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Probability, MET2 2022 VCAA 3

Mika is flipping a coin. The unbiased coin has a probability of \(\dfrac{1}{2}\) of landing on heads and \(\dfrac{1}{2}\) of landing on tails.

Let \(X\) be the binomial random variable representing the number of times that the coin lands on heads.

Mika flips the coin five times.

    1. Find \(\text{Pr}(X=5)\).   (1 mark)
    2. Find \(\text{Pr}(X \geq 2).\) (1 mark)
    3. Find \(\text{Pr}(X \geq 2 | X<5)\), correct to three decimal places.   (2 marks)
    4. Find the expected value and the standard deviation for \(X\).   (2 marks)

The height reached by each of Mika's coin flips is given by a continuous random variable, \(H\), with the probability density function
  

\(f(h)=\begin{cases} ah^2+bh+c         &\ \ 1.5\leq h\leq 3 \\ \\ 0       &\ \ \text{elsewhere} \\ \end{cases}\)
  

where \(h\) is the vertical height reached by the coin flip, in metres, between the coin and the floor, and \(a, b\) and \(c\) are real constants.

    1. State the value of the definite integral \(\displaystyle\int_{1.5}^3 f(h)\,dh\).   (1 mark)
    2. Given that  \(\text{Pr}(H \leq 2)=0.35\)  and  \(\text{Pr}(H \geq 2.5)=0.25\), find the values of \(a, b\) and \(c\).   (3 marks)

    3. The ceiling of Mika's room is 3 m above the floor. The minimum distance between the coin and the ceiling is a continuous random variable, \(D\), with probability density function \(g\).
    4. The function \(g\) is a transformation of the function \(f\) given by \(g(d)=f(rd+s)\), where \(d\) is the minimum distance between the coin and the ceiling, and \(r\) and \(s\) are real constants.
    5. Find the values of \(r\) and \(s\).   (1 mark)

  1. Mika's sister Bella also has a coin. On each flip, Bella's coin has a probability of \(p\) of landing on heads and \((1-p)\) of landing on tails, where \(p\) is a constant value between 0 and 1 .
  2. Bella flips her coin 25 times in order to estimate \(p\).
  3. Let \(\hat{P}\) be the random variable representing the proportion of times that Bella's coin lands on heads in her sample.
    1. Is the random variable \(\hat{P}\) discrete or continuous? Justify your answer.   (1 mark)
    2. If \(\hat{p}=0.4\), find an approximate 95% confidence interval for \(p\), correct to three decimal places.   (1 mark)
    3. Bella knows that she can decrease the width of a 95% confidence interval by using a larger sample of coin flips.
    4. If \(\hat{p}=0.4\), how many coin flips would be required to halve the width of the confidence interval found in part c.ii.?   (1 mark)

Show Answers Only

a. i.    `frac{1}{32}`    ii.    `frac{13}{16}`    iii.    `0.806` (3 d.p.)

a. iv    `text{E}(X)=5/2,  text{sd}(X)=\frac{\sqrt{5}}{2}`

b. i.   `1`   

b. ii.    `a=-frac{4}{5},  b=frac{17}{5},  c=-frac{167}{60}`

b. iii. `r=-1,  s=3`

c. i.   `text{Discrete}`   ii.   `(0.208,  0.592)`   iii.   `n=100`

Show Worked Solution

a.i  `X ~ text{Bi}(5 , frac{1}{2})`

`text{Pr}(X = 5) = (frac{1}{2})^5 = frac{1}{32}`
 

a.ii  By CAS:    `text{binomCdf}(5,0.5,2,5)`     `0.8125`

`text{Pr}(X>= 2) = 0.8125 = frac{13}{16}`

  
a.iii  `\text{Pr}(X \geq 2 | X<5)`

`=frac{\text{Pr}(2 <= X < 5)}{\text{Pr}(X < 5)} = frac{\text{Pr}(2 <= X <= 4)}{text{Pr}(X <= 4)}`

