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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)

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    2. Find \(\text{Pr}(X \geq 2).\) (1 mark)

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    3. Find \(\text{Pr}(X \geq 2 | X<5)\), correct to three decimal places.   (2 marks)

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    4. Find the expected value and the standard deviation for \(X\).   (2 marks)

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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)

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    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)

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    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)

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  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)

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    2. If \(\hat{p}=0.4\), find an approximate 95% confidence interval for \(p\), correct to three decimal places.   (1 mark)

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    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)

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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.

Filed Under: Binomial, Probability density functions Tagged With: Band 2, Band 4, Band 5, Band 6, smc-637-35-Sum probabilities = 1, smc-637-60-Polynomial PDF, smc-638-20-binomCdf (CAS), smc-638-32-Find E(X)/var(X) given n p, smc-638-50-Conditional, smc-719-20-95% confidence intervals, smc-719-40-Find sample size

Probability, MET2 2022 VCAA 18 MC

If `X` is a binomial random variable where `n=20, p=0.88` and `\text{Pr}(X \geq 16|X\geq a)=0.9175`, correct to four decimal places, then `a` is equal to

  1. 11
  2. 12
  3. 13
  4. 14
  5. 15
Show Answers Only

`B`

Show Worked Solution

`X \~ \text{Bi}(20,0.88)`

`\text{Pr}(X \geq 16|X\geq a)=0.9175`

`\frac{\text{Pr}(X \geq 16 \cap X \geq a)}{\text{Pr}(X \geq a)} \approx 0.9175`
 

All options are `< 16`

`:.\ \frac{\text{Pr}(X \geq 16)}{\text{Pr}(X \geq a)} \approx 0.9175`
 

Testing options using CAS

Option A: `\ \frac{\text{Pr}(X \geq 16)}{\text{Pr}(X \geq 11)} \approx 0.9173`

Option B: `\ \frac{\text{Pr}(X \geq 16)}{\text{Pr}(X \geq 12)} \approx 0.9175`

Option C: `\ \frac{\text{Pr}(X \geq 16)}{\text{Pr}(X \geq 13)} \approx 0.9186`

Option D: `\ \frac{\text{Pr}(X \geq 16)}{\text{Pr}(X \geq 14)} \approx 0.9235`

Option E: `\ \frac{\text{Pr}(X \geq 16)}{\text{Pr}(X \geq 15)} \approx 0.9418`
  

`=>B`


♦♦ Mean mark 47%.

Filed Under: Binomial Tagged With: Band 5, smc-638-20-binomCdf (CAS), smc-638-50-Conditional

Probability, MET1 2022 VCAA 4

A card is drawn from a deck of red and blue cards. After verifying the colour, the card is replaced in the deck. This is performed four times.

Each card has a probability of `\frac{1}{2}` of being red and a probability of `\frac{1}{2}` of being blue.

The colour of any drawn card is independent of the colour of any other drawn card.

Let `X` be a random variable describing the number of blue cards drawn from the deck, in any order.

  1. Complete the table below by giving the probability of each outcome.   (2 marks)
     

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  1. Given that the first card drawn is blue, find the probability that exactly two of the next three cards drawn will be red.  (1 mark)

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  2. The deck is changed so that the probability of a card being red is `\frac{2}{3}` and the probability of a card being blue is `\frac{1}{3}`.
  3. Given that the first card drawn is blue, find the probability that exactly two of the next three cards drawn will be red.   (2 marks)

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Show Answers Only

a.   Complete table

\begin{array} {|c|c|c|c|c|c|}
\hline x & 0 & 1 & 2 & 3 & 4 \\
\hline {Pr}(X=x) & 1/16 & 4/16 & 6/16 & 4/16 & 1/16 \\
\hline \end{array}

b.    `3/8`

c.    `4/9`

Show Worked Solution

a.   Using binomial distribution where `n=4` and  `p= 1/2`

`text{Pr}(X=1)` `=\ ^4 C_1*(1/2)^1*(1/2)^3 = 4*1/2*1/8` `=4/16`
`text{Pr}(X=3)` `=\ ^4 C_3*(1/2)^3*(1/2)^1= 4*1/8*1/2` `=4/16`
`text{Pr}(X=4)` `=\ ^4 C_4*(1/2)^4*(1/2)^0= 1*1/16*1` `=1/16`

\begin{array} {|c|c|c|c|c|c|}
\hline x & 0 & 1 & 2 & 3 & 4 \\
\hline {Pr}(X=x) & 1/16 & 4/16 & 6/16 & 4/16 & 1/16 \\
\hline \end{array}

b.   Trials are independent of first trial.

Binomial where `n=3, p=1/2`

`text{Pr}(X=2) = \ ^3 C_2*(1/2)^2*(1/2)^1 = 3*1/8 = 3/8`


♦♦ Mean mark (b) 40%.

c.   Trials are independent.

Binomial where `n=3, p=2/3`

`text{Pr}(X=2) = \ ^3 C_2*(2/3)^2*(1/3)^1 = 3*4/9*1/3 = 4/9`


♦ Mean mark 55%.
MARKER’S COMMENT: Many students wrongly treated this question as conditional probability.

