SmarterEd

Aussie Maths & Science Teachers: Save your time with SmarterEd

  • Login
  • Get Help
  • About

Probability, STD1 S2 2022 HSC 3 MC

A jar contains 12 red, 10 black and 13 white lollies.

Alex picks out a red lolly and eats it. He then randomly picks a second lolly.

What is the probability that the second lolly is also red?

  1. `(11)/(34)`
  2. `(11)/(35)`
  3. `(12)/(34)`
  4. `(12)/(35)`
Show Answers Only

`A`

Show Worked Solution
`P(E)` `=text{favourable outcomes}/text{total outcomes}`  
  `=(12-1)/((12-1)+10+13)`  
  `=11/34`  

 
`=>A`

Filed Under: Single and Multi-Stage Events (Std 1) Tagged With: Band 4, smc-1135-20-Other Multi-Stage Events

Probability, STD1 S2 2021 HSC 20

In a bag, there are six playing cards, 2, 4, 6, 8, Queen and King. The Queen and King are known as picture cards.

Two of these cards are chosen randomly. All the possible outcomes are shown.
 

   
 

  1. What is the probability that the two cards chosen include one or both picture cards?  (1 mark)

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

  2. What is the probability that the two cards chosen do NOT include any picture cards?  (1 mark)

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

Show Answers Only
  1. `9/15`
  2. `6/15`
Show Worked Solution

a.   `P text{(at least 1 picture card)} = 9/15`

 

b.    `P text{(no picture card)}` `= 1 – 9/15`
    `= 6/15`

Filed Under: Probability, Single and Multi-Stage Events (Std 1) Tagged With: Band 4, num-title-ct-core, num-title-qs-hsc, smc-1135-20-Other Multi-Stage Events, smc-1135-30-P(E) = 1 - P(not E), smc-4225-15-Single-stage events, smc-4225-20-Complementary events

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)

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

Show Answers Only
  1.  

     
  2. `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}`

Filed Under: Probability, Single and Multi-Stage Events (Std 1) Tagged With: Band 4, Band 5, num-title-ct-core, num-title-qs-hsc, smc-1135-20-Other Multi-Stage Events, smc-1135-40-Arrays, smc-4225-20-Complementary events, smc-4225-45-Multi-stage events

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)

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

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

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

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

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

Show Answers Only
  1. `5`
  2. `1/2`
  3. `2/3`
Show Worked Solution

i.   `X=3+2=5`
 

ii.   `P(text{score}<4)=6/12=1/2`
 

iii.   `P(3)=2/3`

Filed Under: Multi-Stage Events (Std 2), Single and Multi-Stage Events (Std 1) Tagged With: Band 3, Band 4, smc-1135-20-Other Multi-Stage Events, smc-1135-40-Arrays, smc-829-20-Other Multi-Stage Events, smc-829-50-Arrays

Probability, STD2 S2 2015 HSC 16 MC

The probability of winning a game is `7/10`.

Which expression represents the probability of winning two consecutive games?

  1. `7/10 xx 6/9`
  2. `7/10 xx 6/10`
  3. `7/10 xx 7/9`
  4. `7/10 xx 7/10`
Show Answers Only

`D`

Show Worked Solution

`text{Since the two events are independent:}`

`P text{(W)}` `= 7/10`
`P text{(WW)}` `= 7/10 xx 7/10`

 
`=>D`

Filed Under: Multi-stage Events, Multi-Stage Events, Multi-Stage Events (Std 2), Single and Multi-Stage Events (Std 1) Tagged With: Band 4, num-title-ct-corea, num-title-qs-hsc, smc-1135-20-Other Multi-Stage Events, smc-4238-20-Independent events, smc-829-20-Other Multi-Stage Events

Probability, STD2 S2 2006 HSC 25c

Sonia buys three raffle tickets.

