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Probability, MET1 2009 VCAA 7

The random variable `X` has this probability distribution.
 

vcaa-2009-meth-7a

Find

  1. `text(Pr) (X > 1 | X <= 3)`  (2 marks)
  2. `text(Var)\ (X),` the variance of `X.`  (3 marks)
Show Answers Only
  1. `2/3`
  2. `1.2`
Show Worked Solution

a.   `text(Pr) (X > 1 | X <= 3)`

`= (text{Pr} (X = 2) + text{Pr} (X = 3))/(1 – text{Pr} (X = 4))`

`= (0.4 + 0.2)/(1 – 0.1)`

`= 0.6/0.9`

`= 2/3`

 

b.   `text(E) (X)` `= 0.1 (0) + 1 (0.2) + 2 (0.4) + 3 (0.2) + 4 (0.1)`
  `= 0 + 0.2 + 0.8 + 0.6 + 0.4`
  `= 2`

 

`text(E) (X^2)` `= 0^2 (0.1) + 1^2 (0.2) + 2^2 (0.4) + 3^2 (0.2) + 4^2 (0.1)`
  `= 0 + 0.2 + 1.6 + 1.8 + 1.6`
  `= 5.2`

 

`:.\ text(Var)\ (X)` `= text(E) (X^2) – [text(E) (X)]^2`
  `= 5.2 – (2)^2`
  `= 1.2`

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`