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Probability, MET1 2020 VCAA 2

A car manufacturer is reviewing the performance of its car model X. It is known that at any given six-month service, the probability of model X requiring an oil change is `17/20`, the probability of model X requiring an air filter change is `3/20` and the probability of model X requiring both is `1/20`.

  1. State the probability that at any given six-month service model X will require an air filter change without an oil change.  (1 mark)

  2. The car manufacturer is developing a new model. The production goals are that the probability of model Y requiring an oil change at any given six-month service will be `m/(m + n)`, the probability of model Y requiring an air filter change will be `n/(m + n)` and the probability of model Y requiring both will be `1/(m + n)`, where `m, n ∈ Z^+`.
  3. Determine `m` in terms of `n` if the probability of model Y requiring an air filter change without an oil change at any given six-month service is 0.05.  (2 marks)

Show Answers Only

  1. `1/10`
  2. `m = 19n – 20`

Show Worked Solution

a.   
  `text(Pr)(F ∩ O′)` `= text(Pr)(F) – text(Pr)(F∩ O)`
    `= 3/20 – 1/20`
    `= 1/10`

 

 

b.   
`text(Pr)(F ∩ O′)` `= n/(m + n) – 1/(m + n)`
`1/20` `= (n – 1)/(m + n)`
`m + n` `= 20n – 20`
`m` `= 19n – 20`

Probability, MET1 2017 VCAA 8

For events `A` and `B` from a sample space, `text(Pr)(A text(|)B) = 1/5` and `text(Pr)(B text(|)A) = 1/4`.  Let  `text(Pr)(A nn B) = p`.

  1. Find  `text(Pr)(A)` in terms of `p`.  (1 mark)

  2. Find  `text(Pr)(A^{′} nn B^{′})` in terms of `p`.  (2 marks)

  3. Given that  `text(Pr)(A uu B) <= 1/5`, state the largest possible interval for `p`.  (2 marks)

Show Answers Only

  1. `text(Pr) (A) = 4p,\ \ p > 0`
  2. `1-8p`
  3. `0 < p <= 1/40`

Show Worked Solution

a.    `text(Pr)(A)` `=(text(Pr)(A nn B))/(text(Pr)(B text(|) A))`
    `=p/(1/4)`
    `=4p`

 

b.  `text(Consider the Venn diagram:)`

♦ Mean mark 40%.
MARKER’S COMMENT: The most successful answers used a Venn diagram or table.

`text(Pr)(A^{′} nn B^{′}) = 1-8p`

 

c.  `text(Given  Pr)(A uu B) = 8p`

♦ Mean mark 37%.

`=> 0 < 8p <= 1/5`

`:. 0 < p <= 1/40`

Probability, MET2 2011 VCAA 21 MC

For two events, `P` and `Q`, `text(Pr)(P ∩ Q) = text(Pr)(P′ ∩ Q)`.

`P` and `Q` will be independent events exactly when

  1. `text(Pr)(P′) = text(Pr)(Q)`
  2. `text(Pr)(P ∩ Q′) = text(Pr)(P′ ∩ Q)`
  3. `text(Pr)(P ∩ Q) = text(Pr)(P) + Pr(Q)`
  4. `text(Pr)(P ∩ Q′) = text(Pr)(P ∩ Q)`
  5. `text(Pr)(P) = 1/2`
Show Answers Only

`=> E`

Show Worked Solution

`text(Let)\ \ text(Pr)(P ∩ Q)` `= x = text(Pr)(P′ ∩ Q)`
`text(Let)\ \ text(Pr)(P ∩ Q′)` `= y`

 

`text(If)\ P, Q\ text(independent)`

♦♦♦ Mean mark 15%.
`text(Pr)(P) xx text(Pr)(Q)` `= text(Pr)(P ∩ Q)`
`(y + x)(2x)` `= x`
`:. 2(x + y)` `= 1`
`x + y` `= 1/2`
`text(Pr)(P)` `= 1/2`

`=> E`

Probability, MET1 2007 VCAA 6

Two events, `A` and `B`, from a given event space, are such that  `text(Pr)(A) = 1/5`  and  `text(Pr)(B) = 1/3`.