HSC 2006 25c

  1. What is the probability that Sonia wins first prize?  (1 mark)

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

  2. What is the probability that she wins both prizes?  (2 marks)

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

Show Answers Only
  1. `1/60`
  2. `1/5370`
Show Worked Solution
i.  `text{P (wins 1st prize)}` `= text(# tickets bought) / text(total tickets)`
  `= 3/180`
  `= 1/60`

 

ii.  `text{P (wins both)}` `= text{P (wins 1st)} xx text{P (wins 2nd)}`
  `= 1/60 xx 2/179`
  `= 1/5370`

Filed Under: Multi-stage Events, Multi-Stage Events (Std 2), Single and Multi-Stage Events (Std 1) Tagged With: Band 4, Band 5, smc-1135-20-Other Multi-Stage Events, smc-829-20-Other Multi-Stage Events

Probability, STD2 S2 2006 HSC 10 MC

Kay randomly selected a marble from a bag of marbles, recorded its colour and returned it to the bag. She repeated this process a number of times.
  


  

Based on these results, what is the best estimate of the probability that Kay will choose a green marble on her next selection?

  1.   `5/24`
  2.   `1/24`
  3.   `1/6`
  4.   `1/5`
Show Answers Only

`C`

Show Worked Solution
`text{P(Green)}` `= text(# Green chosen) / text(Total Selections)`
  `= 4/24`
  `= 1/6`

`=>  C`

Filed Under: Multi-stage Events, Multi-Stage Events (Std 2), Probability, Single and Multi-Stage Events (Std 1) Tagged With: Band 4, num-title-ct-core, num-title-qs-hsc, smc-1135-20-Other Multi-Stage Events, smc-4225-35-Relative frequency, smc-829-20-Other Multi-Stage Events

Probability, STD2 S2 2004 HSC 18 MC

Two dice are rolled. What is the probability that only one of the dice shows a six?

  1. `5/36`
  2. `1/6`
  3. `5/18`
  4. `11/36`
Show Answers Only

`C`

Show Worked Solution

`text{P (Only 1 six)}`

`= P text{(6, not 6)} + P text{(not 6, 6)}`

`= 1/6 xx 5/6 + 5/6 xx 1/6`

`= 10/36`

`= 5/18`

`=>  C`

Filed Under: Multi-stage Events, Multi-Stage Events (Std 2), Single and Multi-Stage Events (Std 1) Tagged With: Band 5, smc-1135-20-Other Multi-Stage Events, smc-829-20-Other Multi-Stage Events

Probability, STD2 S2 2007 HSC 10 MC

Each time she throws a dart, the probability that Mary hits the dartboard is  `2/7`.

She throws two darts, one after the other.

What is the probability that she hits the dartboard with both darts?

  1. `1/21` 
  2. `4/49` 
  3. `2/7`
  4. `4/7`
Show Answers Only

`B`

Show Worked Solution

`P text{(hits)} = 2/7`

`P text{(hits twice)}` `= 2/7 xx 2/7`
  `= 4/49`

`=>  B`

Filed Under: Multi-stage Events, Multi-Stage Events (Std 2), Single and Multi-Stage Events (Std 1) Tagged With: Band 4, smc-1135-20-Other Multi-Stage Events, smc-829-20-Other Multi-Stage Events

Probability, STD2 S2 2009 HSC 27c

In each of three raffles, 100 tickets are sold and one prize is awarded.

Mary buys two tickets in one raffle. Jane buys one ticket in each of the other two raffles.

Determine who has the better chance of winning at least one prize. Justify your response using probability calculations.   (4 marks)  

Show Answers Only
`P(text(Mary wins) )` `= 2/100`
  `= 1/50`

 

`P(text(Jane wins at least 1) )` `= 1-P (text(loses both) )`
  `= 1-99/100 xx 99/100`
  `= 1-9801/(10\ 000)`
  `= 199/(10\ 000)`

 
`text{Since}\ \ 1/50 > 199/(10\ 000)`

`=>\ text(Mary has a better chance of winning.)`

Show Worked Solution
`P(text(Mary wins) )` `= 2/100`
  `= 1/50`

 

`P(text(Jane wins at least 1) )` `= 1-P (text(loses both) )`
  `= 1-99/100 xx 99/100`
  `= 1-9801/(10\ 000)`
  `= 199/(10\ 000)`

 
`text{Since}\ \ 1/50 > 199/(10\ 000)`

`=>\ text(Mary has a better chance of winning.)`

♦♦ Mean mark 31%.
MARKER’S COMMENT: Very few students calculated Jane’s chance of winning correctly. Note the use of “at least” in the question. Finding `1-P`(complement) is the best strategy here.