  
By CAS: `frac{text{binomCdf}(5,0.5,2,4)}{text{binomCdf}(5,0.5,0,4)}`
 

`= 0.806452 ~~ 0.806` (3 decimal places)
  

a.iv `X ~ text{Bi}(5 , frac{1}{2})`

`text{E}(X) = n xx p= 5 xx 1/2 = 5/2`

`text{sd}(X) =\sqrt{n p(1-p)}=\sqrt((5/2)(1 – 0.5)) = \sqrt{5/4} =\frac{\sqrt{5}}{2}`

  

b.i   `\int_{1.5}^3 f(h) d h = 1`


♦♦ Mean mark (b.i) 40%.
MARKER’S COMMENT: Many students did not evaluate the integral or evaluated incorrectly.

b.ii  By CAS:

`f(h):= a\·\h^2 + b\·\h +c`
 

`text{Solve}( {(\int_{1.5}^2 f(h) d h = 0.35), (\int_{2.5}^3 f(h) d h = 0.25), (\int_{1.5}^3 f(h) d h = 1):})`

`a = -0.8 = frac{-4}{5}, \ b = 3.4 = frac{17}{5},\  c = =-2.78 \dot{3} = frac{-167}{60}`


♦♦ Mean mark (b.ii) 47%.
MARKER’S COMMENT: Many students did not give exact answers.

b.iii  `h + d = 3`

`:.\  f(h) = f(3  –  d) = f(- d + 3)`

`:.\  r = – 1 ` and ` s = 3`


♦♦♦♦ Mean mark (b.iii) 10%.
MARKER’S COMMENT: Many students did not attempt this question.

c.i  `\hat{p}`  is discrete.

The number of coin flips must be zero or a positive integer so  `\hat{p}`  is countable and therefore discrete.
 

c.ii  `\left(\hat{p}-z \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}, \hat{p}+z \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\right)`

`\left(0.4-1.96 \sqrt{\frac{0.4 \times 0.6}{25}}\ , 0.4-1.96 \sqrt{\frac{0.4 \times 0.6}{25}}\right)`

`\approx(0.208\ ,0.592)`

 

c.iii To halve the width of the confidence interval, the standard deviation needs to be halved.

`:.\  \frac{1}{2} \sqrt{\frac{0.4 \times 0.6}{25}} = \sqrt{\frac{0.4 \times 0.6}{4 xx 25}} = \sqrt{\frac{0.4 \times 0.6}{100}}`

`:.\  n = 100`

She would need to flip the coin 100 times


♦♦♦ Mean mark (c.iii) 30%.
MARKER’S COMMENT: Common incorrect answers were 0, 10, 11, 50 and 101.

Probability, MET1 2023 VCAA 8

Suppose that the queuing time, \(T\) (in minutes), at a customer service desk has a probability density function given by
 

\(f(t) = \begin {cases}
kt(16-t^2)         &\ \ 0 \leq t \leq 4 \\
\\
0 &\ \ \text{elsewhere}
\end{cases}\)

 
for some  \(K \in R\).

  1. Show that  \(k=\dfrac{1}{64}\).   (1 mark)
  2. Find  \(\text{E}(T)\).   (2 marks)
  3. What is the probability that a person has to queue for more than two minutes, given that they have already queued for one minute?   (3 marks)
Show Answers Only
a.    \(\displaystyle \int_{0}^{4} (16-t^2)\,dt\) \(=1\)
  \(k\left[8t^2-\dfrac{t^4}{4}\right]_0^4\) \(=1\)
  \(k\Bigg(8\times16-\dfrac{16\times16}{4}\Bigg)\) \(=1\)
  \(64k\) \(=1\)
  \(\therefore\ k\) \(=\dfrac{1}{64}\)

b.    \(E(T)=\dfrac{32}{15}\)

c.    \(\dfrac{16}{25}=0.64\)