Filed Under: Binomial Tagged With: Band 4, Band 5, smc-638-10-binomial expansion (non-calc), smc-638-50-Conditional

Probability, MET2 2021 VCAA 15 MC

Four fair coins are tossed at the same time.

The outcome for each coin is independent of the outcome for any other coin.

The probability that there is an equal number of heads and tails, given that there is at least one head; is

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

`D`

Show Worked Solution

`X\ ~\ text(Bi) (4, 1/2)`

♦ Mean mark 48%.

`text(By CAS):`

`text(Pr) (X = 2 | X ≥ 1)` `= {text(Pr) (X = 2)}/{text(Pr) (X ≥ 1)}`
  `= 0.375/0.9375`
  `= 2/5`

`=> D`

Filed Under: Binomial, Conditional Probability and Set Notation Tagged With: Band 5, smc-2736-10-Conditional probability, smc-638-50-Conditional

Probability, MET1 2020 VCAA 5

For a certain population the probability of a person being born with the specific gene SPGE1 is `3/5`.

The probability of a person having this gene is independent of any other person in the population having this gene.

  1. In a randomly selected group of four people, what is the probability that three or more people have the SPGE1 gene?  (2 marks)

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  2. In a randomly selected group of four people, what is the probability that exactly two people have the SPGE1 gene, given that at least one of those people has the SPGE1 gene? Express your answer in the form  `(a^3)/(b^4 - c^4)`, where, `a, b, c ∈ Z^+`.  (2 marks)

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Show Answers Only

  1. `297/625`
  2. `(6^3)/(5^4 – 2^4)`

Show Worked Solution

a.   `text(Let)\ \ X =\ text(number of people with gene)`

`X\ ~\ text(Bi) (4, 3/5)`

`P(X >= 3)` `= P(X = 3) + P(X = 4)`
  `= \ ^4C_3(3/5)^3(2/5) + \ ^4C_4(3/5)^4`
  `= (4 xx 27 xx 2)/625 + 81/625`
  `= 297/625`

 

b.    `P(X = 2 | X >= 1)` `= (P(X = 2))/(1 – P(X = 0))`
    `= (\ ^4C_2(3/5)^2(2/5)^2)/(1 – (2/5)^4)`
    `= (6 xx (6^2)/(5^4))/((5^4 – 2^4)/(5^4))`
    `= (6^3)/(5^4 – 2^4)`

Filed Under: Binomial Tagged With: Band 4, smc-638-10-binomial expansion (non-calc), smc-638-50-Conditional

Probability, MET2 2019 VCAA 8 MC

An archer can successfully hit a target with a probability of 0.9. The archer attempts to hit the target 80 times. The outcome of each attempt is independent of any other attempt.

Given that the archer successfully hits the target at least 70 times, the probability that the archer successfully hits the target exactly 74 times, correct to four decimal places, is

A.   0.3635

B.   0.8266

C.   0.1494

D.   0.3005

E.   0.1701

Show Answers Only

`C`

Show Worked Solution

`text(Pr)(H) = 0.9, \ text(Pr)(H prime) = 0.1`

`text(Let)\ \ X = text(number of times archer hits)`

`X\ ~\ text(Bi)(80, 0.9)`

`text(Pr)(X = 74|X >= 70)` `= (text(Pr)(X = 74))/(text(Pr)(X >= 70))`
  `~~ 0.1494`

 
`=>   C`

Filed Under: Binomial Tagged With: Band 4, smc-638-50-Conditional

Probability, MET2 2017 VCAA 3

The time Jennifer spends on her homework each day varies, but she does some homework every day.

The continuous random variable `T`, which models the time, `t,` in minutes, that Jennifer spends each day on her homework, has a probability density function `f`, where

 

`f(t) = {{:(1/625 (t - 20)),(1/625 (70 - t)),(0):}qquad{:(20 <= t < 45),(45 <= t <= 70),(text(elsewhere)):}:}`

 

  1. Sketch the graph of `f` on the axes provided below.  (3 marks)

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  2. Find  `text(Pr)(25 ≤ T ≤ 55)`.  (2 marks)

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  3. Find  `text(Pr)(T ≤ 25 | T ≤ 55)`.  (2 marks)

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  4. Find `a` such that  `text(Pr)(T ≥ a) = 0.7`, correct to four decimal places.  (2 marks)

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  5. The probability that Jennifer spends more than 50 minutes on her homework on any given day is `8/25`. Assume that the amount of time spent on her homework on any day is independent of the time spent on her homework on any other day.

     

    1. Find the probability that Jennifer spends more than 50 minutes on her homework on more than three of seven randomly chosen days, correct to four decimal places.  (2 marks)

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    2. Find the probability that Jennifer spends more than 50 minutes on her homework on at least two of seven randomly chosen days, given that she spends more than 50 minutes on her homework on at least one of those days, correct to four decimal places.  (2 marks)

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Let `p` be the probability that on any given day Jennifer spends more than `d` minutes on her homework.

