Statistics and Probability, NAPX-I4-NC20

Question

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Worked Solution

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Variant 0

DifficultyLevel

627

Question

A group of 85 people were asked if they have their ears pierced.

This table shows the results.


Pierced Not Pierced Total
Men 10 25 35
Women 35 15 50
Total 45 40 85


A man was selected at random.

What is the probability that he does not have his ears pierced?

Worked Solution

PP(man has ears pierced)
= number of men not piercedtotal men\dfrac{\text{number of men not pierced}}{\text{total men}}
= 2535\dfrac{25}{35}
= 0.71

Question Type

Multiple Choice (One Answer)

Variables

Variable nameVariable value
question
A group of 85 people were asked if they have their ears pierced. This table shows the results.
>>| | Pierced| Not Pierced | Total| |:-:|:-:|:-:|:-:| | Men| 10| 25|35| | Women | 35| 15|50| |Total|45|40|85|

A man was selected at random. What is the probability that he does not have his ears pierced?
workedSolution
sm_nogap $P$(man has ears pierced)
>>| | | | ------------- | ---------- | | | \= $\dfrac{\text{number of men not pierced}}{\text{total men}}$ | | | \= $\dfrac{25}{35}$ | | | \= {{{correctAnswer}}} |
correctAnswer
0.71

Answers

Is Correct?Answer
x

0.29

x

0.41

x

0.47

✓

0.71

Variant 1

DifficultyLevel

628

Question

A group of 65 kindergarten students were asked if they watched the cartoon Bluey.

This table shows the results.


Watched Bluey Did Not Watch Bluey Total
Boys 21 11 32
Girls 27 6 33
Total 48 17 65


One of the girls was selected at random.

What is the probability that she watches Bluey?

Worked Solution

PP(watches Bluey)
= number of girls that watch Blueytotal girls\dfrac{\text{number of girls that watch Bluey}}{\text{total girls}}
= 2733\dfrac{27}{33}
= 0.82

Question Type

Multiple Choice (One Answer)

Variables

Variable nameVariable value
question
A group of 65 kindergarten students were asked if they watched the cartoon Bluey. This table shows the results.
>>| | Watched Bluey| Did Not Watch Bluey| Total| |:-:|:-:|:-:|:-:| | Boys| 21| 11|32| | Girls| 27| 6|33| |Total|48|17|65|

One of the girls was selected at random. What is the probability that she watches Bluey?
workedSolution
sm_nogap $P$(watches Bluey)
>>| | | | ------------- | ---------- | | | \= $\dfrac{\text{number of girls that watch Bluey}}{\text{total girls}}$ | | | \= $\dfrac{27}{33}$ | | | \= {{{correctAnswer}}} |
correctAnswer
0.82

Answers

Is Correct?Answer
x

0.42

x

0.56

x

0.74

✓

0.82

Variant 2

DifficultyLevel

629

Question

A group of 55 outback students were asked if they owned a horse.

This table shows the results.


Own a horse Do not own a horse Total
Boys 21 5 26
Girls 23 6 29
Total 44 11 55


One of the boys was selected at random.

What is the probability that he does not own a horse?

Worked Solution

PP(does not own a horse)
= number of boys with no horsetotal boys\dfrac{\text{number of boys with no horse}}{\text{total boys}}
= 526\dfrac{5}{26}
= 0.19

Question Type

Multiple Choice (One Answer)

Variables

Variable nameVariable value
question
A group of 55 outback students were asked if they owned a horse. This table shows the results.
>>| | Own a horse| Do not own a horse| Total| |:-:|:-:|:-:|:-:| | Boys| 21| 5|26| | Girls| 23| 6|29| |Total|44|11|55|

One of the boys was selected at random. What is the probability that he does not own a horse?
workedSolution
sm_nogap $P$(does not own a horse)
>>| | | | ------------- | ---------- | | | \= $\dfrac{\text{number of boys with no horse}}{\text{total boys}}$ | | | \= $\dfrac{5}{26}$ | | | \= {{{correctAnswer}}} |
correctAnswer
0.19

Answers

Is Correct?Answer
x

0.11

✓

0.19

x

0.45

x

0.50

Variant 3

DifficultyLevel

634

Question

A group of 48 mainland students were surveyed if they have visited Tasmania or not.

