Probability of not choosing any from a subset

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Newcastle University Mathematics and Statistics 9 years, 2 months ago

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Question statement

A line-up of $\var{men}$ men is conducted in order that a witness can identify $\var{suspects}$ suspects.

Suppose that all $\var{suspects}$ suspects are in the line-up.

Also suppose that the witness does not recognise any of the suspects but simply chooses $\var{suspects}$ men at random.

Give any introductory information the student needs.

			Name	Type	Generated Value

guilty	integer	3
is	string	are
ans3	number	0.176
suspects	integer	4
noguilty	number	10
q5	integer	1
q4	integer	10
q6	integer	1
is2	string	is
test	number	0.1762237762
ns	integer	9
men	integer	13
singpl	string	man
p6	integer	1
p4	integer	6
p5	integer	1
t4	integer	0
nguilty	number	126
t2	integer	1
t3	integer	0
t1	integer	0
p	string	men
overallnumber	number	715

Automatically regenerate variables?

Name

Data type

Value

xxxxxxxxxx
 
if(suspects>2,suspects-random(1,2),suspects-1)

JME function reference

Description

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`

← Depends on:

suspects

→ Used by:

is
is2
noguilty
p
singpl

Parts

Gap-fill
1. Gap Mathematical expression
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Gap-fill

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

What is the probability that none of the suspects are chosen?

Probability = Unnamed gap?

Input your answer as a fraction and not as a decimal.

Advice

We can work out the probability in two ways:

Method 1.

There are $\var{men-suspects}$ of the $\var{men}$ who are not suspects.

The probability of picking the first who is not a suspect is therefore:

\[\simplify[]{{men-suspects}/{men}}\]

The second choice of a non-suspect will be from $\var{men-suspects-1}$ non-suspects in $\var{men-1}$ with probability:

\[\simplify[]{{men-suspects-1}/{men-1}}\]

Hence the probability of choosing two non-suspects will be .

\[\simplify[]{{men -suspects} / {men}}\times \simplify[]{{men -suspects-1} / {men-1}}\]

Continuing in this way we see that the probability of choosing $\var{suspects}$ non-suspects is:

\[\simplify[zeroFactor,unitFactor,zeroTerm]{{t1} * ({men -suspects} / {men}) * ({men -suspects -1} / {men -1}) * ({men -suspects -2} / {men -2}) + {t2} * ({men -suspects} / {men}) * ({men -suspects -1} / {men -1}) * ({men -suspects -2} / {men -2}) * ({ns -3} / {men -3}) + {t3} * ({men -suspects} / {men}) * ({men -suspects -1} / {men -1}) * ({men -suspects -2} / {men -2}) * ({ns -3} / {men -3}) * ({ns -4} / {men -4}) + {t4} * ({men -suspects} / {men}) * ({men -suspects -1} / {men -1}) * ({men -suspects -2} / {men -2}) * ({ns -3} / {men -3}) * ({ns -4} / {men -4}) * ({ns -5} / {men -5})}=\simplify[std]{{nguilty}/{overallnumber}}\]

on reducing the fraction to its lowest form.

Method 2.

There are $\var{men-suspects}$ of the $\var{men}$ who are not suspects.

Hence there are \[{\var{men-suspects}\choose \var{suspects}}=\var{comb(men-suspects,suspects)}\] ways of choosing $\var{suspects}$ non-suspects.

In total there are \[{\var{men}\choose \var{suspects}}=\var{comb(men,suspects)}\] ways of choosing $\var{suspects}$ from all present.

Hence the probability is \[\frac{\var{comb(men-suspects,suspects)}}{\var{comb(men,suspects)}}= \simplify[std]{{nguilty}/{overallnumber}} \]

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Probability of not choosing any from a subset

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Bill Foster 6 years, 2 months ago

Newcastle University Mathematics and Statistics 9 years, 2 months ago

Error in variable testing condition

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Generated value: `integer`

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Probability of not choosing any from a subset

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

Bill Foster 6 years, 2 months ago

Newcastle University Mathematics and Statistics 9 years, 2 months ago

Error in variable testing condition

Error

Warning

Generated value: integer

Parts

Gap-fill

Generated value: `integer`