Binomial Distribution

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neslihan suzen was given access to the Binomial Distribution 6 years, 10 months ago

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Hemali Patel was given access to the Binomial Distribution 7 years, 6 months ago

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Edward Hall 7 years, 7 months ago

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Edward Hall 7 years, 7 months ago

Gave some feedback: Has some problems

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Tom Bennison was given access to the Binomial Distribution 7 years, 7 months ago

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Edward Hall 7 years, 7 months ago

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Question statement

A bag contains $\var{x}$ white balls and $\var{y}$ red balls.

Give any introductory information the student needs.

			Name	Type	Generated Value

x	integer	59
y	integer	86
p	rational	59/145
wballs	integer	5
rballs	integer	0
prob_zero_white	number	0.0733925952
nballs	integer	5
prob_n_red	number	0.0111537279
prob_white_less_than	number	0.9888462721
total_balls	integer	145
oneminusp	rational	86/145

Automatically regenerate variables?

Name

Data type

Value

A random number between and (inclusive) with step size

Description

Number of white balls

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

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Generated value: `integer`

This variable doesn't seem to be used anywhere.

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1. Step Step 0. Number entry
2. Step Step 1. Number entry
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A ball is chosen at random from the bag.

Advice

Part (a)

There are $\var{x}$ white balls and $\var{y}$ red balls. Thus, there are a total of $\var{x}+\var{y}=\var{total_balls}$ balls.

i) The probability of picking a white ball is $\frac{Number White Balls}{TotalNumberBalls} = \frac{\var{x}}{\var{total_balls}} = \var{p}$.

ii) The probability of picking a red ball is $\frac{Number Red Balls}{TotalNumberBalls} = \frac{\var{y}}{\var{total_balls}} = \var{1-p}$

Part (b)

As $\var{nballs}$ balls are selected with replacement and there are only two colours of ball, we have a binomial distibution. Let $X$ be the number of white balls selected, then $X\sim B(\var{nballs},p)$, where $p=\var{p}$. Similarly, let $Y$ be the number of red balls selected, then $Y\sim B(\var{nballs},q)$, where $q=\var{oneminusp}$.

i) The probability that exactly $\var{rballs}$ of the selected balls are red is thus given by:

${\var{nballs} \choose \var{rballs}} q^\var{rballs}(1-q)^{\var{nballs}-\var{rballs}} = \var{prob_n_red}$.

ii) The probability that none of the selected balls are white is given by:

${\var{nballs} \choose \var{0}} p^\var{0}(1-p)^{\var{nballs}} = (1-p)^{\var{nballs}} = \var{prob_zero_white}$.

iii) For this part we need to find the cumulative binomial probability, the probability that less than $\var{wballs}$ ball are white is given by.

$\sum_{i=0}^{\var{wballs}-1}{\var{nballs} \choose \var{i}} p^\var{i}(1-p)^{\var{nballs}-i} = \var{prob_white_less_than}$.

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Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

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Checkpoint description

neslihan suzen was given access to the Binomial Distribution 6 years, 10 months ago

Hemali Patel was given access to the Binomial Distribution 7 years, 6 months ago

Edward Hall 7 years, 7 months ago

Edward Hall 7 years, 7 months ago

Tom Bennison was given access to the Binomial Distribution 7 years, 7 months ago

Edward Hall 7 years, 7 months ago

Error in variable testing condition

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Warning

Generated value: `integer`

Parts

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Binomial Distribution

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

neslihan suzen was given access to the Binomial Distribution 6 years, 10 months ago

Hemali Patel was given access to the Binomial Distribution 7 years, 6 months ago

Edward Hall 7 years, 7 months ago

Edward Hall 7 years, 7 months ago

Tom Bennison was given access to the Binomial Distribution 7 years, 7 months ago

Edward Hall 7 years, 7 months ago

Error in variable testing condition

Error

Warning

Generated value: integer

Parts

Information only

Generated value: `integer`