Filed Under: Multi-stage Events, Multi-Stage Events, Multi-Stage Events (Std 2), Single and Multi-Stage Events (Std 1) Tagged With: Band 5, num-title-ct-pathb, num-title-qs-hsc, smc-1135-20-Other Multi-Stage Events, smc-1135-30-P(E) = 1 - P(not E), smc-4238-70-Complementary events, smc-4238-80-"at least", smc-829-20-Other Multi-Stage Events, smc-829-30-P(E) = 1 - P(not E)

Probability, STD2 S2 2013 HSC 26c

The probability that Michael will score more than 100 points in a game of bowling is `31/40`. 

  1. A commentator states that the probability that Michael will score less than 100 points in a game of bowling is  `9/40`.

     

    Is the commentator correct? Give a reason for your answer.   (1 mark)

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

  2. Michael plays two games of bowling. What is the probability that he scores more than 100 points in the first game and then again in the second game?   (1 mark)

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

Show Answers Only
  1. `text{Incorrect. Less than “or equal to 100” is correct.}`
  2. `961/1600`
Show Worked Solution
♦♦♦ Mean mark 11%

i.   `text(The commentator is incorrect. The correct)`

`text(statement is)\ Ptext{(score} <=100 text{)} =9/40`

`text{(i.e. less than “or equal to 100” is the correct statement)}`

 

♦ Mean mark 34%
ii. `\ \ \ P(text{score >100 in both})` `= 31/40 xx 31/40` 
    `= 961/1600`

Filed Under: Fundamental understanding, Fundamental Understanding (Std 1), Fundamental Understanding (Std 2), Multi-stage Events, Multi-Stage Events, Multi-Stage Events (Std 2), Single and Multi-Stage Events (Std 1) Tagged With: Band 5, Band 6, num-title-ct-corea, num-title-qs-hsc, smc-1135-20-Other Multi-Stage Events, smc-4238-20-Independent events, smc-4238-70-Complementary events, smc-829-20-Other Multi-Stage Events

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

`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.

Filed Under: Multi-stage Events, Multi-Stage Events (Std 2), Single and Multi-Stage Events (Std 1) Tagged With: Band 5, num-title-ct-pathb, num-title-qs-hsc, smc-1135-20-Other Multi-Stage Events, smc-1135-30-P(E) = 1 - P(not E), smc-4238-70-Complementary events, smc-4238-80-"at least", smc-829-20-Other Multi-Stage Events, smc-829-30-P(E) = 1 - P(not E)

Probability, STD2 S2 2013 HSC 18 MC

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

What is the probability of obtaining a sum of 6?

  1. `1/6`
  2. `1/12`
  3. `5/12`
  4. `5/36`
Show Answers Only

`D`

Show Worked Solution

`text(Total outcomes)=6xx6=36`

`text{Outcomes that sum to 6}=text{(1,5) (5,1) (2,4) (4,2) (3,3)} =5`

`:.\ P\text{(sum of 6)} =5/36`

`=>\ D`

♦♦ Mean mark 35%.

Filed Under: Multi-stage Events, Multi-Stage Events, Multi-Stage Events (Std 2), Single and Multi-Stage Events (Std 1) Tagged With: Band 5, num-title-ct-corea, num-title-qs-hsc, smc-1135-20-Other Multi-Stage Events, smc-4238-20-Independent events, smc-829-20-Other Multi-Stage Events

Copyright © 2014–2025 SmarterEd.com.au · Log in