SmarterEd

Aussie Maths & Science Teachers: Save your time with SmarterEd

  • Login
  • Get Help
  • About

Probability, STD2 S2 2023 HSC 23

One hundred tickets are sold in a raffle which offers two prizes. Hazel buys five of the tickets.

A ticket is drawn at random for the first prize. A second ticket is drawn from the remaining tickets for the other prize.

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

Show Answers Only

`1/495`

Show Worked Solution

`P(W_1, W_2) = 5/100 xx 4/99=1/495`

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

Probability, STD2 S2 2020 HSC 15 MC

The top of a rectangular table is divided into 8 equal sections as shown.
 

A standard die with faces labelled 1 to 6 is rolled onto the table. The die is equally likely to land in any of the 8 sections of the table. If the die does not land entirely in one section of the table, it is rolled again.

A score is calculated by multiplying the value shown on the top face of the die by the number shown in the section of the table where the die lands.

What is the probability of getting a score of 6?

  1. `frac{1}{48}`
  2. `frac{1}{12}`
  3. `frac{1}{8}`
  4. `frac{1}{6}`
Show Answers Only

`B`

Show Worked Solution

`text{Strategy 1}`

♦ Mean mark 40%.

`text{Combinations that score 6:}`

`4 \ text{combinations score} \ 6`

`therefore \ P(6)` `= 4 xx frac{1}{8} xx frac{1}{6}`
  `= frac{1}{12}`

 

`text{Strategy 2}`

`Ptext{(6)}` `= Ptext{(T1)}Ptext{(D6)} + Ptext{(T2)}Ptext{(D3)} + Ptext{(T3)}Ptext{(D2)} + Ptext{(T6)}Ptext{(D1)}`
  `= frac{1}{8} xx frac{1}{6} + frac{1}{8} xx frac{1}{6} + frac{1}{8} xx frac{1}{6} + frac{1}{8} xx frac{1}{6}`
  `= 4 xx frac{1}{8} xx frac{1}{6}`
  `= frac{1}{12}`

 
`=> \ B`

Filed Under: Multi-Stage Events (Std 2) Tagged With: Band 5, smc-829-20-Other Multi-Stage Events, smc-829-50-Arrays

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 2017 HSC 24 MC

A deck of 52 playing cards contains 12 picture cards. Two cards from the deck are drawn at random and placed on a table.

What is the probability, correct to four decimal places, that exactly one picture card is on the table?

A.     `0.0498`

B.     `0.1810`

C.     `0.3550`

D.     `0.3620`

Show Answers Only

`text(D)`

Show Worked Solution

`P(text(exactly 1 picture card))`

`= P(text(picture)) xx P(text(no picture)) + P(text(no picture)) xx P(text(picture))`

`= 12/52 xx 40/51quad+quad40/52 xx 12/51`

♦♦♦ Mean mark 21%.

`= 2((12 xx 40)/(52 xx 51))`

`= 0.36199…`

`=>\ text(D)`

Filed Under: Multi-stage Events, Multi-Stage Events (Std 2) Tagged With: Band 6, smc-829-20-Other Multi-Stage Events

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 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 2014 HSC 16 MC

In Mathsville, there are on average eight rainy days in October.

Which expression could be used to find a value for the probability that it will rain on two consecutive days in October in Mathsville?

  1. `8/31 xx 7/30`
  2. `8/31 xx 7/31`
  3. `8/31 xx 8/30`
  4. `8/31 xx 8/31`
Show Answers Only

`D`

Show Worked Solution

`P text{(rains)} = 8/31\ \ \text{(independent event for each day)}`

`text{Since each day has same probability:}`

`P(R_1 R_2) = 8/31 xx 8/31`

`=>  D`

♦♦♦ Mean mark 16%.
Lowest mark of any MC question in 2014!

Filed Under: Multi-stage Events, Multi-Stage Events, Multi-Stage Events (Std 2) Tagged With: Band 6, num-title-ct-corea, num-title-qs-hsc, smc-4238-20-Independent 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