Probability -- choosing digits with replacement - highest digit

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Lois Rollings

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Lois Rollings 8 years, 6 months ago

Published this.

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Lois Rollings 9 years, 6 months ago

Created this as a copy of IP1.4.

Name	Status	Author	Last Modified
Probability -- choosing digits with replacement - highest digit	draft	Lois Rollings	02/09/2016 17:22
Michael's copy of Probability -- choosing digits with replacement - highest digit	draft	Michael Hoffmann	23/07/2018 09:46

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Question statement

Suppose a sequence of $\var{r}$ digits is picked at random (with replacement) from the set

$\{0,\;1,\;2,\ldots,\;\var{d}\}$

What is the probability that:

Give any introductory information the student needs.

			Name	Type	Generated Value

d	integer	9
ans1	decimal	dec("0.020")
k	integer	6
r	integer	11
t	integer	63
tol	number	0.001
tans1	rational	1977326743/100000000000
ans2	decimal	dec("0.016")
tans2	rational	1614529687/100000000000

Automatically regenerate variables?

Name

Data type

Value

xxxxxxxxxx
 
random(7,8,9)

JME function reference

Description

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

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Generated value: `integer`

→ Used by:

k
tans1
tans2

Statement
Advice

Parts

Part a) Gap-fill
1. Gap Number entry
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Part b) Gap-fill
1. Gap Number entry
  Add

Gap-fill

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No digit in the sequence exceeds $\var{k}$ ?

Probability =Unnamed gap? (to $3$ decimal places)

Advice

a) There are $\var{d+1}$ different digits and so there are $\var{d+1}^{\var{r}}=\var{(d+1)^r}$ different seqences of length $\var{r}$ .

However, if we consider sequences in which the digits do not exceed $\var{k}$ ,

then there are $\var{k+1}$ different possible digits in the sequence, and so there are $\var{k+1}^{\var{r}}=\var{(k+1)^r}$ possible sequences.

$\begin{eqnarray*} \mbox{Probability} &=& \frac{\mbox{number of successes}}{\mbox{number of trials}}\\ &=& \frac{\var{(k+1)^r}}{\var{(d+1)^r}}\\ &=& \var{ans1} \end{eqnarray*}$ to

$3$ decimal places.

b) Number $N$ of sequences with largest digit $\var{k}$ is given by:

$\begin{eqnarray*} N &=&\mbox{number with largest digit}\;\le \var{k} - \mbox{number with largest digit}\;\le\var{k-1} \\ &=& \var{k+1}^{\var{r}}-\var{k}^{\var{r}}\\ &=& \var{(k+1)^r}-\var{k^r}\\ &=& \var{(k+1)^r-k^r}\\ \mbox{So probability} &=& \frac{\mbox{number of successes}}{\mbox{number of trials}} =\frac{N}{\mbox{number of trials}}\\ &=& \frac{\var{(k+1)^r-k^r}}{\var{(d+1)^r}}\\ &=& \var{ans2} \end{eqnarray*}$ to

$3$ decimal places.

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Probability -- choosing digits with replacement - highest digit

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Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

History

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Checkpoint description

Lois Rollings 8 years, 6 months ago

Lois Rollings 9 years, 6 months ago

Error in variable testing condition

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Generated value: `integer`

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Gap-fill

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Error

Probability -- choosing digits with replacement - highest digit

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

History

Comment

Checkpoint description

Lois Rollings 8 years, 6 months ago

Lois Rollings 9 years, 6 months ago

Error in variable testing condition

Error

Warning

Generated value: integer

Parts

Gap-fill

Generated value: `integer`