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Probability, 2ADV S1 2025 MET1 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}{|ca|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)

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

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    2. Find \(E (X)\).   (1 mark)

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Show Answers Only

a.    \(\text{See Worked Solution}\)

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.  \(E(X)\) \(=0 \times \dfrac{4}{15}+1 \times \dfrac{30}{75}+2 \times \dfrac{15}{75}+3 \times \dfrac{2}{15}\)
    \(=\dfrac{6}{5}\)

Filed Under: Discrete Probability Distributions, Discrete Random Variables Tagged With: Band 4, smc-7136-10-Sum of Probabilities = 1, smc-7136-20-\(E(X)\) / Mean, smc-7136-70-Other Probability, smc-992-10-Sum of Probabilities = 1, smc-992-20-E(X) / Mean, smc-992-70-Other Probability

Probability, 2ADV S1 2012 MET1 4

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

\begin{array} {|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & \ \ \ 0\ \ \  & \ \ \ 1\ \ \  & \ \ \ 2\ \ \  & \ \ \ 3\ \ \  \\
\hline
\rule{0pt}{2.5ex} P(X=x) \rule[-1ex]{0pt}{0pt} & 0.2 & 0.2 & 0.5 & 0.1 \\
\hline
\end{array}

  1. Find the mean of  `X`.   (2 marks)

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  2. What is the probability that Daniel receives only one telephone call on each of three consecutive days?   (1 mark)

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  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)

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Show Answers Only

a.    `1.5`

b.    `0.008`

c.    `29/64`

Show Worked Solution
a.     `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.    
`P(1, 1, 1)= 0.2 xx 0.2 xx 0.2= 0.008`
  

c.    `text(Conditional Probability:)`

♦ Mean mark (c) 36%.

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

`= (P(1, 3) + P(2, 2) + P(3, 1))/(P(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`

Filed Under: Discrete Probability Distributions, Discrete Random Variables Tagged With: Band 3, Band 4, Band 5, smc-7136-20-\(E(X)\) / Mean, smc-7136-60-Conditional Probability, smc-7136-70-Other Probability, smc-992-20-E(X) / Mean, smc-992-60-Conditional Probability, smc-992-70-Other Probability

Probability, 2ADV S1 2008 MET1 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

\begin{array} {|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & \ \ \ 0\ \ \  & \ \ \ 1\ \ \  & \ \ \ 2\ \ \  & \ \ \ 3\ \ \  \\
\hline
\rule{0pt}{2.5ex} P(X=x) \rule[-1ex]{0pt}{0pt} & 0.1 & 0.2 & 0.3 & 0.4 \\
\hline
\end{array}

  1. What is the mode of  `X`?   (1 mark)

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  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)

    --- 3 WORK AREA LINES (style=lined) ---

Show Answers Only

a.    `3`

b.    `0.3`

Show Worked Solution

a.    `3`

b.    `P(0,0) + P(1,1) + P(2,2) + P(3,3)`

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

`= 0.3`

Filed Under: Discrete Probability Distributions, Discrete Random Variables Tagged With: Band 4, smc-7136-40-Median and Mode, smc-7136-70-Other Probability, smc-992-40-Median and Mode, smc-992-70-Other Probability

Probability, 2ADV S1 2016 MET2 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.

\begin{array} {|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & \ \ \ \ 0\ \ \ \  & \ \ \ \ 1\ \ \ \  & \ \ \ \ 2\ \ \ \  & \ \ \ \ 3\ \ \ \  \\
\hline
\rule{0pt}{2.5ex} P(X=x) \rule[-1ex]{0pt}{0pt} & 0.5 & 0.25 & 0.2 & 0.05 \\
\hline
\end{array}

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`
Show Answers Only

`C`

Show Worked Solution

`P (0, 0)+ P (1, 1)+ P (2, 2)+ P (3, 3)`

`= 0.5^2+0.25^2+0.2^2+0.05^2`

`=0.355`

 
`=>   C`

Filed Under: Discrete Probability Distributions, Discrete Random Variables Tagged With: Band 3, smc-7136-70-Other Probability, smc-992-70-Other Probability

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