Show Worked Solution
a.    \(\displaystyle \int_{0}^{4} kt(16-t^2)\,dt\) \(=1\)
  \(k\left[8t^2-\dfrac{t^4}{4}\right]_0^4\) \(=1\)
  \(k\Bigg(8\times16-\dfrac{16\times16}{4}\Bigg)\) \(=1\)
  \(64k\) \(=1\)
  \(\therefore\ k\) \(=\dfrac{1}{64}\)

♦ Mean mark (a) 43%.
b.    \(E(T)\) \(=\dfrac{1}{64}\displaystyle \int_{0}^{4} (16t^2-t^4)\,dt\)
    \(=\dfrac{1}{64}\left[\dfrac{16t^3}{3}-\dfrac{t^5}{5}\right]_0^4\)
    \(=\dfrac{1}{64}\Bigg(\dfrac{1024}{3}-\dfrac{1024}{5}-0\Bigg)\)
    \(=\dfrac{1}{64}\times\dfrac{2048}{15}\)
    \(=\dfrac{32}{15}\)

♦♦ Mean mark (b) 38%.
c.    \(\text{Pr}(2<T<4|T>1)\) \(=\dfrac{\text{Pr}(2<T<4)}{\text{Pr}(T>1)}\)
    \(=\dfrac{\dfrac{1}{64}\displaystyle \int_{2}^{4} (16t-t^3)\,dt}{\dfrac{1}{64}\displaystyle \int_{1}^{4} (16t-t^3)\,dt}\)
    \(=\dfrac{\left[8t^2-\dfrac{t^4}{4}\right]_2^4}{\left[8t^2-\dfrac{t^4}{4}\right]_1^4}\)
    \(=\dfrac{(64-(32-4))}{\bigg(64-\bigg(8-\dfrac{1}{4}\bigg)\bigg)}\)
    \(=\dfrac{36}{\bigg(\dfrac{225}{4}\bigg)}=\dfrac{144}{225}=\dfrac{16}{25}=0.64\)

♦♦♦ Mean mark (c) 24%.
MARKER’S COMMENT: Simplifying fractions caused problems. Cancelling factors will assist with calculations.

Probability, MET1 2021 VCAA 7

A random variable  `X`  has the probability density function  `f`  given by

`f(x) = {{:(k/(x^2)),(0):}\ \ \ \ {:(1 <= x <= 2),(text{elsewhere}):}:}`

where  `k`  is a positive real number.

  1. Show that  `k = 2`.  (1 mark)

  2. Find  `E(X)`.  (2 marks)

Show Answers Only

  1. `text(See Worked Solution)`
  2. `2log_e2`

Show Worked Solution

♦ Mean mark part (a) 48%.

a.    `int_1^2 k/x^2\ dx` `=1`
  `k[- 1/x]_1^2` `=1`
  `k(-1/2+1)` `=1`
  `k/2` `=1`
  `:.k` `=2\ \ text{… as required}`

 

♦ Mean mark part (b) 44%.
MARKER’S COMMENT: Common error not recognising  `log_e1=0`

b.    `E(X)` `=int_1^2 x * 2/x^2\ dx`
    `=int_1^2 2/x\ dx`
    `=[2log_e x]_1^2`
    `=2log_e2-2log_e1`
    `=2log_e2`

Probability, MET2-NHT 2019 VCAA 10 MC

Let  `f`  be the probability density function  `f : [ 0, (2)/(3)] → R, \ f(x) = kx(2x + 1)(3x -2)(3x + 2)`.
The value of  `k`  is

  1.     `(308)/(405)`
  2.  `–(308)/(405)`
  3.  `–(405)/(308)`
  4.     `(405)/(308)`
  5.     `(960)/(133)`
Show Answers Only

`C`

Show Worked Solution

`text(Solve for)\ k :`

`k int_0^((2)/(3)) x (2x + 1)(3x – 2)(3x + 2)\ dx = 1`

`k = -(405)/(308)`
 

`=> \ C`

Probability, MET1 2010 VCAA 7

The continuous random variable `X` has a distribution with probability density function given by
 

`f(x) = {(ax(5 - x), \ text(if)\ \ 0 <= x <= 5), (0,\ text (if)\ \ x < 0\ \ text(or if)\ \ x > 5):}`
 

where `a` is a positive constant.

