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Probability, 2ADV S1 2023 HSC 2 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`

Filed Under: Multi-Stage Events (Adv-2027), Multi-Stage Events (Y11) Tagged With: 2adv-std2-common, Band 3, smc-6469-40-Arrays, smc-989-40-Arrays

Probability, 2ADV S1 2018 HSC 16b

A game involves rolling two six-sided dice, followed by rolling a third six-sided die. To win the game, the number rolled on the third die must lie between the two numbers rolled previously. For example, if the first two dice show 1 and 4, the game can only be won by rolling a 2 or 3 with the third die.

  1. What is the probability that a player has no chance of winning before rolling the third die?   (2 marks)

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  2. What is the probability that a player wins the game?   (2 marks)

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Show Answers Only
  1. `4/9`
  2. `5/27`
Show Worked Solution

i.   `text(Construct a sample space of the number)`

♦ Mean mark 40%.
COMMENT: Constructing the full sample space is a critical step here..

`text(of possible winning rolls:)`
 

`text{P(no chance)}` `= text(number of pairs with no gap)/text(total possibilities)`
  `= 16/36`
  `= 4/9`

 

ii.   `text(The sample space in the table shows:)`

♦♦♦ Mean mark 7%.

`text(→ 8 combinations leave a gap for a single winning number,)`

`text(→ 6 combinations leave a gap for two winning numbers,)`

`vdots`

`:.\ text{P(winning)}` `= 1/36 [8 xx 1/6 + 6 xx 2/6 + 4 xx 3/6 + 2 xx 4/6]`
  `= 1/36 (8/6 + 12/6 + 12/6 + 8/6)`
  `= 1/36 (40/6)`
  `= 5/27`

Filed Under: 3. Probability, Multi-Stage Events (Adv-2027), Multi-Stage Events (Y11) Tagged With: Band 5, Band 6, smc-6469-40-Arrays, smc-989-40-Arrays

Probability, 2ADV S1 2004 HSC 6c

In a game, a turn involves rolling two dice, each with faces marked  0, 1, 2, 3, 4 and 5. The score for each turn is calculated by multiplying the two numbers uppermost on the dice.

  1. What is the probability of scoring zero on the first turn?  (2 marks)

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  2. What is the probability of scoring `16` or more on the first turn?  (1 mark)

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  3. What is the probability that the sum of the scores in the first two turns is less than 45?  (2 marks)

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Show Answers Only
  1. `(11)/(36)`
  2. `1/9`
  3. `1291/1296`
Show Worked Solution
MARKER’S COMMENT: Students who drew up the table for the sample space were “overwhelmingly” more successful in all parts of this question.
i.    Probability, 2UA 2004 HSC 6c

`:. P(0) = (11)/(36)`

 

ii.  `P(≥ 16)= 4/36=1/9`

 

iii.  `Ptext{(Sum} < 45) = 1 − Ptext{(Sum} ≥ 45)`

`Ptext{(Sum} ≥ 45)` `=P(20,25)+P(25,20)+P(25,25)`
  `=(2/36 xx 1/36) + (2/36 xx 1/36)+(1/36 xx 1/36)`
  `=2/1296 + 2/1296+ 1/1296`
  `=5/1296`

 

`:.Ptext{(Sum} < 45)` `= 1 − 5/1296`
  `= 1291/1296`

Filed Under: 3. Probability, Multi-Stage Events (Adv-2027), Multi-Stage Events (Y11) Tagged With: Band 4, Band 5, smc-6469-30-Complementary Probability, smc-6469-40-Arrays, smc-989-30-Complementary Probability, smc-989-40-Arrays

Probability, 2ADV S1 2007 HSC 4b

Two ordinary dice are rolled. The score is the sum of the numbers on the top faces.

  1. What is the probability that the score is 10?  (2 marks)

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  2. What is the probability that the score is not 10?  (1 mark)

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Show Answers Only
  1. `1/12`
  2. `11/12`
Show Worked Solution
i. 2UA HSC 2007 4b

`text{P (score = 10)}`

`= 3/36`

`= 1/12`

 

ii.  `text{P (score is not ten)}`

`= 1 – text{P (score is ten)}`

`= 1 – 1/12`

`= 11/12`

Filed Under: 3. Probability, Multi-Stage Events (Adv-2027), Multi-Stage Events (Y11) Tagged With: Band 3, Band 4, smc-6469-30-Complementary Probability, smc-6469-40-Arrays, smc-989-30-Complementary Probability, smc-989-40-Arrays

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