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Probability, 2ADV S1 2024 HSC 2 MC

In a group of 60 students, 38 play basketball, 35 play hockey and 5 do not play either basketball or hockey.

How many students play both basketball and hockey?

  1. 55
  2. 18
  3. 13
  4. 8
Show Answers Only

\(B\)

Show Worked Solution

\(\text{Method 1:}\)

 
\(\text{Method 2:}\)

\(n(B \cup H)\) \(=n(B) + n(H)-n(B \cap H) \)  
\(55\) \(=38+35-n(B \cap H) \)  
\(n(B \cap H) \) \(=18\)  

 
\( \Rightarrow B \)

Probability, 2ADV S1 2012 MET1 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^+`.
     
    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, 2ADV S1 2020 HSC 14

History and Geography are two of the subjects students may decide to study. For a group of 40 students, the following is known.

    • 7 students study neither History nor Geography
    • 20 students study History
    • 18 students study Geography
  1. A student is chosen at random. By a using a Venn diagram, or otherwise, find the probability that the student studies both History and Geography.  (2 marks)

  2. A students is chosen at random. Given that the student studies Geography, what is the probability that the student does NOT study History?  (1 mark)

  3. Two different students are chosen at random, one after the other. What is the probability that the first student studies History and the second student does NOT study History?  (2 marks)

Show Answers Only
  1.  
  2. `13/18`
  3. `10/39`
Show Worked Solution
a.   
`P(text(H and G))` `= 5/40`
  `= 1/8`

 

♦ Mean mark (b) 49%.
b.    `P(bartext(H) | text(G))` `= (P(bartext(H) ∩ text(G)))/(Ptext{(G)})`
    `= 13/18`

 

c.    `P(text(H), bartext(H))` `= 20/40 xx 20/39`
    `= 10/39`

Probability, 2ADV S1 SM-Bank 3

In a workplace of 25 employees, each employee speaks either French or German, or both.

If 36% of the employees speak German, and 20% speak both French and German.

  1. Calculate the probability one person chosen could speak German if they could speak French. Give your answer to the nearest percent.  (1 mark)

  2. Calculate the probability one person chosen could not speak French if they could speak German. Give your answer to the nearest percent.  (1 mark)

Show Answers Only
  1. `24text(%)`
  2. `44text(%)`
Show Worked Solution

i.   `text(Expressing in a Venn diagram:)`
 


 

`P(G|F)` `= (P(G ∩ F))/(P(F))`
  `= 0.20/0.84`
  `= 0.238…`
  `= 24text(%)`

 

ii.    `P(text(not)\ F|G)` `= (P(text(not)\ F ∩ G))/(P(G))`
    `= 0.16/0.36`
    `= 0.444…`
    `= 44text(%)`

Probability, 2ADV S1 SM-Bank 4 MC

In a classroom, students are asked what sports club they are members of and the results are shown in the Venn diagram.
 


 

A student who is a member of a soccer club is chosen at random. What is the probability that he/she is also a member of a surf club?

  1. `2/5`
  2. `4/11`
  3. `2/9`
  4. `7/18`
Show Answers Only

`D`

Show Worked Solution
`P(text(Surf | Soccer))` `= (n(text(Surf) ∩ text(Soccer)))/(n(text(Soccer)))`
  `= (3 + 4)/(3 + 4 + 5 + 6)`
  `= 7/18`

 
`=>\ D`

Probability, 2ADV S1 2007 MET1 6

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

  1. Calculate  `P(A′ ∩ B)`  when  `P(A ∩ B) = 1/8`.  (1 mark)

  2. Calculate  `P(A′ ∩ B)`  when `A` and `B` are mutually exclusive events.  (1 mark)

Show Answers Only
  1. `5/24`
  2. `1/3`
Show Worked Solution

i.   `text(Sketch Venn diagram:)`

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

met1-2007-vcaa-q6-answer3

`:.P(A′ ∩ B)` `=P(B) – P(A ∩B)`
  `=1/3 – 1/8`
  `=5/24`

 

♦♦ Mean mark 23%.
ii.    met1-2007-vcaa-q6-answer4

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

`:. P(A′ ∩ B) = 1/3`

Probability, 2ADV S1 2017 MET1 8

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

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

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

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

Show Answers Only
  1. `P (A) = 4p,\ \ p > 0`
  2. `1-8p`
  3. `0 < p <= 1/40`
Show Worked Solution
i.    `P\ (A)` `=(P\ (A nn B))/(P\ (B text(|) A))`
    `=p/(1/4)`
    `=4p`

 

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

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

`P\ (A^{′} nn B^{′}) = 1-8p`

 

iii.  `text(Given)\ P(A uu B) = 8p`

♦ Mean mark 37%.

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

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

Probability, 2ADV S1 2011 MET1 8

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

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

  1. `P (A uu B) = 3/4`  (2 marks)

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

Show Answers Only
  1. `3/20`
  2. `1/4`
Show Worked Solution

i.   `text(Sketch Venn Diagram)`

vcaa-2011-meth-8i

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

 

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

 

ii.   vcaa-2011-meth-8ii

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

`:.\ P (A^{′} nn B) = P (B) = 1/4`

Probability, 2ADV S1 2015 MET1 8

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

  1.  Calculate  `P(A ∩ B)`.  (1 mark)

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

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

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

i.   `text(Using Conditional Probability:)`

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

 

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

 

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

♦♦ Mean mark 28%.
MARKER’S COMMENT: A lack of understanding of independent events was clearly evident.
`P(A ∩ B)` `= P(A) xx P(B)`
`1/4` `= P(A) xx 1/3`
`:. P(A)` `= 3/4`

 

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

Probability, 2ADV S1 2012 MET2 13 MC

`A` and `B` are events of a sample space `S.`

`P(A nn B) = 2/5`  and  `P(A nn B^(′)) = 3/7`

`P(B^(′) | A)`  is equal to

  1. `6/35`
  2. `15/29`
  3. `14/35`
  4. `29/35`
Show Answers Only

`B`

Show Worked Solution

met2-2012-vcaa-13-mc-answer

`P(B ^(′) | A)` `= (P(B^(′) nn A))/(P(A))`
  `= (P(B^(′) nn A))/(P(B^(′) nn A)+ P(A nn B))`
  `= (3/7)/(3/7 + 2/5)`
  `= 15/29`

`=>   B`