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Probability, SMB-015

On a tray there are 12 hard‑centred chocolates `(H)` and 8 soft‑centred chocolates `(S)`. Two chocolates are selected at random. A partially completed probability tree is shown.
 


 

What is the probability of selecting at least one soft-centred chocolate?  (3 marks)

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`62/95`

Show Worked Solution

`P(text{at least one}\ S)`

`= 1-P(HH)`

`= 1-(12/20 xx 11/19)`

`= 1-33/95`

`= 62/95`

♦ Mean mark 45%.

Filed Under: Multi-Stage Events Tagged With: num-title-ct-pathb, smc-4238-10-Dependent events, smc-4238-50-Probability trees, smc-4238-70-Complementary events, smc-4238-80-"at least"

Probability, SMB-014

A game consists of two tokens being drawn at random from a barrel containing 20 tokens. There are 17 red tokens and 3 black tokens. The player keeps the two tokens drawn.

  1.  Complete the probability tree by writing the missing probabilities in the boxes.  (2 marks)
     
     
       

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  2.  What is the probability that a player draws at least one red token? Give your answer in exact form.  (2 marks)

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  1.  
  2. `187/190`
Show Worked Solution
i.   

 

ii.   `P(text(at least one red))`

`= 1-P(BB)`

`= 1-3/20 xx 2/19`

`= 187/190`

Filed Under: Multi-Stage Events Tagged With: num-title-ct-pathb, smc-4238-10-Dependent events, smc-4238-50-Probability trees, smc-4238-70-Complementary events, smc-4238-80-"at least"

Probability, SMB-012

Two dice are rolled. What is the probability that only one of the dice shows a three?   (2 marks)

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`5/18`

Show Worked Solution

`text{Each dice roll is an independent event.}`

`P(3) = 1/6, \ P\text{(not 3)} = 1-1/6=5/6`

`text{P (Only one 3)}`

`= P text{(3, not 3)} + P text{(not 3, 3)}`

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

`= 10/36`

`= 5/18`

Filed Under: Multi-Stage Events Tagged With: num-title-ct-corea, smc-4238-20-Independent events, smc-4238-70-Complementary events

Probability, SMB-010

Each time she throws a dart, the probability that Gaga hits the dartboard is  `4/7`.

She throws two darts, one after the other.

What is the probability that she misses the dartboard with both darts?   (2 marks)

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`9/49`

Show Worked Solution

`P text{(hits)} = 4/7\ \ =>\ \ P\text{(misses)} = 1-4/7=3/7`

`P text{(misses twice)}` `= 3/7 xx 3/7`
  `= 9/49`

Filed Under: Multi-Stage Events Tagged With: num-title-ct-corea, smc-4238-20-Independent events, smc-4238-70-Complementary events

Probability, STD2 S2 2005 HSC 23a

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.
 

  1. Justine claims that each of her friends is equally likely to win first prize.

     

    Give a reason why Justine’s statement is NOT correct.   (1 mark)

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  2. What is the probability that first prize is NOT won by Khalid or Herman?   (2 marks)

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  1. `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.)`
  2. `69/100`
Show Worked Solution

i.    `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.)`

 

ii.  `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`

Filed Under: Combinations and Single Stage Events (Std 2), Fundamental understanding, Fundamental Understanding (Std 1), Fundamental Understanding (Std 2), Multi-Stage Events, Single and Multi-Stage Events (Std 1), Single stage events Tagged With: Band 3, Band 4, num-title-ct-corea, num-title-qs-hsc, smc-1135-05-Simple Probability, smc-4238-70-Complementary events, smc-828-10-Simple Probability

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)

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  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)

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  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)

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