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Probability, STD2 S2 2025 HSC 8 MC

A spinner made up of 4 colours is spun 100 times. The frequency histogram shows the results.
 

Which of these spinners is most likely to give the results shown?
 

Show Answers Only

\(A\)

Show Worked Solution
\(P(\text{White})\) \(=\dfrac{50}{100}=\dfrac{1}{2}\)
\(P(\text{Red})\) \(=\dfrac{25}{100}=\dfrac{1}{4}\)  
\(P(\text{Yellow})\) \(=\dfrac{15}{100}=\dfrac{3}{20}\)
\(P(\text{Green})\) \(=\dfrac{10}{100}=\dfrac{2}{20}=\dfrac{1}{10}\)

 
\(\text{Eliminate Options B and D as white}\ \neq \dfrac{1}{2}\ \text{of spinner.}\)

\(\text{Eliminate Option C as red}\ \neq \dfrac{1}{4}\ \text{of spinner.}\)

\(\Rightarrow A\)

Filed Under: Combinations and Single Stage Events (Std 2), Relative Frequency (Std 2) Tagged With: Band 4, smc-827-20-Games of Chance, smc-828-10-Simple Probability

Probability, STD2 S2 2025 HSC 13 MC

A ten-sided die has faces numbered 1 to 10 .

The die is constructed so that the probability of obtaining the number 1 is greater than the probability of obtaining any of the other numbers. The numbers 2 to 10 are equally likely to occur.

When the die is rolled 153 times, a 1 is obtained 72 times.

By using the relative frequency of rolling a 1, which of the following is the best estimate for the probability of rolling a 10 ?

  1. \(\dfrac{1}{17}\)
  2. \(\dfrac{1}{11}\)
  3. \(\dfrac{1}{10}\)
  4. \(\dfrac{1}{9}\)
Show Answers Only

\(A\)

Show Worked Solution

\(P(1) = \dfrac{72}{153}=\dfrac{8}{17} \)

\(\text{Let}\ \ p=P(2)=P(3) = … =P(10) \)

\(\dfrac{8}{17}+9p\) \(=1\)  
\(9p\) \(=1-\dfrac{8}{17}\)  
\(p\) \(=\dfrac{1}{17}\)  

 
\(\Rightarrow A\)

Filed Under: Relative Frequency (Std 2) Tagged With: 2adv-std2-common, Band 5, smc-827-20-Games of Chance, smc-827-40-Expected Frequency (np)

Probability, STD2 S2 2019 HSC 20

A roulette wheel has the numbers 0, 1, 2, …, 36 where each of the 37 numbers is equally likely to be spun.
 

 
If the wheel is spun 18 500 times, calculate the expected frequency of spinning the number 8.  (2 marks)

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Show Answers Only

`500`

Show Worked Solution

`P(8) = 1/37`

`:.\ text(Expected Frequency (8))`

`= 1/37 xx 18\ 500`

`= 500`

Filed Under: Probability, Relative Frequency (Std 1), Relative Frequency (Std 2), Relative Frequency (Y11) Tagged With: Band 4, common-content, num-title-ct-core, num-title-qs-hsc, smc-1133-20-Games of Chance, smc-4225-35-Relative frequency, smc-827-20-Games of Chance, smc-827-40-Expected Frequency (np), smc-990-20-Games of Chance, smc-990-40-Expected Frequency (np)

Probability, STD2 S2 2018 HSC 26a

Jeremy rolled a biased 6-sided die a number of times. He recorded the results in a table.
  

\begin{array} {|l|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} \text{Number} \rule[-1ex]{0pt}{0pt} & \ \ 1 \ \ & \ \ 2 \ \  & \ \ 3 \ \  & \ \ 4 \ \  & \ \ 5 \ \  & \ \ 6 \ \ \\
\hline
\rule{0pt}{2.5ex} \text{Frequency} \rule[-1ex]{0pt}{0pt} & \ \ 23 \ \ & \ \ 19 \ \  & \ \ 48 \ \  & \ \ 20 \ \  & \ \ 21 \ \  & \ \ 19 \ \ \\
\hline
\end{array} 

What is the relative frequency of rolling a 3?  (1 mark)

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Show Answers Only

\(\dfrac{8}{25}\)

Show Worked Solution
♦ Mean mark 40%.

