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Probability, 2ADV S1 2025 HSC 19

Three girls, Amara, Bala and Cassie, have nominated themselves for the local soccer team. Exactly one of the girls will be selected. The chances of their selection are in the ratio \(1:2:3\), respectively.

The probability that the team wins when:

  • Amara is selected is 0.5
  • Bala is selected is 0.4
  • Cassie is selected is 0.2.

Given that the team wins, find the probability that Amara was selected.   (3 marks)

Show Answers Only

\(P(A |W)=\dfrac{5}{19}\)

Show Worked Solution

\(P(A |W)=\dfrac{P(A \cap W)}{P(W)}\)

\(P(W)=\dfrac{1}{6} \times \dfrac{1}{2}+\dfrac{1}{3} \times \dfrac{2}{5}+\dfrac{1}{2} \times \dfrac{1}{5}=\dfrac{19}{60}\)

\(P(A \cap W)=\dfrac{1}{6} \times \dfrac{1}{2}=\dfrac{1}{12}\)

\(\therefore P(A |W)=\dfrac{1}{12} \times \dfrac{60}{19}=\dfrac{5}{19}\)

Probability, 2ADV S1 EQ-Bank 1 MC

Consider the independent events \(A\) and \(B\) where the following is true:

  • \(P(A)=p, \ P(B)=p^2\)  and
  • \(P(A)+P(B)=1\)

What represents \(P\left(A \cup B^{\prime}\right)\) ?

  1. \(p \)
  2. \(1-p^2+p^3 \)
  3. \(1+p-p^2 \)
  4. \(1-p^3\)
Show Answers Only

\(B\)

Show Worked Solution

\(P\left(A \cup B^{\prime}\right)=P(A)+P\left(B^{\prime}\right)-P\left(A \cap B^{\prime}\right)\)

\(\text{Calculate each value:}\)

\(P(A)=p \ \ \text{(given)}\)

\(P(B^{\prime})=1-P(B)=1-p^2\)

\(\text {Since}\ P(A)\ \text{and}\ P(B)\ \text {are independent, so are}\ P(A)\ \text {and}\ P\left(B^{\prime}\right)\)

\(P\left(A \cap B^{\prime}\right)=p\left(1-p^2\right)=p-p^3\)

\(\therefore P\left(A \cup B^{\prime}\right)\) \(=p+1-p^2-\left(p-p^3\right)\)
  \(=1-p^2+p^3\)

\(\Rightarrow B\)

Probability, 2ADV S1 EQ-Bank 1

Arnold has a manbag that contains three coins.

Two are fair coins where  \(P(\text{tails})=P(\text{heads})\).

The third coin is biased where  \(P(\text{tails})=\dfrac{2}{3}\).

Initially, Arnold tosses all 3 coins.

  1. Determine the probability that at least 1 coin lands on tails.   (1 mark)

Arnold then randomly selects one coin and tosses it.

  1. Show the probability it is a head = \(\dfrac{4}{9}\).   (1 mark)

  2. Given that the coin lands on a tail, what is the probability that the coin is biased?   (2 marks)

Show Answers Only

a.   \(\dfrac{11}{12}\)

b.   \(\text{See Worked Solution}\)

c.   \(\dfrac{2}{5}\)

Show Worked Solution

a.    \(\text{Biased coin} \ \ \Rightarrow \  P(H)=\dfrac{1}{3}\)

\(P(\text{at least} \ 1 \ T)\) \(=1-P(\text{All heads})\)
  \(=1-\left[\dfrac{1}{2} \times \dfrac{1}{2} \times \dfrac{1}{3}\right]\)
  \(=\dfrac{11}{12}\)

 

b.    \(\text{1 coin tossed only:}\)

\(P(H)=\dfrac{1}{3} \times \dfrac{1}{2}+\dfrac{1}{3} \times \dfrac{1}{2}+\dfrac{1}{3} \times \dfrac{1}{3}=\dfrac{4}{9}\)
  

c.
       