  1. Calculate  `text(Pr)(A^{′} ∩ B)`  when  `text(Pr)(A ∩ B) = 1/8`.  (1 mark)

  2. Calculate  `text(Pr)(A^{′} ∩ B)`  when `A` and `B` are mutually exclusive events.  (1 mark)

Show Answers Only

  1. `5/24`
  2. `1/3`

Show Worked Solution

a.   `text(Sketch Venn diagram:)`

♦♦ Mean mark (a) 31%.
MARKER’S COMMENTS: Students who drew a Venn diagram or Karnaugh map were the most successful.

met1-2007-vcaa-q6-answer3

`:. text(Pr)(A^{′} ∩ B)` `=text(Pr)(B)-text(Pr)(A ∩B)`
  `=1/3-1/8`
  `=5/24`

 

♦♦ Mean mark (b) 23%.

b.    met1-2007-vcaa-q6-answer4

`text(Mutually exclusive means)\ \ text(Pr)(A ∩ B)=0,`

`:. text(Pr)(A^{′} ∩ B) = 1/3`

Probability, MET1 2011 VCAA 8

Two events, `A` and `B`, are such that  `text(Pr) (A) = 3/5`  and  `text(Pr) (B) = 1/4.`

If `A^{′}` denotes the compliment of `A`, calculate  `text(Pr) (A^{′} nn B)` when

  1. `text(Pr) (A uu B) = 3/4`  (2 marks)

  2. `A` and `B` are mutually exclusive.  (1 mark)

Show Answers Only

  1. `text(Pr) (A^{′} nn B) = 3/20`
  2. `text(Pr) (A^{′} nn B) = 1/4`

Show Worked Solution

a.   `text(Sketch Venn Diagram)`

vcaa-2011-meth-8i

`text(Pr) (A uu B)` `= text(Pr) (A) + text(Pr) (B)-text(Pr) (A nn B)`
`3/4` `= 3/5 + 1/4-text(Pr) (A nn B)`
`text(Pr) (A nn B)` `= 1/10`

 

 `:.\ text(Pr) (A^{′} nn B) = 1/4-1/10 = 3/20`

 

b.   vcaa-2011-meth-8ii

`text(Pr) (A∩ B)=0\ \ text{(mutually exclusive)},`

`:.\ text(Pr) (A^{′} nn B) = text(Pr) (B) = 1/4`

Probability, MET1 2015 VCAA 8

For events `A` and `B` from a sample space, `text(Pr)(A | B) = 3/4`  and  `text(Pr)(B) = 1/3`.

  1. Calculate  `text(Pr)(A ∩ B)`.   (1 mark)

  2. Calculate  `text(Pr)(A^{′} ∩ B)`, where `A^{′}` denotes the complement of `A`.   (1 mark)

  3. If events `A` and `B` are independent, calculate  `text(Pr)(A ∪ B)`.   (1 mark)

Show Answers Only

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

Show Worked Solution

a.   `text(Using Conditional Probability:)`

`text(Pr)(A | B)` `= (text(Pr)(A ∩ B))/(text(Pr)(B))`
`3/4` `= (text(Pr)(A ∩ B))/(1/3)`
`:. text(Pr)(A ∩ B)` `= 1/4`

 

b.    met1-2015-vcaa-q8-answer
`text(Pr)(A^{′} ∩ B)` `= text(Pr)(B)-text(Pr)(A ∩B)`
  `= 1/3-1/4`
  `= 1/12`

 

c.   `text(If)\ A, B\ text(independent)`

♦♦ Mean mark 28%.
MARKER’S COMMENT: A lack of understanding of independent events was clearly evident.

`text(Pr)(A ∩ B)` `= text(Pr)(A) xx Pr(B)`
`1/4` `= text(Pr)(A) xx 1/3`
`:. text(Pr)(A)` `= 3/4`

 

`text(Pr)(A ∪ B)` `= text(Pr)(A) + text(Pr)(B)-text(Pr)(A ∩ B)`
  `= 3/4 + 1/3-1/4`
`:. text(Pr)(A ∪ B)` `= 5/6`