\(\text{Rel Freq}\) \(=\dfrac{\text{number of 3’s rolled}}{\text{total rolls}}\)
  \(=\dfrac{48}{150}\)
  \(=\dfrac{8}{25}\)

Filed Under: Probability, Relative Frequency (Std 1), Relative Frequency (Std 2), Relative Frequency (Y11) Tagged With: Band 5, common-content, num-title-ct-core, num-title-qs-hsc, smc-1133-20-Games of Chance, smc-4225-35-Relative frequency, smc-827-20-Games of Chance, smc-990-20-Games of Chance

Probability, STD2 S2 2015 HSC 26e

The table shows the relative frequency of selecting each of the different coloured jelly beans from packets containing green, yellow, black, red and white jelly beans.

\begin{array} {|c|c|}
\hline
\rule{0pt}{2.5ex} \textit{Colour} \rule[-1ex]{0pt}{0pt} & \textit{Relative frequency} \\
\hline
\rule{0pt}{2.5ex} \text{Green} \rule[-1ex]{0pt}{0pt} & 0.32 \\
\hline
\rule{0pt}{2.5ex} \text{Yellow} \rule[-1ex]{0pt}{0pt} & 0.13 \\
\hline
\rule{0pt}{2.5ex} \text{Black} \rule[-1ex]{0pt}{0pt} & 0.14 \\
\hline
\rule{0pt}{2.5ex} \text{Red} \rule[-1ex]{0pt}{0pt} &  \\
\hline
\rule{0pt}{2.5ex} \text{White} \rule[-1ex]{0pt}{0pt} & 0.24 \\
\hline
\end{array}

  1. What is the relative frequency of selecting a red jelly bean?  (1 mark)

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  2. Based on this table of relative frequencies, what is the probability of NOT selecting a black jelly bean?  (1 mark)

    --- 2 WORK AREA LINES (style=lined) ---

Show Answers Only
  1. \(0.17\)
  2. \(0.86\)
Show Worked Solution

i.  \(\text{Relative frequency of red}\)

\(= 1-(0.32 + 0.13 + 0.14 + 0.24)\)

\(= 1-0.83\)

\(= 0.17\)

 

ii.  \(P\text{(not selecting black)}\)

\(= 1-P\text{(selecting black)}\)

\(= 1-0.14\)

\(= 0.86\)

Filed Under: Combinations and Single Stage Events (Std 2), Probability, Relative Frequency, Relative Frequency (Std 1), Relative Frequency (Std 2), Relative Frequency (Y11), Single and Multi-Stage Events (Std 1), Single stage events Tagged With: Band 3, Band 4, common-content, num-title-ct-core, num-title-qs-hsc, smc-1133-20-Games of Chance, smc-1135-05-Simple Probability, smc-4225-20-Complementary events, smc-4225-35-Relative frequency, smc-827-20-Games of Chance, smc-828-10-Simple Probability, smc-990-20-Games of Chance

Probability, STD2 S2 2006 HSC 6 MC

Marcella is planning to roll a standard six-sided die 60 times.

How many times would she expect to roll the number 4?

  1.   6
  2.   10
  3.   15
  4.   20
Show Answers Only

`B`

Show Worked Solution

`P(4) = 1/6`

`:.\ text(Expected times to roll 4)`

`= 1/6 xx text(number of rolls)`

`= 1/6 xx 60`

`= 10`

`=>  B`

Filed Under: Probability, Relative Frequency, Relative Frequency (Std 1), Relative Frequency (Std 2), Relative Frequency (Y11) Tagged With: Band 4, common-content, num-title-ct-core, num-title-qs-hsc, smc-1133-20-Games of Chance, smc-1133-30-Expected Frequency (np), smc-4225-35-Relative frequency, smc-827-20-Games of Chance, smc-827-40-Expected Frequency (np), smc-990-20-Games of Chance, smc-990-40-Expected Frequency (np)

Probability, STD2 S2 2008 HSC 22 MC

A die has faces numbered 1 to 6. The die is biased so that the number 6 will appear more often than each of the other numbers. The numbers 1 to 5 are equally likely to occur.

The die was rolled 1200 times and it was noted that the 6 appeared 450 times.

Which statement is correct?