\(P(T)=1-\dfrac{4}{9}=\dfrac{5}{9}\)

\(P\left(T_{\text {bias}} \big | T\right)\) \(=\dfrac{P\left(T_{\text {bias}}\right) \cap P(T)}{P(T)}\)
  \(=\dfrac{\frac{1}{3} \times \frac{2}{3}}{\frac{5}{9}}\)
  \(=\dfrac{2}{5}\)

Probability, 2ADV S1 2024 MET2 9*

At a Year 12 formal, 45% of the students travelled to the event in a hired limousine, while the remaining 55% were driven to the event by a parent.

Of the students who travelled in a hired limousine, 30% had a professional photo taken.

Of the students who were driven by a parent, 60% had a professional photo taken.

Given that a student had a professional photo taken, what is the probability that the student travelled to the event in a hired limousine?   (3 marks)

Show Answers Only

\(\dfrac{9}{31}\)

Show Worked Solution

\(\text{Pr(Limo|PP)}\) \(=\dfrac{\text{Pr(Limo)}\ \cap\ \text{Pr(PP)}}{\text{Pr(PP)}}\)
  \(=\dfrac{0.45\times 0.30}{(0.45\times 0.30)+(0.55\times 0.6)}\)
  \(=\dfrac{0.135}{0.465}\)
  \(=\dfrac{9}{31}\)

Probability, 2ADV S1 2024 HSC 9 MC

A bag contains 2 red and 3 white marbles. Jovan randomly selects two marbles at the same time from this bag. The probability tree diagram shows the probabilities for each of the outcomes.
 

Given that one of the marbles that Jovan has selected is red, what is the probability that the other marble that he has selected is also red?

  1. \(\dfrac{1}{10}\)
  2. \(\dfrac{1}{7}\)
  3. \(\dfrac{1}{4}\)
  4. \(\dfrac{7}{10}\)
Show Answers Only

\(B\)

Show Worked Solution
\(P(R_2|_{\text{other is red}})\) \(=\dfrac{P(R_1 \cap R_2)}{P\text{(at least 1 red)}}\)  
  \(=\dfrac{\frac{2}{5} \times \frac{1}{4}}{1-P(WW)}\)  
  \(=\dfrac{\frac{2}{5} \times \frac{1}{4}}{1-(\frac{3}{5} \times \frac{2}{4})}\)  
  \(=\dfrac{1}{7}\)  

 
\(\Rightarrow B\)

♦♦♦ Mean mark 14%.

Probability, 2ADV S1 2023 HSC 31

Four Year 12 students want to organise a graduation party. All four students have the same probability, \(P(F)\), of being available next Friday. All four students have the same probability, \(P(S)\), of being available next Saturday.

It is given that  \(P(F)=\dfrac{3}{10}, P(S\mid F)=\dfrac{1}{3}\), and \(P(F\mid S)=\dfrac{1}{8}\).

Kim is one of the four students.

  1. Is Kim's availability next Friday independent from his availability next Saturday? Justify your answer.  (1 mark)

  2. Show that the probability that Kim is available next Saturday is \(\dfrac{4}{5}\).  (2 marks)

  3. What is the probability that at least one of the four students is NOT available next Saturday?  (2 marks)

Show Answers Only

a.    \(P(F) \neq P(F|S)\ \text{which is not the case}\ \ (\dfrac{3}{10} \neq \dfrac{1}{8}) \)

\(\therefore\ \text{Kim’s availability on Friday is not independent of Saturday}\)

b.    \(\text{See Worked Solutions} \)

c.    \(\dfrac{369}{625} \)

Show Worked Solution

a.    \(P(F) \neq P(F|S)\ \text{which is not the case}\ \ (\dfrac{3}{10} \neq \dfrac{1}{8}) \)

\(\therefore\ \text{Kim’s availability on Friday is not independent of Saturday}\)