Let `q` be the probability that on two or three days out of seven randomly chosen days she spends more than `d` minutes on her homework.

  1. Express `q` as a polynomial in terms of `p`.  (2 marks)

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    1. Find the maximum value of `q`, correct to four decimal places, and the value of `p` for which this maximum occurs, correct to four decimal places.  (2 marks)

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    2. Find the value of `d` for which the maximum found in part g.i. occurs, correct to the nearest minute.  (2 marks)

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Show Answers Only

  1.  

  2. `4/5`
  3. `1/41`
  4. `39.3649`
    1. `0.1534`
    2. `0.7626`
  5. `q =7p^2(1-p)^4(2p+3)`
    1. `p = 0.3539quadtext(and)quadq = 0.5665`
    2. `49\ text(min)`

Show Worked Solution

a.   

MARKER’S COMMENT: Many did not draw graph along `t`-axis between 0 and 20 and for  `t>70`.

 

b.   `text(Pr)(25 <= T <= 55)`

`= int_25^45 1/625 (t – 20)\ dt + int_45^55 1/625 (70 – t)\ dt`

`= 4/5`

 

c.   `text(Pr)(T ≤ 25 | T ≤ 55)`

`=(text(Pr)(T <= 25))/(text(Pr)( T <= 55))`

`= (int_20^25 1/625(t – 20)\ dt)/(1 – int_55^70 1/625(70 – t)\ dt)`

`= (1/50)/(1 – 9/50)`

`= 1/41`

 

d.   `text(Pr)(T ≥ a) = 0.7`

♦ Mean mark part (d) 36%.

`=>\ text(Pr)(T <= a) = 0.3`

`text(Solve:)`

`int_20^a 1/625(t – 20)\ dt` `= 0.3quadtext(for)quada ∈ (20, 45)`

 

`:. a == 39.3649`

 

e.i.   `text(Let)\ X =\ text(Number of days Jenn studies more than 50 min)`

`X ~\ text(Bi) (7, 8/25)`

`text(Pr)(X >= 4) = 0.1534`

 

e.ii.    `text(Pr)(X >= 2 | X >= 1)` `= (text(Pr)(X >= 2))/(text(Pr)(X >= 1))`
    `= (0.7113…)/(0.9327…)`
    `= 0.7626\ \ text{(to 4 d.p.)}`

 

f.   `text(Let)\ Y =\ text(Number of days Jenn spends more than)\ d\ text(min)`

`Y ~\ text(Bi)(7,p)`

♦ Mean mark part (f) 36%.

`q` `= text(Pr)(Y = 2) + text(Pr)(Y = 3)`
  `= ((7),(2))p^2(1 – p)^5 + ((7),(3))p^3(1 – p)^4`
  `= 21p^2(1 – p)^5 + 35p^3(1 – p)^4`
   `=7p^2(1-p)^4[3(1-p)+5p]`
  `=7p^2(1-p)^4(2p+3)`

 

g.i.   `text(Solve)\ \ q′(p) = 0,`

♦♦ Mean mark part (g)(i) 30%.

`p` `=0.35388…`
  `=0.3539\ \ text{(to 4 d.p.)}`

`:. q_text(max)= 0.5665\ \ text{(to 4 d.p.)}`

 

g.ii.   `text(Pr)(T > d) = p= 0.35388…`

♦♦♦ Mean mark part (g)(ii) 8%.

  `text(Solve:)`

`int_d^70 (1/625(70 – t))dt` `= 0.35388… quadtext(for)quadd ∈ (45,70)`

 

`:. d` `=48.967…`
  `=49\ text(mins)`

Filed Under: Binomial, Probability density functions Tagged With: Band 4, Band 5, Band 6, smc-637-40-Conditional probability, smc-637-45-Other probability, smc-637-50-Linear PDF, smc-638-20-binomCdf (CAS), smc-638-50-Conditional

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)

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    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)

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  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)

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    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)

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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)`

Filed Under: Binomial, Probability density functions Tagged With: Band 4, Band 5, smc-637-10-E(X), smc-637-60-Polynomial PDF, smc-638-20-binomCdf (CAS), smc-638-50-Conditional

Probability, MET2 2013 VCAA 9 MC

Harry is a soccer player who practises penalty kicks many times each day.

Each time Harry takes a penalty kick, the probability that he scores a goal is 0.7, independent of any other penalty kick.

One day Harry took 20 penalty kicks.

Given that he scored at least 12 goals, the probability that Harry scored exactly 15 goals is closest to

A.   `0.1789`

B.   `0.8867`

C.   `0.8`

D.   `0.6396`

E.   `0.2017`

Show Answers Only

`E`

Show Worked Solution

`text(Let)\ \ X =\ text(number of goals,)`

`X ∼\ text(Bi)(20, 0.7)`

`text(Pr) (X = 15 | X >= 12)`

`= (text(Pr) (X = 15))/(text(Pr) (X >= 12))`

`~~ 0.2017`

`=>   E`

Filed Under: Binomial Tagged With: Band 4, smc-638-50-Conditional

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