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Probability, MET1 2025 VCAA 4

The probability distribution for the discrete random variable \(X\) is given in the table below, where \(k\) is a positive real number.

\begin{array}{|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}x \rule[-1ex]{0pt}{0pt}& \quad \quad 0 \quad \quad & \quad \quad 1 \quad \quad & \quad \quad 2 \quad \quad & \quad \quad 3 \quad \quad \\
\hline
\rule{0pt}{2.5ex}\operatorname{Pr}(X=x) \rule[-1ex]{0pt}{0pt}& \dfrac{4}{k} &\dfrac{2 k}{75} &\dfrac{k}{75} & \dfrac{2}{k} \\
\hline
\end{array}

  1. Show that  \(k=10\)  or  \(k=15\).   (2 marks)

  2. Let  \(k=15\).
    1. Find \(\operatorname{Pr}(X>1)\).   (1 mark)

    2. Find \(E (X)\).   (1 mark)

Show Answers Only

a.    \(\text{Sum of probabilities\(=1\):}\)

\(\dfrac{4}{k}+\dfrac{2 k}{75}+\dfrac{k}{75}+\dfrac{2}{k}\) \(=1\)
\(\dfrac{6}{k}+\dfrac{3 k}{75}\) \(=1\)
\(3 k^2+450\) \(=75k\)
\(3 k^2-75 k+450\) \(=0\)
\(k^2-25 k+150\) \(=0\)
\((k-10)(k-15)\) \(=0\)

  
\(\therefore k=10 \ \ \text{or 15}\)

b.i.   \(\operatorname{Pr}(X>1)=\dfrac{1}{3}\) 

b.ii.  \(E(X)=\dfrac{90}{75}\)

Show Worked Solution

a.    \(\text{Sum of probabilities\(=1\):}\)

\(\dfrac{4}{k}+\dfrac{2 k}{75}+\dfrac{k}{75}+\dfrac{2}{k}\) \(=1\)
\(\dfrac{6}{k}+\dfrac{3 k}{75}\) \(=1\)
\(3 k^2+450\) \(=75k\)
\(3 k^2-75 k+450\) \(=0\)
\(k^2-25 k+150\) \(=0\)
\((k-10)(k-15)\) \(=0\)

 

\(\therefore k=10 \ \ \text{or 15}\)
 

b.i.   \(\operatorname{Pr}(X>1)\) \(=\operatorname{Pr}(X=2)+\operatorname{Pr}(X=3)\)
    \(=\dfrac{15}{75}+\dfrac{2}{15}\)
    \(=\dfrac{1}{3}\)

 

b.ii.   \(E(X)\) \(=0 \times \dfrac{4}{15}+1 \times \dfrac{30}{75}+2 \times \dfrac{15}{75}+3 \times \dfrac{2}{15}\)
    \(=\dfrac{30}{75}+\dfrac{30}{75}+\dfrac{30}{75}\)
    \(=\dfrac{90}{75}\)

Probability, MET2 2024 VCAA 14 MC

Let \(h\) be the probability density function for a continuous random variable \(X\), where

\(h(x)=\left\{
\begin{array} {c}
\rule{0pt}{2.5ex} \ \ \ \ \ \dfrac{x}{6}+k \rule[-1ex]{0pt}{0pt} & -3 \leq x<0 \\
\rule{0pt}{2.5ex} \ \ -\dfrac{x}{2}+k \rule[-1ex]{0pt}{0pt} & 0 \leq x \leq 1 \\
\rule{0pt}{2.5ex} 0 \rule[-1ex]{0pt}{0pt} & \text { elsewhere } \\
\end{array}\right.\)

and \(k\) is a positive real number.

The value of  \(\text{Pr}(X<0.5)\)  is

  1. \(\dfrac{1}{2}\)
  2. \(\dfrac{15}{16}\)
  3. \(\dfrac{3}{16}\)
  4. \(\dfrac{49}{48}\)
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\(B\)

Show Worked Solution

\(\text{Using CAS}:\)

♦ Mean mark 53%.

\(\text{Alternatively, given that}\ \ k=\dfrac{1}{2}\ \text{by CAS:}\)

\(\text{Pr}\left(X<\dfrac{1}{2}\right)\) \(=\displaystyle \int_{-3}^0 \dfrac{x}{6}+\dfrac{1}{2}\,dx +\int_0^{0.5} \dfrac{1}{2}-\dfrac{x}{2}\,dx\)
  \(=\dfrac{1}{2}{\left[\dfrac{x^2}{6}+x\right] _{-3}^0} +\dfrac{1}{2}{\left[x-\dfrac{x^2}{2}\right] _0^{0.5}}\) 
  \(=\dfrac{1}{2}\left[0-\left(\dfrac{9}{6}-3\right)\right]+\dfrac{1}{2}\left[\left(0.5-\dfrac{0.5^2}{2}\right)-0\right]\)
  \(=\dfrac{15}{16}\)

  
\(\Rightarrow B\)

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`
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`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 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`
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`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`
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`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 2010 VCAA 8

The discrete random variable `X` has the probability distribution

vcaa-2010-meth-8a

Find the value of `p.`   (3 marks)

Show Answers Only

`1/2`

Show Worked Solution
`text(Sum of probabilities)` `= 1`
`p^2 + p^2 + p/4 + (4p + 1)/8` `= 1`
`16p^2 + 2p + 4p + 1` `= 8`
`16p^2 + 6p – 7` `= 0`
`(2p – 1) (8p + 7)` `= 0`

 

`:. p = 1/2,\ \ \ (p>0)`

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`