♦♦♦ Mean mark (a) 18%.
b.     \(P(S|F) \) \(= \dfrac{P(S) \cap P(F)}{P(F)} \)
  \(\dfrac{1}{3}\) \(= \dfrac{P(S) \cap P(F)}{\frac{3}{10}} \)
  \(\dfrac{1}{10}\) \(=P(S) \cap P(F) \)
♦ Mean mark (b) 44%.
\(P(F|S)\)  \(= \dfrac{P(F) \cap P(S)}{P(S)} \)  
\(\dfrac{1}{8}\) \(=\dfrac{\frac{1}{10}}{P(S)} \)  
\(\dfrac{1}{8} \times P(S) \) \(=\dfrac{1}{10} \)  
\(P(S)\) \(=\dfrac{4}{5} \)  

 

c.     \(P\text{(at least 1 not available)}\) \(=1-P\text{(all are available)} \)
    \(=1-\left(\dfrac{4}{5}\right)^4 \)
    \(=\dfrac{369}{625} \)
♦♦ Mean mark (c) 34%.

Probability, 2ADV S1 2022 HSC 15

In a bag there are 3 six-sided dice. Two of the dice have faces marked  1, 2, 3, 4, 5, 6. The other is a special die with faces marked  1, 2, 3, 5, 5, 5.

One die is randomly selected and tossed.

  1. What is the probability that the die shows a 5?  (1 mark)

  2. Given that the die shows a 5, what is the probability that it is the special die?  (1 mark)

Show Answers Only
  1. `5/18`
  2. `3/5`
Show Worked Solution
a.    `P(5)` `=1/3 xx 1/6 + 1/3 xx 1/6 + 1/3 xx 1/2`
    `=1/18+1/18+1/6`
    `=5/18`

 

b.   `Ptext{(special die given a 5 is rolled)}`

`=(Ptext{(special die)} ∩  P(5))/(P(5))`

`=(1/3 xx 1/2)/(5/18)`

`=1/6 xx 18/5`

`=3/5`


Mean mark (a) 53%.
 
♦ Mean mark (b) 40%.

Probability, 2ADV S1 2019 MET2 11 MC

`A` and `B` are events from a sample space such that  `P(A) = p`, where  `p > 0, \ P(B\ text{|}\ A) = m`  and  `P(B\ text{|}\ A^{′}) = n`.

`A` and `B` are independent events when

  1. `m = n`
  2. `m = 1-p`
  3. `m + n = 1`
  4. `m = p`
Show Answers Only

`A`

Show Worked Solution

`text(S) text(ince)\ A and B\ text(are independent:)`

`P(B\ text{|}\ A) = P(B\ text{|}\ A^{′})`

`:. m = n`
 

`=>   A`

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 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 2014 MET1 9

Sally aims to walk her dog, Mack, most mornings. If the weather is pleasant, the probability that she will walk Mack is `3/4`, and if the weather is unpleasant, the probability that she will walk Mack is `1/3`.

Assume that pleasant weather on any morning is independent of pleasant weather on any other morning.

  1. In a particular week, the weather was pleasant on Monday morning and unpleasant on Tuesday morning.

     

     

    Find the probability that Sally walked Mack on at least one of these two mornings.  (2 marks)

  2. In the month of April, the probability of pleasant weather in the morning was `5/8`.
    1. Find the probability that on a particular morning in April, Sally walked Mack.  (2 marks)

    2. Using your answer from part b.i., or otherwise, find the probability that on a particular morning in April, the weather was pleasant, given that Sally walked Mack that morning.  (2 marks)

Show Answers Only
  1. `5/6`
  2. i. `19/32`
     
    ii. `15/19`
Show Worked Solution
a.    `Ptext{(at least 1 walk)}` `= 1 – Ptext{(no walk)}`
    `= 1 – 1/4 xx 2/3`
    `= 5/6`

 

b.i.   `text(Construct tree diagram:)`
 

met1-2014-vcaa-q9-answer1 
 

`P(PW) + P(P′W)` `= 5/8 xx 3/4 + 3/8 xx 1/3`
  `= 19/32`

 

♦ Part (b)(ii) mean mark 38%.
b.ii.    `P(P | W)` `= (P(P ∩ W))/(P(W))`
    `= (5/8 xx 3/4)/(19/32)`
    `= 15/32 xx 32/19`
    `= 15/19`

Probability, 2ADV S1 2007 MET1 11

There is a daily flight from Paradise Island to Melbourne. The probability of the flight departing on time, given that there is fine weather on the island, is 0.8, and the probability of the flight departing on time, given that the weather on the island is not fine, is 0.6.

