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Statistics, EXT1 S1 2024 HSC 7 MC

A driver's knowledge test contains 30 multiple-choice questions, each with 4 options. An applicant must get at least 29 correct to pass.

If an applicant correctly answers the first 25 questions and randomly guesses the last 5 questions, what is the probability that the applicant will pass the test?

  1. \(\dfrac{1}{256}\)
  2. \(\dfrac{15}{1024}\)
  3. \(\dfrac{1}{64}\)
  4. \(\dfrac{21}{256}\)
Show Answers Only

\(C\)

Show Worked Solution

\(\text{For each of the last 5 questions:}\)

\(P(C) = \dfrac{1}{4}, \ \ P(\bar{C}) = \dfrac{3}{4}\)

\(P(\text{at least 4 correct})\) \(=\ ^5C_4 \Bigg(\dfrac{1}{4}\Bigg)^{4} \Bigg(\dfrac{3}{4}\Bigg)^{1} + \ ^5C_5 \Bigg(\dfrac{1}{4}\Bigg)^{5}\Bigg(\dfrac{3}{4}\Bigg)^{0}\)  
 

\(=5 \times \dfrac{1}{256} \times \dfrac{3}{4}+1\times \dfrac{1}{1024} \times 1\)

  
  \(=\dfrac{1}{64}\)  

 
\(\Rightarrow C\)

♦ Mean mark 44%.

Statistics, EXT1 S1 2023 HSC 12c

A gym has 9 pieces of equipment: 5 treadmills and 4 rowing machines.

On average, each treadmill is used 65% of the time and each rowing machine is used 40% of the time.

  1. Find an expression for the probability that, at a particular time, exactly 3 of the 5 treadmills are in use.  (2 marks)

  2. Find an expression for the probability that, at a particular time, exactly 3 of the 5 treadmills are in use and no rowing machines are in use.  (1 mark)

Show Answers Only

  1. \(P\text{(3 of 5 treadmills in use)}\ =\ ^5C_3 (0.65)^3(0.35)^2 \)
  2. \(P\text{(3 treadmills and no rowing)}\ = \ ^5C_3 (0.65)^3(0.35)^2 \times (0.6)^4 \)

Show Worked Solution

i.    \(P(T)=0.65, \ \ P(\overline{T})=1-0.65=0.35 \)

\(P\text{(3 of 5 treadmills in use)}\ =\ ^5C_3 (0.65)^3(0.35)^2 \)
 

ii.  \(P(R)=0.4, \ \ P(\overline{R})=1-0.4=0.6 \)

\(P\text{(no rowing machines in use)}\ =\ ^4C_0 (0.6)^4(0.4)^0=(0.6)^4 \)

\(\text{Since 2 events are independent:} \)

\(P\text{(3 treadmills and no rowing)}\ = \ ^5C_3 (0.65)^3(0.35)^2 \times (0.6)^4 \)

Statistics, EXT1 S1 2016 MET1 4

A paddock contains 10 tagged sheep and 20 untagged sheep. Four times each day, one sheep is selected at random from the paddock, placed in an observation area and studied, and then returned to the paddock.

  1. What is the probability that the number of tagged sheep selected on a given day is zero?  (1 mark)

  2. What is the probability that at least one tagged sheep is selected on a given day?  (1 mark)

  3. What is the probability that no tagged sheep are selected on each of six consecutive days?

     

    Express your answer in the form `(a/c)^c`, where `a`, `b` and `c` are positive integers.  (1 mark)

Show Answers Only
  1. `16/81`
  2. `65/81`
  3. `(2/3)^24`
Show Worked Solution

a.   `text(Let)\ \ X =\ text(Number of tagged sheep,)`

`X ~\ text(Bin)(n,p)\ ~\ text(Bin)(4,1/3)`

`P(X = 0)` `= ((4),(0)) xx (1/3)^0 xx (2/3)^4`
  `= 16/81`

 

b.    `P(X >= 1)` `= 1 – P(X = 0)`
    `= 1 – 16/81`
    `= 65/81`

 

c.   `text(Let)\ \ Y =\ text(Days that no tagged sheep selected,)`

`Y ~\ text(Bin)(6,16/81)`

`P(Y = 6)` `= ((6),(6)) xx (16/81)^6 xx (65/81)^0`
  `= (16/81)^6`
  `=(2/3)^24`

Statistics, EXT1 S1 2005 HSC 6a

There are five matches on each weekend of a football season. Megan takes part in a competition in which she earns one point if she picks more than half of the winning teams for a weekend, and zero points otherwise. The probability that Megan correctly picks the team that wins any given match is `2/3`.

