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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 2021 HSC 6 MC

The random variable  `X`  represents the number of successes in 10 independent Bernoulli trials. The probability of success is  `p = 0.9`  in each trial.

Let  `r = P(X ≥ 1)`.

Which of the following describes the value of `r`?

  1. `r > 0.9`
  2. `r = 0.9`
  3. `0.1 < r < 0.9`
  4. `r <= 0.1`
Show Answers Only

`A`

Show Worked Solution

`p = 0.9, \ \ barp = 0.1`

♦ Mean mark 50%.
`P(X >= 1)` `= 1 – P(X = 0)`
  `= 1 – (0.1)^10`
  `= 0.999…`

 
`:. r > 0.9`

`=> A`

Statistics, EXT1 S1 SM-Bank 3

In a chocolate factory the material for making each chocolate is sent to a machines.

The time, `X` seconds, taken to produce a chocolate by machine is a binomial distribution where it can be shown that  `P(X <= 3) = 9/32`.

A random sample of 10 chocolates is chosen. Find the probability, correct to two decimal places, that exactly 4 of these 10 chocolates took 3 or less seconds to produce.  (2 marks)

Show Answers Only

`0.18`

Show Worked Solution

`text(Let)\ \ C = \ text(number of chocolates that take less than 3 seconds)`

COMMENT: Take care as  `P(X <= 3) = 9/32`  provides the equivalent of  `p` here.

`C ∼\ text(Bin)(n, p) ∼\ text(Bin)(10, 9/32)`

`P(C = 4)`  `=((10),(4)) (9/32)^4 (23/32)^6` 
  `=0.181…`
  `=0.18\ \ \ text{(2 d.p.)}`

Statistics, EXT1 S1 SM-Bank 2

A school has a class set of 22 new laptops kept in a recharging trolley. Provided each laptop is correctly plugged into the trolley after use, its battery recharges.

On a particular day, a class of 22 students uses the laptops. All laptop batteries are fully charged at the start of the lesson. Each student uses and returns exactly one laptop. The probability that a student does not correctly plug their laptop into the trolley at the end of the lesson is 10%. The correctness of any student’s plugging-in is independent of any other student’s correctness.

Determine the probability that at least one of the laptops is not correctly plugged into the trolley at the end of the lesson. Give your answer correct to three decimal places.  (2 marks)

Show Answers Only

`0.902`

Show Worked Solution

`text(Let)\ \ X = text(number not correctly plugged)`

`X\ ~\ text(Bin) (n,p)\ ~\ text(Bin) (22, 0.1)`

`P(X>=1)` `=1 – P(X=0)`
  `=1-((n),(0)) (0.1^0)(0.9^22)`
  `=1 – 0.9^22`
  `=0.9015…`
  `=0.902\ \ text{(3 d.p.)}`

Statistics, EXT1 S1 2007 MET1 5

It is known that 50% of the customers who enter a restaurant order a cup of coffee. If four customers enter the restaurant, what is the probability that more than two of these customers order coffee? (Assume that what any customer orders is independent of what any other customer orders.)  (2 marks)

Show Answers Only

`5/16`

Show Worked Solution

`X~\ text(Bin)(n,p)\ ~\ text(Bin)(4, 1/2)`

`P(X > 2)` `= P(X = 3) + P(X = 4)`
  `= ((4),(3))(1/2)^3(1/2) + ((4),(4))(1/2)^4(1/2)^0`
  `= 4 xx 1/16 + 1 xx 1/16` 
  `=5/16`

Statistics, EXT1 S1 2018 HSC 12d

A group of 12 people sets off on a trek. The probability that a person finishes the trek within 8 hours is 0.75.

Find an expression for the probability that at least 10 people from the group complete the trek within 8 hours.  (2 marks)

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`text(See Worked Solutions)`

Show Worked Solution

`text{P(finishes within 8 hours)} = 0.75`

`text{P(not finish within 8 hours)} = 0.25`

`text(Let)\ \ X = text(number who finish below 8 hours)`
 

`:.\ text{P(at least 10 finish within 8 hours)}`

`=\ text{P(X=10) + P(X=11) + P(X=12)}`

`= ((12), (10)) (0.75)^10 (0.25)^2 + ((12), (11)) (0.75)^11 (0.25)^1 + ((12), (12)) (0.75)^12`

Statistics, EXT1 S1 2017 HSC 11g

The probability that a particular type of seedling produces red flowers is  `1/5`.

Eight of these seedlings are planted.

  1. Write an expression for the probability that exactly three of the eight seedlings produce red flowers.  (1 mark)

  2. Write an expression for the probability that none of the eight seedlings produces red flowers.  (1 mark)

  3. Write an expression for the probability that at least one of the eight seedlings produces red flowers.  (1 mark)

Show Answers Only
  1. `\ ^8C_3 · (1/5)^3 · (4/5)^5`
  2. `(4/5)^8`
  3. `1 – (4/5)^8`
Show Worked Solution

i.   `P(text{Red}) = 1/5,\ P(text{Not Red}) = 4/5`

`P(text(exactly 3 are red))`

`= \ ^8C_3 · (1/5)^3 · (4/5)^5`

 

ii.   `P(text(none are red))`

`=\ ^8C_0 * (1/5)^0 * (4/5)^8`

`= (4/5)^8`

 

iii.   `P(text(at least 1 is red))`

`= 1 – P(text(none are red))`

`= 1 – (4/5)^8`

Statistics, EXT1 S1 2007 HSC 4a

In a large city, 10% of the population has green eyes.