In March the probability of a day being fine is 0.4.

Find the probability that on a particular day in March

  1. the flight from Paradise Island departs on time  (2 marks)

  2. the weather is fine on Paradise Island, given that the flight departs on time.  (2 marks)

Show Answers Only
  1. `0.68`
  2. `8/17`
Show Worked Solution
i.   

 
`P(FT) +\ P(F^{′}T)`

`= 0.4 xx 0.8 + 0.6 xx 0.6`

`= 0.32 + 0.36`

`= 0.68`

 

ii.   `text(Conditional probability:)`

♦♦ Mean mark 29%.
MARKER’S COMMENT: Students continue to struggle with conditional probability. Attention required here.
`P(F\ text(|)\ T)` `= (P(F ∩ T))/(P(T))`
  `= 0.32/0.68`

 

`:. P(F\ text(|)\ T) = 8/17`

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 2009 MET1 5

Four identical balls are numbered 1, 2, 3 and 4 and put into a box. A ball is randomly drawn from the box, and not returned to the box. A second ball is then randomly drawn from the box.

  1. What is the probability that the first ball drawn is numbered 4 and the second ball drawn is numbered 1?  (1 mark)

  2. What is the probability that the sum of the numbers on the two balls is 5?  (1 mark)

  3. Given that the sum of the numbers on the two balls is 5, what is the probability that the second ball drawn is numbered 1?  (2 marks)

Show Answers Only
  1. `1/12`
  2. `1/3`
  3. `1/4`
Show Worked Solution
i.    `P (4, 1)` `= 1/4 xx 1/3`
    `= 1/12`

 

ii.   `P (text(Sum) = 5)`

`= P (1, 4) + P (2, 3) + P (3, 2) + P (4, 1)`

`= 4 xx (1/4 xx 1/3)`

`= 1/3`

 

iii.   `text(Conditional Probability)`

♦ Mean mark 46%.

`P (2^{text(nd)} = 1\ |\ text(Sum) = 5)`

`= (P (4, 1))/(P (text(Sum) = 5))`

`= (1/12)/(1/3)`

`= 1/4`

Probability, 2ADV S1 2014 MET2 14 MC

If  `X`  is a random variable such that  `P(X > 5) = a`  and  `P(X > 8) = b`, then  `P(X < 5 | X < 8)`  is

A.   `a/b`

B.   `(a - b)/(1 - b)`

C.   `(1 - b)/(1 - a)`

D.   `(a - 1)/(b - 1)`

Show Answers Only

`D`

Show Worked Solution

`P(X < 5) | P(X < 8)`

♦ Mean mark 45%.

`=(P(X<5 ∩ X<8))/(P(X < 8))`

`= (P(X < 5))/(P(X < 8))`

`= (1 – a)/(1 – b)`

`= (a – 1)/(b – 1)`

 
`=>   D`

Probability, 2ADV S1 2013 MET2 17 MC

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

Given that  `P(A | B) = p,\ \ P(B) = p^2`  and  `P(A) = p^(1/3),\ P(B | A)`  is equal to

A.   `p^3`

B.   `p^(4/3)`

C.   `p^(7/3)`

D.   `p^(8/3)`

Show Answers Only

`D`

Show Worked Solution
`P(A | B)` `= (P(A ∩ B))/(P(B)`
`p` `= (P(A ∩ B))/(p^2)`
`:. P(A ∩ B)` `= p^3`
♦ Mean mark 49%.

 

`P(B | A)` `= (P(A ∩ B))/(P(A))`
  `= (p^3)/(p^(1/3))`
`:. P(B | A)` `= p^(8/3)`

 
`=>   D`