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Probability, MET2 2023 VCAA 12 MC

The probability mass function for the discrete random variable \(X\) is shown below.

\begin{array} {|c|c|c|c|c|}
\hline X &\ \ \ \ \ \ -1\ \ \ \ \ \ &\ \ \ \ \ \ 0\ \ \ \ \ \ &\ \ \ \ \ \ 1\ \ \ \ \ \ &  2 \\
\hline \text{Pr}(X=x) & k^2 & 3k & k & -k^2-4k+1 \\
\hline \end{array}

The maximum possible value for the mean of \(X\) is:

  1. \(0\)
  2. \(\dfrac{1}{3}\)
  3. \(\dfrac{2}{3}\)
  4. \(1\)
  5. \(2\)
Show Answers Only

\(E\)

Show Worked Solution
\(E(X)\) \(  =(-1)\times k^2+1\times k+2\times(-k^2-4k+1)\)
  \(=-3k^2-7k+2\)

 

\(\text{Probabilities in table must be}\ \geqslant 0\ \ \Rightarrow \ k \geqslant 0.\)

\(E(X)_{\max}\ \text{occurs when}\ k=0\)

\(E(X)_{\max}\ =2\)

 
\(\Rightarrow E\)


♦♦ Mean mark 29%.
MARKER’S COMMENT: 26% of students incorrectly chose C.

Probability, MET2-NHT 2019 VCAA 5 MC

Consider the probability distribution for the discrete random variable `X` shown in the table below.

The value of  `text(E)(X)`  is

  1.  `(76)/(65)`
  2.  `1`
  3.  `0`
  4.  `(2)/(13)`
  5.  `(86)/(65)`
Show Answers Only

`A`

Show Worked Solution
`1` `= b + b + b + (3)/(5) – b + (3b)/(5)`
`(2)/(5)` `= (13b)/(5)`
`b` `= (2)/(13)`

 

`text(E)(X)` `= -(2)/(13) + 0 + (2)/(13) + 2((3)/(5) – (2)/(13)) +3 ((6)/(65))`
  `= (76)/(65)`

Probability, MET2 2019 VCAA 7 MC

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

  `qquad x` `qquad 0 qquad` `qquad 1 qquad` `qquad 2 qquad` `qquad 3 qquad`
  `qquad Pr(X = x) qquad` `a` `3a` `5a` `7a`

 
The mean of `X` is

  1. `1/16`
  2. `1`
  3. `35/16`
  4. `17/8`
  5. `2`
Show Answers Only

`D`

Show Worked Solution

`16a = 1 qquad => qquad a = 1/16`

`text(E)(X)` `= 3/16 xx 1 + 5/16 xx 2 + 7/16 xx 3`
  `=34/16`
  `= 17/8`

 
`=>   D`

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 2007 VCAA 19 MC

The discrete random variable `X` has probability distribution as given in the table. The mean of `X` is 5.

VCAA 2007 19mc

The values of `a` and `b` are

  1. `{:(a = 0.05, and b = 0.25):}`
  2. `{:(a = 0.1­, and b = 0.29):}`
  3. `{:(a = 0.2­, and b = 0.9):}`
  4. `{:(a = 0.3­, and b = 0):}`
  5. `{:(a = 0­­­, and b = 0.3):}`
Show Answers Only

`A`

Show Worked Solution

`text(Sum of probabilities) = 1`

`a + 0.2 + 0.2 + 0.3 + b = 1`

 

`text(S)text{ince}\ \ text(E)(X) = 5,`

`5` `=(0 xx a) + (2 xx 0.2) + (4 xx 0.2) + (6 xx 0.3) + 8b`
`8b` `=2`
`:. b` `=0.25`

 

`:. a = 0.05,\ \  b = 0.25`

`=>   A`

Probability, MET2 2010 VCAA 15 MC

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

VCAA 2010 15mc

 
If the mean of `X` is 1 then

  1. `a = 0.3 and b = 0.1`
  2. `a = 0.2 and b = 0.2`
  3. `a = 0.4 and b = 0.2`
  4. `a = 0.1 and b = 0.5`
  5. `a = 0.1 and b = 0.3`
Show Answers Only

`C`

Show Worked Solution

`text(E)(X) = 1,`

`1 xx b + 2 xx 0.4` `=1`
`b` `=0.2`

 

`text(Sum of probabilities) = 1,`

`a + 0.2 + 0.4` `= 1`
`a` `=0.4`

`=>   C`

Probability, MET2 2016 VCAA 19 MC

Consider the discrete probability distribution with random variable `X` shown in the table below.
 

 
The smallest and largest possible values of  `text(E)(X)`  are respectively

  1. `−0.8 and 1`
  2. `−0.8 and 1.6`
  3. `0 and 2.4`
  4. `0.2125 and 1`
  5. `0 and 1`
Show Answers Only

`E`

Show Worked Solution

`text(Smallest)\ text(E)(X)\ \ text(occurs when)\ \ a=0.8,`

♦♦♦ Mean mark 15%.
`:.\ text(Smallest)\ text(E)(X)` `=0.8 xx -1 + 0.2 xx 4`
  `=0`

 

`text(Consider the value of)\ b,`

`text(Sum of probabilities) = 1`

`:. 0 <= 4b <= 0.8 \ => \ 0 <= b <= 0.2`

 

`text(Largest)\ text(E)(X)\ \ text(occurs when)\ \ a=0, and b=0.2,`

`:.\ text(Largest)\ text(E)(X)`

`=0.2 xx 0 + 0.2 xx 0.2+(2xx0.2)xx(2xx0.2)+0.2 xx 4`

`=0.04 + 0.16 + 0.8`

`=1`

`=>   E`

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`

Probability, MET2 2015 VCAA 14 MC

Consider the following discrete probability distribution for the random variable `X.`
 

VCAA 2015 14mc 

 
The mean of this distribution is

  1. `2`
  2. `3`
  3. `7/2`
  4. `11/3`
  5. `4`
Show Answers Only

`D`

Show Worked Solution

`text(Find)\ p:`

`p + 2p + 3p + 4p + 5p` `= 1`
`:. p` `= 1/15`

 

`text(E)(X)` `= 1 xx p + 2(2p) + 3(3p) + 4(4p) + 5(5p)`
  `= 55p`
  `= 55 xx (1/15)`
  `= 11/3`

`=>   D`