  1. What is the probability that two randomly chosen people both have green eyes?  (1 mark)

  2. What is the probability that exactly two of a group of 20 randomly chosen people have green eyes? Give your answer correct to three decimal places.  (1 mark)

  3. What is the probability that more than two of a group of 20 randomly chosen people have green eyes? Give your answer correct to two decimal places.  (2 marks)

Show Answers Only
  1. `0.01`
  2. `0.285\ \ \ text{(to 3 d.p.)}`
  3. `0.32\ \ \ text{(to 2 d.p.)}`
Show Worked Solution
i.       `P(text(G))` `= 0.1`
  `P(text(GG))` `= 0.1 xx 0.1`
    `= 0.01`

 

ii.  `P(text(not G)) = 1 − 0.1 = 0.9`

`:. P(text(2 out of 20 have green eyes))`

`= \ ^(20)C_2 · (0.1)^2 · (0.9)^(18)`

`= 0.2851…`

`= 0.285\ \ \ text{(to 3 d.p.)}`

 

iii. `P(text(more than 2 have green eyes))`

`= 1 − [P(0) + P(1) + P(2)]`

`= 1 − [0.9^20 + \ ^20C_1(0.1)^1(0.9)^19 + \ ^20C_2(0.1)^2(0.9)^(18)]`

`= 1 − [0.1215… + 0.2701… + 0.2851…]`

`= 1 − 0.6769…`

`= 0.3230`

`= 0.32\ \ \ text{(to 2 d.p.)}`

Statistics, EXT1 S1 2014 HSC 11b

The probability that it rains on any particular day during the 30 days of November is 0.1.

Write an expression for the probability that it rains on fewer than 3 days in November.   (2 marks)

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`\ ^30 C_2 (0.1)^2 (0.9)^28 +\ ^30C_1 (0.1) (0.9)^29 + (0.9)^30`

Show Worked Solution
`P (R)` `= 0.1`
`P (bar R)` `= 1 – 0.1 = 0.9`

 
`text(Over 30 days:)`

`P (R<3)`

`= P (R=2) + P(R=1) + P (R=0)`

  
  `=\ ^30 C_2 (0.1)^2 (0.9)^28 +\ ^30C_1 (0.1)^1 (0.9)^29`  
  `qquad  qquad +\ ^30C_0 (0.1)^0 (0.9)^30`  
  `=\ ^30C_2 (0.1)^2 (0.9)^28 +\ ^30C_1 (0.1)(0.9)^29 + (0.9)^30`  

Statistics, EXT1 S1 2009 HSC 4a

A test consists of five multiple-choice questions. Each question has four alternative answers. For each question only one of the alternative answers is correct.

Huong randomly selects an answer to each of the five questions. 

  1. What is the probability that Huong selects three correct and two incorrect answers?   (2 marks)

  2. What is the probability that Huong selects three or more correct answers?    (2 marks)

  3. What is the probability that Huong selects at least one incorrect answer?  (1 mark)

Show Answers Only
  1. `45/512`
  2. `53/512`
  3. `1023/1024`
Show Worked Solution

i.   `P text{(correct)} = 1/4`

`P text{(wrong)} = 3/4`

`P text{(3 correct, 2 wrong)}`

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

`= (5!)/(3!2!) * (1/64) * (9/16)`

`= 90/1024`

`= 45/512`
 

ii.  `P text{(3 or more correct)}`

`= P text{(3 correct)} + P text{(4 correct)} + P text{(5 correct)}`

`=\ ^5C_3 * (1/4)^3 (3/4)^2 +\ ^5C_4 * (1/4)^4 (3/4)^1 +\ ^5C_5 (1/4)^5 (3/4)^0`

`= 90/1024 + 15/1024 + 1/1024`

`= 53/512`
 

iii.  `P text{(at least 1 incorrect)}`

TIP: The use of “at least” should flag a good chance of applying `1-P text{(complement)}` to solve.

`= 1\ – P text{(0 incorrect)}`

`= 1 -\ ^5C_5 (1/4)^5 (3/4)^0`

`= 1\ – 1/1024`

`= 1023/1024`

Statistics, EXT1 S1 2013 HSC 11c

An examination has 10 multiple-choice questions, each with 4 options. In each question, only one option is correct. For each question a student chooses one option at random.

Write an expression for the probability that the student chooses the correct option for exactly 7 questions.   (2 marks)

Show Answers Only

`\ ^10C_7 (1/4)^7 (3/4)^3`

Show Worked Solution
`P text{(Correct)}` `= 1/4`
`P text{(Incorrect)}` `= 3/4`

 
`:. P text{(exactly 7 correct)} =\ ^10C_7 (1/4)^7 (3/4)^3`