  1. Find the value of `a`.  (3 marks)

  2. Express  `text(Pr) (X < 3)`  as a  definite integral. (Do not evaluate the definite integral.)  (1 mark)

Show Answers Only

  1. `6/125`
  2. `int_0^3 f(x)\ dx`

Show Worked Solution

a.   `text(Total Area under curve)` `= 1`
  `a int_0^5 (5x – x^2)\ dx` `= 1`
  `a [5/2 x^2 – 1/3 x^3]_0^5` `= 1`
  `a [(125/2 – 125/3) – (0)]` `= 1`
  `125/6 a` `= 1`
  `:. a` `= 6/125`

 

b.   `text(Pr) (X < 3) = int_0^3 f(x)\ dx`

Probability, MET2 2015 VCAA 3

Mani is a fruit grower. After his oranges have been picked, they are sorted by a machine, according to size. Oranges classified as medium are sold to fruit shops and the remainder are made into orange juice.

The distribution of the diameter, in centimetres, of medium oranges is modelled by a continuous random variable, `X`, with probability density function
 

`f(x) = {(3/4(x-6)^2(8-x), 6 <= x <= 8), (\ \ \ \ \ \ \ 0, text(otherwise)):}`
 

    1. Find the probability that a randomly selected medium orange has a diameter greater than 7 cm.   (2 marks)

    2. Mani randomly selects three medium oranges.
    3. Find the probability that exactly one of the oranges has a diameter greater than 7 cm.
    4. Express the answer in the form `a/b`, where `a` and `b` are positive integers.   (2 marks)

  2. Find the mean diameter of medium oranges, in centimetres.   (1 mark)

For oranges classified as large, the quantity of juice obtained from each orange is a normally distributed random variable with a mean of 74 mL and a standard deviation of 9 mL.

  1. What is the probability, correct to three decimal places, that a randomly selected large orange produces less than 85 mL of juice, given that it produces more than 74 mL of juice?  (2 marks)

Mani also grows lemons, which are sold to a food factory. When a truckload of lemons arrives at the food factory, the manager randomly selects and weighs four lemons from the load. If one or more of these lemons is underweight, the load is rejected. Otherwise it is accepted.

It is known that 3% of Mani’s lemons are underweight.

    1. Find the probability that a particular load of lemons will be rejected. Express the answer correct to four decimal places.   (2 marks)

    2. Suppose that instead of selecting only four lemons, `n` lemons are selected at random from a particular load.
    3. Find the smallest integer value of `n` such that the probability of at least one lemon being underweight exceeds 0.5  (2 marks)

Show Answers Only

    1. `11/16`
    2. `825/4096`
  2. `36/5\ text(cm)`
  3. `0.778`
    1. `0.1147`
    2. `23`

Show Worked Solution

a.i.    `text(Pr)(X > 7)` `= int_7^8 f(x)\ dx`
    `= int_7^8 (3/4(x-6)^2(8-x))\ dx`
    `= 11/16`

 

a.ii.   `text(Let)\ \ Y =\ text(number with diameter > 7cm)`

  `Y ∼\ text(Bi)(3,11/6)`

`text(Pr)(Y = 1)` `= ((3),(1))(11/16)^1 xx (5/16)^2`
  `= 825/4096`

 

b.   `text{E(X)}` `= int_6^8 (x xx f(x))\ dx`
    `= 36/5`
    `=7.2\ text(cm)`

 

c.   `text(Let)\ \ L = text(Large juice quantity)`

`L ∼\ N(74,9^2)`

`text(Pr)(L < 85 | L > 74)` `= (text(Pr)(L < 85 ∩ L>74))/(text(Pr)(L > 74))`
  `= (text(Pr)(74 < L < 85))/(text(Pr)(L > 74))`
  `= (0.3891…)/0.5`
  `= 0.7783…` 
  `=0.778\ \ text{(3 d.p.)}`