  1. The probability of rolling the number 5 is expected to be  `1/7`.
  2. The number 6 is expected to appear 2 times as often as any other number.
  3. The number 6 is expected to appear 3 times as often as any other number.
  4. The probability of rolling an even number is expected to be equal to the probability of rolling an odd number.
Show Answers Only

`C`

Show Worked Solution

`P(6) = 450/1200 = 3/8`
 

`text(Numbers 1-5 are rolled) = 1200-450=750\ text(times)` 

`:.\ text(Each number is expected to appear)`

`750/5 = 150\ text(times)`
 

`:.P text{(specific number ≠ 6)}`

`= 150/1200`

 `= 1/8`
 

`=>  C`

Filed Under: Multi-stage Events, Multi-Stage Events (Std 2), Relative Frequency (Std 2) Tagged With: Band 5, smc-827-20-Games of Chance, smc-829-20-Other Multi-Stage Events

Probability, STD2 S2 2011 HSC 24b

A die was rolled 72 times. The results for this experiment are shown in the table.
  

\begin{array} {|c|c|}
\hline
\rule{0pt}{2.5ex} \textit{Number obtained} \rule[-1ex]{0pt}{0pt} & \textit{Frequency} \\
\hline
\rule{0pt}{2.5ex} \ 1 \rule[-1ex]{0pt}{0pt} & 16 \\
\hline
\rule{0pt}{2.5ex} \ 2 \rule[-1ex]{0pt}{0pt} & 11 \\
\hline
\rule{0pt}{2.5ex} \ 3 \rule[-1ex]{0pt}{0pt} & \textbf{A} \\
\hline
\rule{0pt}{2.5ex} \ 4 \rule[-1ex]{0pt}{0pt} & 8 \\
\hline
\rule{0pt}{2.5ex} \ 5 \rule[-1ex]{0pt}{0pt} & 12 \\
\hline
\rule{0pt}{2.5ex} \ 6 \rule[-1ex]{0pt}{0pt} & 15 \\
\hline
\end{array}

  1. Find the value of  `A`.   (1 mark)

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  2. What was the relative frequency of obtaining a 4.   (1 mark)

    --- 1 WORK AREA LINES (style=lined) ---

  3. If the die was unbiased, which number was obtained the expected number of times?   (1 mark)

    --- 1 WORK AREA LINES (style=lined) ---

Show Answers Only
  1. \(10\)
  2. \(\dfrac{1}{9}\)
  3. \(5\)
Show Worked Solution
i.     \(\text{Since die rolled 72 times}\)
\(\therefore\ A\) \(=72-(16+11+8+12+15)\)
  \(=72-62\)
  \(=10\)
♦ Mean mark 38%
IMPORTANT: Many students confused ‘relative frequency’ with ‘frequency’ and incorrectly answered 8.
ii.     \(\text{Relative frequency of 4}\) \(=\dfrac{8}{72}\)
  \(=\dfrac{1}{9}\)

 

iii.  \(\text{Expected frequency of any number}\)
\(=\dfrac{1}{6}\times 72\)
\(=12\)
 
\(\therefore\ \text{5 was obtained the expected number of times.}\)

Filed Under: Probability, Relative Frequency, Relative Frequency (Std 1), Relative Frequency (Std 2), Relative Frequency (Y11) Tagged With: Band 2, Band 4, Band 5, common-content, num-title-ct-core, num-title-qs-hsc, smc-1133-20-Games of Chance, smc-1133-30-Expected Frequency (np), smc-4225-35-Relative frequency, smc-827-20-Games of Chance, smc-827-40-Expected Frequency (np), smc-990-20-Games of Chance, smc-990-40-Expected Frequency (np)

Probability, STD2 S2 2009 HSC 28d

In an experiment, two unbiased dice, with faces numbered  1, 2, 3, 4, 5, 6  are rolled 18 times.

The difference between the numbers on their uppermost faces is recorded each time. Juan performs this experiment twice and his results are shown in the tables.

 2009 28d

Juan states that Experiment 2 has given results that are closer to what he expected than the results given by Experiment 1.

