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Probability, SMB-009

Jon spins each pointer 50 times.

 

Each time he added the numbers that the pointers landed on.

His results are shown below.

\begin{array} {|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}\textbf{Sum of numbers}\rule[-1ex]{0pt}{0pt} & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & \textbf{Total}\\
\hline
\rule{0pt}{2.5ex}\textbf{Number of spins}\rule[-1ex]{0pt}{0pt} & 1 & 2 & 3 & 6 & 8 & 9 & 7 & 5 & 4 & 4 & 1 & \textbf{50} \\
\hline
\end{array}

What percentage of the spins resulted in a sum of 9?   (2 marks)

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Show Answers Only

`text(10%)`

Show Worked Solution
`text(Percentage)` `= text(number of 9’s)/text(number of spins) xx 100`
  `=5/50 xx 100`
  `=10 text(%)`

Filed Under: Multi-Stage Events Tagged With: num-title-ct-corea, smc-4238-20-Independent events, smc-4238-60-Arrays

Probability, SMB-006

Bromley rolls two standard dice at the same time and adds up the total.

 An incomplete table of the possible outcomes is below.

\begin{array} {|c|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}\textbf{Face}  \rule[-1ex]{0pt}{0pt} & \textbf{1} & \textbf{2} & \textbf{3} & \textbf{4} & \textbf{5} & \textbf{6}\\
\hline
\rule{0pt}{2.5ex}\textbf{1}\rule[-1ex]{0pt}{0pt} & 2 & 3 & 4 & 5 & 6 & 7 \\
\hline
\rule{0pt}{2.5ex}\textbf{2}\rule[-1ex]{0pt}{0pt} & 3 & 4 & 5 & & & 8 \\
\hline
\rule{0pt}{2.5ex}\textbf{3}\rule[-1ex]{0pt}{0pt} & 4 & 5 & 6 &  &  & 9 \\
\hline
\rule{0pt}{2.5ex}\textbf{4}\rule[-1ex]{0pt}{0pt} & 5 &  & 7 & 8 & 9 & 10 \\
\hline
\rule{0pt}{2.5ex}\textbf{5}\rule[-1ex]{0pt}{0pt} & 6 &  & 8 &  & 10 & 11 \\
\hline
\rule{0pt}{2.5ex}\textbf{6}\rule[-1ex]{0pt}{0pt} & 7 & 8 & 9 & 10 & 11 & 12 \\
\hline
\end{array}

  1. Complete the table.   (2 marks)

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  2. What is the most likely total that Bromley will roll?   (1 mark)

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i.

\begin{array} {|c|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}\textbf{Face}  \rule[-1ex]{0pt}{0pt} & \textbf{1} & \textbf{2} & \textbf{3} & \textbf{4} & \textbf{5} & \textbf{6}\\
\hline
\rule{0pt}{2.5ex}\textbf{1}\rule[-1ex]{0pt}{0pt} & 2 & 3 & 4 & 5 & 6 & 7 \\
\hline
\rule{0pt}{2.5ex}\textbf{2}\rule[-1ex]{0pt}{0pt} & 3 & 4 & 5 & \colorbox{lightblue}{6} & \colorbox{lightblue}{7} & 8 \\
\hline
\rule{0pt}{2.5ex}\textbf{3}\rule[-1ex]{0pt}{0pt} & 4 & 5 & 6 & \colorbox{lightblue}{7} & \colorbox{lightblue}{8} & 9 \\
\hline
\rule{0pt}{2.5ex}\textbf{4}\rule[-1ex]{0pt}{0pt} & 5 & \colorbox{lightblue}{6}  & 7 & 8 & 9 & 10 \\
\hline
\rule{0pt}{2.5ex}\textbf{5}\rule[-1ex]{0pt}{0pt} & 6 & \colorbox{lightblue}{7} & 8 & \colorbox{lightblue}{9} & 10 & 11 \\
\hline
\rule{0pt}{2.5ex}\textbf{6}\rule[-1ex]{0pt}{0pt} & 7 & 8 & 9 & 10 & 11 & 12 \\
\hline
\end{array}

ii.    `7`

Show Worked Solution

i.

\begin{array} {|c|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}\textbf{Face}  \rule[-1ex]{0pt}{0pt} & \textbf{1} & \textbf{2} & \textbf{3} & \textbf{4} & \textbf{5} & \textbf{6}\\
\hline
\rule{0pt}{2.5ex}\textbf{1}\rule[-1ex]{0pt}{0pt} & 2 & 3 & 4 & 5 & 6 & 7 \\
\hline
\rule{0pt}{2.5ex}\textbf{2}\rule[-1ex]{0pt}{0pt} & 3 & 4 & 5 & \colorbox{lightblue}{6} & \colorbox{lightblue}{7} & 8 \\
\hline
\rule{0pt}{2.5ex}\textbf{3}\rule[-1ex]{0pt}{0pt} & 4 & 5 & 6 & \colorbox{lightblue}{7} & \colorbox{lightblue}{8} & 9 \\
\hline
\rule{0pt}{2.5ex}\textbf{4}\rule[-1ex]{0pt}{0pt} & 5 & \colorbox{lightblue}{6}  & 7 & 8 & 9 & 10 \\
\hline
\rule{0pt}{2.5ex}\textbf{5}\rule[-1ex]{0pt}{0pt} & 6 & \colorbox{lightblue}{7} & 8 & \colorbox{lightblue}{9} & 10 & 11 \\
\hline
\rule{0pt}{2.5ex}\textbf{6}\rule[-1ex]{0pt}{0pt} & 7 & 8 & 9 & 10 & 11 & 12 \\
\hline
\end{array}

 
ii.
    `text{The most frequent total = 7 (6 times)}`

`:.\ text{7 is the most likely to be rolled.}`

Filed Under: Multi-Stage Events Tagged With: num-title-ct-corea, smc-4238-60-Arrays

Probability, SMB-001

A coin is tossed 3 times. There are 8 possible outcomes.

What is the probability of getting 2 heads and 1 tail in any order?   (2 marks)

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`3/8`

Show Worked Solution

`text(Strategy 1)`

`text{The table (array) below lists the possible outcomes:}`
 

  1st toss H H H H T T T T
  2nd toss H H T T H H T T
  3rd toss H T H T H T H T
       ✓  ✓    ✓      

`text(From table,)\ Ptext{(2H, 1T)} = 3/8`
  

`text(Strategy 2)`

`text(Using a probability tree:)`
 

Filed Under: Multi-Stage Events Tagged With: num-title-ct-corea, smc-4238-20-Independent events, smc-4238-50-Probability trees, smc-4238-60-Arrays

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