  1. Show that the probability that Megan earns one point for a given weekend is  0.7901, correct to four decimal places.  (2 marks)

  2. Hence find the probability that Megan earns one point every week of the eighteen-week season. Give your answer correct to two decimal places.  (1 mark)

  3. Find the probability that Megan earns at most 16 points during the eighteen-week season. Give your answer correct to two decimal places.  (2 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `0.01\ \ text{(to 2 d.p.)}`
  3. `0.92\ \ text{(to 2 d.p.)}`
Show Worked Solution

i.  `P\ text{(earns a point)}`

`= P\ text{(picks 3, 4 or 5 winners)}`

`=\ ^5 C_3  (2/3)^3 * (1/3)^2 + \ ^5 C_4  (2/3)^4 * (1/3)^1`

`+\ ^5 C_5 (2/3)^5*(1/3)^0`

`= 10 * 8/27 * 1/9 + 5 * 16/81 * 1/3 + 1 * 32/243*1`

`= 80/243 + 80/243 + 32/243`

`= 192/243`

`= 0.790123…`

`= 0.7901\ \ text{(to 4 d.p.)  …  as required.}`

 

ii.  `P\ text{(earns a point 18 weeks in a row)}`

`= (0.7901…)^18`

`= 0.01440…`

`= 0.01\ \ text{(to 2 d.p.)}`

 

iii.  `P\ text{(earns at most 16 points)}`

`= 1 – P\ text{(earns 17 or 18 points)}`
 

`P\ text{(earns 17)}`

`=\ ^18 C_17 * (0.7901…)^17 xx (1 – 0.7901…)`

`= 0.0688…`

`P\ text{(earns 18)} = 0.01440…\ \ \ \ \ text{(from (ii))}`
 

`:.\ P\ text{(earns at most 16 points)}`

`= 1 – (0.0688… + 0.0144…)`

`= 0.916…`

`= 0.92\ \ \ text{(to 2 d.p.)}`

Statistics, EXT1 S1 2012 HSC 12c

Kim and Mel play a simple game using a spinner marked with the numbers  1, 2, 3, 4 and 5.
 

2012 12c
 

The game consists of each player spinning the spinner once. Each of the five numbers is equally likely to occur.

The player who obtains the higher number wins the game.

If both players obtain the same number, the result is a draw.

  1. Kim and Mel play one game. What is the probability that Kim wins the game?   (1 mark)

  2. Kim and Mel play six games. What is the probability that Kim wins exactly three games?    (2 marks)

Show Answers Only
  1. `2/5`
  2. `864/3125`
Show Worked Solution

i.  `text(Method 1)`

`P text{(Kim wins game)}`

`= P text{(K spins 5)} xx P text{(M<5)} + P text{(K spins 4)} xx P text{(M<4)} +\ …`

`= (1/5 xx 4/5) + (1/5 xx 3/5) + (1/5 xx 2/5) + (1/5 xx 1/5)`

`= 10/25`

`= 2/5`

 

♦ Mean mark 37%

`text(Method 2)`

`P text{(Draw)} = 1/5`

`:.\ P text{(Not a draw)}= 1\ – 1/5=4/5`

 
`text(S)text(ince Kim and Mel have equal chance)`

`P text{(K wins)}` `= 1/2 xx 4/5`
  `= 2/5`

 

ii.  `P text{(Kim wins)} = 2/5`

`P text{(Kim doesn’t win)} = 3/5`
 

`text(After 6 games,)`

`P text{(Kim wins exactly 3)}`

`=\ ^6C_3 (2/5)^3 (3/5)^3`

`= (6!)/(3!3!) xx 8/125 xx 27/125`

`= 864/3125`