 

d.i.  `text{Solution 1 [by CAS]}`

`text(Let)\ \ W =\ text(number of lemons under weight)`

 `W ∼\ text(Bi)(4,0.03)`

`text(Pr)(W >= 1) = 0.1147qquad[text(CAS: binomCdf)\ (4,0.03,1,4)]`

 

`text(Solution 2)`

`text(Pr)(W>=1)` `=1-text(Pr)(W=0)`
  `=1-(0.97)^4`
  `=0.11470…`
  `=0.1147\ \ text{(4 d.p.)}`

 

d.ii.  `W ∼\ text(Bi)(n,0.03)`

♦ Mean mark 44%.

`text(Pr)(W >= 1)` `> 1/2`
`1-text(Pr)(W = 0)` `> 1/2`
`1/2` `> (0.97)^n`
`n` `> 22.8`
`:. n_text(min)` `= 23`

Probability, MET2 2013 VCAA 2

FullyFit is an international company that owns and operates many fitness centres (gyms) in several countries. At every one of FullyFit’s gyms, each member agrees to have his or her fitness assessed every month by undertaking a set of exercises called `S`. There is a five-minute time limit on any attempt to complete `S` and if someone completes `S` in less than three minutes, they are considered fit.

  1. At FullyFit’s Melbourne gym, it has been found that the probability that any member will complete `S` in less than three minutes is `5/8.` This is independent of any member.
  2. In a particular week, 20 members of this gym attemp `S.`
    1. Find the probability, correct to four decimal places, that at least 10 of these 20 members will complete `S` in less than three minutes (2 marks)

    2. Given that at least 10 of these 20 members complete `S` in less than three minutes, what is the probability, correct to three decimal places, that more than 15 of them complete `S` in less than three minutes?  (3 marks)

  3. When FullyFit surveyed all its gyms throughout the world, it was found that the time taken by members to complete `S` is a continuous random variable `X`, with a probability density function `g`, as defined below.
     
    `qquad qquad qquad g(x) = {(((x-3)^3 + 64)/256 ,  1 <= x <= 3),((x + 29)/128 ,  3 < x <= 5),(0 ,  text(elsewhere)):}`

    1. Find `text(E)(X)`, correct to four decimal places.  (2 marks)

    2. In a random sample of 200 FullyFit members, how many members would be expected to take more than four minutes to complete `S?` Give your answer to the nearest integer.  (2 marks)

Show Answers Only

a.i.   `0.9153`

a.ii.   `0.086`

b.i.   `3.0458`

b.ii.  `52\ text(people)`

Show Worked Solution

a.i.   `text(Let)\ \ y =\ text(number who complete in less than 3 min)`

`Y ∼\ text(Bi)(20, 5/8)`

`text(Pr)(Y >= 10)` `= 0.91529…`
  `= 0.9153\ \ text{(4 d.p.)}`

 

a.ii.    `text(Pr)(Y> 15 | Y >= 10)` `= (text(Pr)(Y > 15))/(text(Pr)(Y >= 10))`
    `= (0.079041…)/(0.915292…)`
    `= 0.086\ \ text{(3 d.p.)}`

 

b.i.    `text(E)(X)` `=int_-oo^oo (x xx g(x))\ dx`
    `= int_1^3 x[((x-3)^3 + 64)/256]dx + int_3^5 x((x + 29)/128) dx`
    `= 3.04583…`
    `= 3.0458\ \ text{(4 d.p.)}`

 

b.ii.  `text(Let)\ \ W =\ text(number who take more than 4 min)`

♦ Mean mark 48%.

`W ∼\ text(Bi)(200, int_4^5 (x + 29)/128\ dx)`

`W ∼\ text(Bi)(200, 67/256)`

`text(E)(W)` `= np`
  `= 200 xx 67/256`
  `= 1675/32`
  `= 52.3437…`
  `= 52\ text(people)`