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Probability, MET2 2018 VCAA 12 MC

The discrete random variable `X` has the following probability distribution.
 

Let `mu` be the mean of `X`.

`text(Pr) (X < mu)`  is

  1. `1/2`
  2. `1/4`
  3. `17/20`
  4. `4/5`
  5. `7/10`
Show Answers Only

`E`

Show Worked Solution
`mu` `= (1 xx 9/20) + (2 xx 1/10) + (3 xx 1/20) + (6 xx 3/20)`
  `= 17/10`

 

`text(Pr) (X < 17/10)` `= text(Pr) (X = 0) + text(Pr) (X = 1)`
  `= 7/10`

`=>   E`

Probability, MET2 2016 VCAA 7 MC

The number of pets, `X`, owned by each student in a large school is a random variable with the following discrete probability distribution.
 

  

 
If two students are selected at random, the probability that they own the same number of pets is

  1. `0.3`
  2. `0.305`
  3. `0.355`
  4. `0.405`
  5. `0.8`
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`C`

Show Worked Solution
`text(Pr) (0, 0)` `+ text(Pr) (1, 1)` `+ text(Pr) (2, 2)` `+ text(Pr) (3, 3)`
`= .5^2 ` `+ qquad .25^2` `+ qquad .2^2` `+ qquad .05^2`
`= .355`      

 
`=>   C`

Probability, MET1 2008 VCAA 7

Jane drives to work each morning and passes through three intersections with traffic lights. The number `X` of traffic lights that are red when Jane is driving to work is a random variable with probability distribution given by
 

met1-2008-vcaa-q7
 

  1. What is the mode of `X`?  (1 mark)
  2. Jane drives to work on two consecutive days. What is the probability that the number of traffic lights that are red is the same on both days?  (2 marks)
Show Answers Only
  1. `3`
  2. `0.3`
Show Worked Solution

a.   `3`
 

b.   `text(Pr)(0,0) + text(Pr)(1,1) + text(Pr)(2,2) + text(Pr)(3,3)`

`= 0.1^2 + 0.2^2 + 0.3^2 + 0.4^2`

`= 0.3`

Probability, MET1 2012 VCAA 4

On any given day, the number `X` of telephone calls that Daniel receives is a random variable with probability distribution given by
 

vcaa-2012-meth-4
 

  1. Find the mean of `X`.(2 marks)
  2. What is the probability that Daniel receives only one telephone call on each of three consecutive days?  (1 mark)
  3. Daniel receives telephone calls on both Monday and Tuesday.

     

    What is the probability that Daniel receives a total of four calls over these two days?  (3 marks)

Show Answers Only
  1. `1.5`
  2. `0.008`
  3. `29/64`
Show Worked Solution
a.    `text(E) (X)` `= 0 xx 0.2 + 1 xx 0.2 + 2 xx 0.5 + 3 xx 0.1`
    `= 0 + .2 + 1 + 0.3`
    `= 1.5`

 

b.   `text(Pr) (1, 1, 1)`

MARKER’S COMMENT: Many students understood that the calculation of 0.2³ was required but were not able evaluate correctly.

`= 0.2 xx 0.2 xx 0.2`

`= 0.008`

 

c.   `text(Conditional Probability:)`

♦ Mean mark 36%.

`text(Pr) (x = 4 | x >= 1\ text{both days})`

`= (text{Pr} (1, 3) + text{Pr} (2, 2) + text{Pr} (3, 1))/(text{Pr}(x>=1\ text{both days}))`

`= (0.2 xx 0.1 + 0.5 xx 0.5 + 0.1 xx 0.2)/(0.8 xx 0.8)`

`= (0.02 + 0.25 + 0.02)/0.64`

`= 0.29/0.64`

`= 29/64`

Probability, MET1 2013 VCAA 7

The probability distribution of a discrete random variable, `X`, is given by the table below.
 

vcaa-2013-meth-7
 

  1. Show that  `p = 2/3 or p = 1`.  (3 marks)
  2. Let  `p = 2/3`.

    1. Calculate  `text(E) (X)`.  (2 marks)
    2. Find  `text(Pr) (X >= text(E) (X))`.  (1 mark)
Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
    1. `28/15`
    2. `8/15`
Show Worked Solution

a.   `text(S)text(ince probabilities must sum to 1:)`

`0.2 + 0.6p^2 + 0.1 + 1 – p + 0.1` `= 1`
`0.6p^2 – p + 0.4` `= 0`
`6p^2 – 10p + 4` `= 0`
`3p^2 – 5p + 2` `= 0`
`(p – 1) (3p – 2)` `= 0`

`:. p = 1 or p = 2/3`

 

b.i.   `text(E) (X)` `= sum x text(Pr) (X = x)`
    `= 1 xx (3/5 xx 2^2/3^2) + 2 (1/10) + 3 (1-2/3) + 4 (1/10)`
    `= 4/15 + 1/5 + 1 + 2/5`
    `= 28/15`

 

♦♦ Part (b)(ii) mean mark 32%.
  ii.   `text(Pr) (X >= 28/15)` `= text(Pr) (X = 2) + text(Pr) (X = 3) + text(Pr) (X = 4)`
    `= 1/10 + 1/3 + 1/10`
    `= 8/15`