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Probability, STD2 S2 2023 HSC 8 MC

A game involves throwing a die and spinning a spinner.

The sum of the two numbers obtained is the score.

The table of scores below is partially completed.
 

What is the probability of getting a score of 7 or more?

  1. `1/6`
  2. `1/4`
  3. `5/18`
  4. `5/12`

Show Answers Only

`D`

Show Worked Solution

`Ptext{(score 7+)} = 10/24=5/12`

`=>D`

Probability, STD1 S2 2020 HSC 26

Barbara plays a game of chance, in which two unbiased six-sided dice are rolled. The score for the game is obtained by finding the difference between the two numbers rolled. For example, if Barbara rolls a 2 and a 5, the score is 3.

The table shows some of the scores.
 


 

  1. Complete the six missing values in the table to show all possible scores for the game.   (1 mark)
  2. What is the probability that the score for a game is NOT 0?   (2 marks)

Show Answers Only

 a.

b.    `frac{5}{6}`

Show Worked Solution

a.     

♦ Mean mark part (b) 47%.
b.      `Ptext{(not zero)}` `= frac{text(numbers) ≠ 0}{text(total numbers)}`
    `= frac{30}{36}= frac{5}{6}`

 
\(\text{Alternate solution (b)}\)

b.      `Ptext{(not zero)}` `= 1-Ptext{(zero)}`
    `= 1-frac{6}{36}`
    `= frac{5}{6}`

Probability, STD2 S2 2011 HSC 26a

The two spinners shown are used in a game.

2UG 2011 26a1

Each arrow is spun once. The score is the total of the two numbers shown by the arrows.
A table is drawn up to show all scores that can be obtained in this game.

2UG 2011 26a2

  1. What is the value of `X` in the table?   (1 mark)

  2. What is the probability of obtaining a score less than 4?   (1 mark)

  3. On Spinner `B`, a 2 is obtained. What is the probability of obtaining a score of 3?   (1 mark)

Show Answers Only

a.    `5`

b.    `1/2`

c.    `2/3`

Show Worked Solution

a.    `X=3+2=5`

b.    `P(text{score}<4)=6/12=1/2`

c.    `P(3)=2/3`

Probability, STD2 S2 2012 HSC 12 MC

Two unbiased dice, each with faces numbered 1, 2, 3, 4, 5, 6, are rolled. 

What is the probability of a 6 appearing on at least one of the dice? 

  1. `1/6`  
  2. `11/36` 
  3. `25/36`  
  4. `5/6`  
Show Answers Only

`B`

Show Worked Solution

`text(Method 1: Using an array`

`P text{(at least 1 six)}=11/36`

\begin{align}
\textbf{Die B }
\begin{array}{c}
\textbf{Die A}  \\
\begin{array}{c|c|c|c|c|c|c}
\ & 1 & 2 & 3 & 4 & 5 & 6  \\
\hline
\ 1 & 1,1  & 1,2 & 1,3 & 1,4 & 1,5 & \fcolorbox{red}{white}{1,6} \\
\hline
\ 2 & 2,1 & 2,2 & 2,3 & 2,4 & 2,5 & \fcolorbox{red}{white}{2,6}  \\
\hline
\ 3 & 3,1 & 3,2 & 3,3 & 3,4 & 3,5 & \fcolorbox{red}{white}{3,6}  \\
\hline
\ 4 & 4,1 & 4,2 & 4,3 & 4,4 & 4,5 & \fcolorbox{red}{white}{4,6}  \\
\hline
\ 5 & 5,1 & 5,2 & 5,3 & 5,4 & 5,5 & \fcolorbox{red}{white}{5,6}  \\
\hline
\ 6 & \fcolorbox{red}{white}{6,1} & \fcolorbox{red}{white}{6,2} & \fcolorbox{red}{white}{6,3} & \fcolorbox{red}{white}{6,4} & \fcolorbox{red}{white}{6,5} & \fcolorbox{red}{white}{6,6}  \\
\end{array}
\end{array}
\end{align}

 

`text(Method 2: Using )P text{(E)} = 1-P\text{(not E)}`
  

`P text{(at least 1 six)}`

`=1-P text{(no six)} xx P text{(no six)} `

`=1-5/6 xx 5/6=11/36`

`=>  B`

♦♦♦ Mean mark 25%
COMMENT: The term “at least” should flag that calculating the probability of `1-P text{(event not happening)}` is likely to be the most efficient way to solve.