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?
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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?
\(A\)
| \(P(\text{White})\) | \(=\dfrac{50}{100}=\dfrac{1}{2}\) |
| \(P(\text{Red})\) | \(=\dfrac{25}{100}=\dfrac{1}{4}\) |
| \(P(\text{Yellow})\) | \(=\dfrac{15}{100}=\dfrac{3}{20}\) |
| \(P(\text{Green})\) | \(=\dfrac{10}{100}=\dfrac{2}{20}=\dfrac{1}{10}\) |
\(\text{Eliminate Options B and D as white}\ \neq \dfrac{1}{2}\ \text{of spinner.}\)
\(\text{Eliminate Option C as red}\ \neq \dfrac{1}{4}\ \text{of spinner.}\)
\(\Rightarrow A\)
The faces on a twenty-sided die are labelled $0.05, $0.10, $0.15, … , $1.00.
The die is rolled once.
What is the probability that the amount showing on the upper face is more than 50 cents but less than 80 cents?
`A`
`text(Possible faces that satisfy are:)`
`55text(c),60text(c),65text(c),70text(c),75text(c)`
`:.\ text(Probability)= 5/20= 1/4`
`=>A`
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}
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a. \(0.17\)
b. \(0.86\)
a. \(\text{Relative frequency of red}\)
\(= 1-(0.32 + 0.13 + 0.14 + 0.24)= 1-0.83= 0.17\)
b. \(P\text{(not selecting black)}\)
\(= 1-P\text{(selecting black)}= 1-0.14= 0.86\)
The diagram shows a spinner.
The arrow is spun and will stop in one of the six sections.
What is the probability that the arrow will stop in a section containing a number greater
than 4?
`D`
`P\ text((number greater than 4))`
`= P(7) + P (9)`
`= 2/6 + 1/6= 1/2`
`=> D`
There are 100 tickets sold in a raffle. Justine sold all 100 tickets to five of her friends. The number of tickets she sold to each friend is shown in the table.
Give a reason why Justine’s statement is NOT correct. (1 mark)
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a. `text(The claim is incorrect because each of her friends)`
`text(bought a different number of tickets and therefore)`
`text(their chances of winning are different.)`
b. `69/100`
a. `text(The claim is incorrect because each of her friends bought)`
`text(a different number of tickets and therefore their chances of)`
`text(winning are different.)`
b. `text(Number of tickets not sold to K or H)= 45 + 10 + 14= 69`
`:.\ text(Probability 1st prize NOT won by K or H)= 69/100`
A bag contains some marbles. The probability of selecting a blue marble at random from this bag is `3/8`.
Which of the following could describe the marbles that are in the bag?
`D`
`P(B) = 3/8`
| `text(In)\ A,\ \ ` | `P(B) = 3/11` |
| `text(In)\ B,\ \ ` | `P(B) = 6/17 ` |
| `text(In)\ C,\ \ ` | `P(B) = 3/11` |
| `text(In)\ D,\ \ ` | `P(B) = 6/16 = 3/8` |
`=> D`
On Saturday, Jonty recorded the colour of T-shirts worn by the people at his gym. The results are shown in the graph.
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a. `34`
b. `15/34`
a. `text(# People)=5+15+10+3+1=34`
b. `P (B\ text{or}\ G)=P(B)+P(G)=5/34+10/34=15/34`
A bag contains red, green, yellow and blue balls.
\begin{array} {|c|c|}
\hline
\rule{0pt}{2.5ex} \textit{Colour} \rule[-1ex]{0pt}{0pt} & \textit{Probability} \\
\hline
\rule{0pt}{2.5ex} \text{Red} & \dfrac{1}{3} \\
\hline
\rule{0pt}{2.5ex} \text{Green} & \dfrac{1}{4} \\
\hline
\rule{0pt}{2.5ex} \text{Yellow} & \text{?} \\
\hline
\rule{0pt}{2.5ex} \text{Blue} & \dfrac{1}{6} \\
\hline
\end{array}
The table shows the probability of choosing a red, green, or blue ball from the bag.
If there are 12 yellow balls in the bag, how many balls are in the bag altogether
\(C\)
| \(P(R)+P(G)+P(Y)+P(B)\) | \(=1\) |
| \(\dfrac{1}{3}+\dfrac{1}{4}+P(Y)+\dfrac{1}{6}\) | \(=1\) |
| \(P(Y)\) | \(= 1-(\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{6})\) |
| \(=1-\dfrac{9}{12}\) | |
| \(=\dfrac{1}{4}\) |
| \(P(Y)\) | \(=\dfrac{\text{Yellow balls}}{\text{Total balls}}\) |
| \(\dfrac{1}{4}\) | \(=\dfrac{12}{\text{Total balls}}\) |
\(\therefore\ \text{ Total balls}=48\)
\(\Rightarrow C\)
Which of the following events would be LEAST likely to occur?
`C`
`P(A)=1/2,\ \ P(B)=1/6`
`P(C)=1/26,\ \ P(D)=4/100=1/25`
`=> C`