This table shows the results.


Visited Tasmania Not Visited Tasmania Total
Boys 12 6 18
Girls 18 12 30
Total 30 18 48


One of the girls was selected at random.

What is the probability that she has not visited Tasmania?

Worked Solution

PP(not travelled to Tasmania)
= number of girls not visitedtotal girls\dfrac{\text{number of girls not visited}}{\text{total girls}}
= 1230\dfrac{12}{30}
= 0.4

Question Type

Multiple Choice (One Answer)

Variables

Variable nameVariable value
question
A group of 48 mainland students were surveyed if they have visited Tasmania or not. This table shows the results.
>>| | Visited Tasmania| Not Visited Tasmania| Total| |:-:|:-:|:-:|:-:| | Boys| 12| 6|18| | Girls| 18| 12|30| |Total|30|18|48|

One of the girls was selected at random. What is the probability that she has not visited Tasmania?
workedSolution
sm_nogap $P$(not travelled to Tasmania)
>>| | | | ------------- | ---------- | | | \= $\dfrac{\text{number of girls not visited}}{\text{total girls}}$ | | | \= $\dfrac{12}{30}$ | | | \= {{{correctAnswer}}} |
correctAnswer
0.4

Answers

Is Correct?Answer
x

0.25

✓

0.4

x

0.6

x

0.67

Variant 4

DifficultyLevel

637

Question

A researcher collected 60 live insect specimens for a study. The insects were either cicadas or wasps.

The researcher counted how many females and males of each type, and recorded the results in the table below.


Cicada Wasp Total
Male 17 19 36
Female 14 10 24
Total 31 29 60


One of the wasps was selected at random.

What is the probability that it was a female?

Worked Solution

PP(wasp is a female)
= number of female waspstotal females\dfrac{\text{number of female wasps}}{\text{total females}}
= 1029\dfrac{10}{29}
= 0.34

Question Type

Multiple Choice (One Answer)

Variables

Variable nameVariable value
question
A researcher collected 60 live insect specimens for a study. The insects were either cicadas or wasps. The researcher counted how many females and males of each type, and recorded the results in the table below.
>>| | Cicada| Wasp| Total| |:-:|:-:|:-:|:-:| | Male| 17| 19|36| | Female| 14| 10|24| |Total|31|29|60|

One of the wasps was selected at random. What is the probability that it was a female?
workedSolution
sm_nogap $P$(wasp is a female)
>>| | | | ------------- | ---------- | | | \= $\dfrac{\text{number of female wasps}}{\text{total females}}$ | | | \= $\dfrac{10}{29}$ | | | \= {{{correctAnswer}}} |
correctAnswer
0.34

Answers

Is Correct?Answer
x

0.17

✓

0.34

x

0.42

x

0.48

Variant 5

DifficultyLevel

640

Question

Joe owns a private zoo that has 52 big cats made up of tigers and lions.

Joe counted how many females and males of each type, and recorded the results in the table below.


Lions Tigers Total
Male 8 5 13
Female 25 14 39
Total 33 19 52


One of the female big cats was selected at random.

What is the probability that it was a tiger?

Worked Solution

PP(female is a tiger)
= number of female tigerstotal females\dfrac{\text{number of female tigers}}{\text{total females}}
= 1439\dfrac{14}{39}
= 0.36

Question Type

Multiple Choice (One Answer)

Variables

Variable nameVariable value
question
Joe owns a private zoo that has 52 big cats made up of tigers and lions. Joe counted how many females and males of each type, and recorded the results in the table below.
>>| | Lions| Tigers| Total| |:-:|:-:|:-:|:-:| | Male| 8| 5|13| | Female| 25| 14|39| |Total|33|19|52|

One of the female big cats was selected at random. What is the probability that it was a tiger?
workedSolution
sm_nogap $P$(female is a tiger)
>>| | | | ------------- | ---------- | | | \= $\dfrac{\text{number of female tigers}}{\text{total females}}$ | | | \= $\dfrac{14}{39}$ | | | \= {{{correctAnswer}}} |
correctAnswer
0.36

Answers

Is Correct?Answer
x

0.27

✓

0.36

x

0.74

x

0.76

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