Is he correct? Explain your answer by finding the sample space for the dice differences and using theoretical probability.    (4 marks)

--- 8 WORK AREA LINES (style=lined) ---

 

Show Answers Only

 `text{Juan is correct (See Worked Solutions)}`

Show Worked Solution
♦♦♦ Mean mark 7%. Toughest question in the 2009 exam.
MARKER’S COMMENT: This question guides students by asking for an explanation using the sample space for the dice differences. This step alone received 2 full marks. Note that instructions to explain your answer requires mathematical calculations to support an argument.

`text(Sample space for dice differences)`

2UG-2009-28d1

2UG-2009-28d2_1

2UG-2009-28d3_1

`text(Juan is correct.  The table shows Experiment 1)`

`text(has greater total differences to the expected)`

`text(frequencies than Experiment 2)`

Filed Under: Multi-stage Events, Multi-Stage Events (Std 2), Relative Frequency, Relative Frequency (Std 2), Relative Frequency (Y11) Tagged With: Band 6, common-content, smc-827-20-Games of Chance, smc-827-40-Expected Frequency (np), smc-829-50-Arrays

Probability, STD2 S2 2009 HSC 9 MC

A wheel has the numbers 1 to 20 on it, as shown in the diagram. Each time the wheel is spun, it stops with the marker on one of the numbers.
 


 

 The wheel is spun 120 times.

 How many times would you expect a number less than 6 to be obtained?

  1.   `20` 
  2.   `24` 
  3.   `30` 
  4.   `36` 
Show Answers Only

`C`

Show Worked Solution

`P(text(number < 6) ) = 5/20 = 1/4`

`:.\ text(Expected times)` `= 1/4 xx text(times spun)`
  `= 1/4 xx 120`
  `= 30`

`=>  C`

Filed Under: Probability, Relative Frequency, Relative Frequency (Std 1), Relative Frequency (Std 2), Relative Frequency (Y11) Tagged With: Band 4, common-content, num-title-ct-core, num-title-qs-hsc, smc-1133-20-Games of Chance, smc-1133-30-Expected Frequency (np), smc-4225-35-Relative frequency, smc-827-20-Games of Chance, smc-827-40-Expected Frequency (np), smc-990-20-Games of Chance, smc-990-40-Expected Frequency (np)

Probability, STD2 S2 2012 HSC 17 MC

A spinner with different coloured sectors is spun 40 times. The results are recorded in the table.

 What is the relative frequency of obtaining the colour orange? 

  1.    `3/20`  
  2.    `1/5`  
  3.    `6`  
  4.    `8` 
Show Answers Only

`A`

Show Worked Solution
♦♦ Mean mark 34%
COMMENT: Note that relative frequency is the frequency of an event divided by the total frequencies (i.e. the probability).
`text(Total frequency)` `= 40\ text(spins)`
`text(Orange freq.)` `= 40\-(2 + 4 + 6 + 10 +12)`
  `=6`
`:.\ text(Relative freq.)` `= 6/40 = 3/20`

 
`=>  A`

Filed Under: Probability, Relative Frequency, Relative Frequency (Std 1), Relative Frequency (Std 2), Relative Frequency (Y11) Tagged With: Band 5, common-content, num-title-ct-core, num-title-qs-hsc, smc-1133-20-Games of Chance, smc-4225-35-Relative frequency, smc-827-20-Games of Chance, smc-990-20-Games of Chance

Probability, STD2 S2 2013 HSC 7 MC

In an experiment, a standard six-sided die was rolled 72 times. The results are shown in the table.
 

Which number on the die was obtained the expected number of times?

  1.    1
  2.    2
  3.    3
  4.    6
Show Answers Only

`B`

Show Worked Solution

`text(Probability of rolling a specific number)=1/6`

`:.\ text(After 72 rolls, a specific number is expected)`

 `1/6xx72=12\ text(times.)`

`=>\ B`

Filed Under: Probability, Relative Frequency, Relative Frequency (Std 1), Relative Frequency (Std 2), Relative Frequency (Y11) Tagged With: Band 3, common-content, num-title-ct-core, num-title-qs-hsc, smc-1133-20-Games of Chance, smc-1133-30-Expected Frequency (np), smc-4225-35-Relative frequency, smc-827-20-Games of Chance, smc-827-40-Expected Frequency (np), smc-990-20-Games of Chance, smc-990-40-Expected Frequency (np)

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