspace.
Probability by Elliott Fletcher in Elliott's workspace.
Transition Probability by Chris Templet in Chris's workspace.

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Checkpoint description

Christian Lawson-Perfect 7 years, 8 months ago

Christian Lawson-Perfect 7 years, 8 months ago

Elliott Fletcher 7 years, 8 months ago

Elliott Fletcher 7 years, 8 months ago

Elliott Fletcher 7 years, 8 months ago

Aiden McCall commented 7 years, 8 months ago

Elliott Fletcher 7 years, 9 months ago

Chris Graham commented 7 years, 9 months ago

Vicky Hall 7 years, 9 months ago

Vicky Hall commented 7 years, 9 months ago

Elliott Fletcher 7 years, 9 months ago

Elliott Fletcher 7 years, 9 months ago

Error in variable testing condition

Error

Warning

Generated value: `integer`

Parts

Gap-fill

a)

i)

ii)

b)

c)

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Basic Probability Questions

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

Christian Lawson-Perfect 7 years, 8 months ago

Christian Lawson-Perfect 7 years, 8 months ago

Elliott Fletcher 7 years, 8 months ago

Elliott Fletcher 7 years, 8 months ago

Elliott Fletcher 7 years, 8 months ago

Aiden McCall commented 7 years, 8 months ago

Elliott Fletcher 7 years, 9 months ago

Chris Graham commented 7 years, 9 months ago

Vicky Hall 7 years, 9 months ago

Vicky Hall commented 7 years, 9 months ago

Elliott Fletcher 7 years, 9 months ago

Elliott Fletcher 7 years, 9 months ago

Error in variable testing condition

Error

Warning

Generated value: integer

Parts

Gap-fill

a)

i)

ii)

b)

c)

Generated value: `integer`