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Statistics, 2ADV 2026 HSC 7 MC

A biased six-sided die has faces numbered \(-4,\ {-3},\ 0,\ 1,\ 2\) and \(6\).

The table shows the probability distribution for \(X\), the number which lands face up after the die is rolled.

\(X\) \(-4\) \(-3\) \(0\) \(1\) \(2\) \(6\)
\(P(X)\) \(\dfrac{3}{18}\) \(\dfrac{4}{18}\) \(\dfrac{3}{18}\) \(\dfrac{1}{18}\) \(\dfrac{1}{18}\) \(\dfrac{6}{18}\)

The die is rolled twice.

What is the probability that the sum of the two numbers rolled is \(-3\)?

  1. \(\dfrac{1}{54}\)
  2. \(\dfrac{5}{54}\)
  3. \(\dfrac{4}{18}\)
  4. \(\dfrac{6}{18}\)
Show Answers Only

\(B\)

Show Worked Solution

\(\text{Sum of}\ {-3}\ \text{occurs with}\ (-4, 1), \ (1, -4), \ (-3, 0), \ (0, -3)\)

\(P(\text{sum}=-3)=2 \times \dfrac{3}{18} \times \dfrac{1}{18}+2 \times \dfrac{4}{18} \times \dfrac{3}{18}\)

\(\phantom{P(\text{sum}=-3)}=\dfrac{6}{324}+\dfrac{24}{324}\)

\(\phantom{P(\text{sum}=-3)}=\dfrac{5}{54}\)

\(\Rightarrow B\)

Probability, 2ADV S1 EQ-Bank 16

The table shows the probability distribution for the discrete random variable \(X\).

\begin{array}{|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt}& \quad 1 \quad & \quad 2 \quad & \quad 3 \quad & \quad 4 \quad \\
\hline
\rule{0pt}{2.5ex} P(X=x) \rule[-1ex]{0pt}{0pt}& 0.3 & 0.4 & 0.2 & 0.1 \\
\hline
\end{array}

Calculate \(\operatorname{Var}(X)\).   (2 marks)

Show Answers Only

\(\operatorname{Var}(X)=0.89\)

Show Worked Solution
\(\operatorname{Var}(X)\) \(=E\left(X^2\right)-\mu^2\)
  \(=1^2(0.3)+2^2(0.4)+3^2(0.2)+4^2(0.1)-[1(0.3)+2(0.4)+3(0.2)+4(0.1)]^2\)
  \(=5.3-2.1^2\)
  \(=0.89\)

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)

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

    2. Find \(E (X)\).   (1 mark)

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

Probability, 2ADV S1 2025 HSC 1 MC

The probability distribution table for a discrete random variable \(X\) is shown.

\begin{array}{|c|c|}
\hline \quad \quad x \quad \quad & \quad P(X=x) \quad\\
\hline 1 & 0.4 \\
\hline 2 & 0.2 \\
\hline 3 & \\
\hline
\end{array}

What is the value of \(P(X=3)\) ?

  1. \(0.2\)
  2. \(0.4\)
  3. \(1.2\)
  4. \(2.0\)
Show Answers Only

\(B\)

Show Worked Solution

\(\text{Probabilities sum to 1:}\)

\(0.4+0.2+P(X=3) = 1\ \ \rightarrow\ \ P(X=3) = 0.4\)

\(\Rightarrow B\)

Probability, 2ADV S1 2024 MET2 3*

A discrete random variable \(X\) is defined using the probability distribution below, where \(k\) is a positive real number.

\begin{array} {|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & \ \ \ \ \text{0}\ \ \ \  \rule[-1ex]{0pt}{0pt} & \ \ \ \ \text{1}\ \ \ \ \rule[-1ex]{0pt}{0pt} & \ \ \ \ \text{2}\ \ \ \ \rule[-1ex]{0pt}{0pt} & \ \ \ \ \text{3}\ \ \ \ \rule[-1ex]{0pt}{0pt} & \ \ \ \ \text{4}\ \ \ \ \\
\hline
\rule{0pt}{2.5ex} \text{Pr} \  (X = x) \rule[-1ex]{0pt}{0pt} &  2k \rule[-1ex]{0pt}{0pt} & 3k \rule[-1ex]{0pt}{0pt} & 5k \rule[-1ex]{0pt}{0pt} & 3k \rule[-1ex]{0pt}{0pt} & 2k \\
\hline
\end{array}

Find \(\operatorname{Pr}(X<4 \mid X>1)\)   (3 marks)

Show Answers Only

\(C\)

Show Worked Solution

\(2k+3k+5k+3k+2k=1\ \Rightarrow\ k=\dfrac{1}{15}\)

\(\operatorname{Pr}(X<4 \mid X>1)\) \(=\dfrac{\operatorname{Pr}(X>1)\ \cap \ \operatorname{Pr}(X<4)}{\operatorname{Pr}(X>1)}\)
  \(=\dfrac{\operatorname{Pr}(1<X<4)}{\operatorname{Pr}(X>1)}\)
  \(=\dfrac{\frac{5}{15}+\frac{3}{15}}{\frac{5}{15}+\frac{3}{15}+\frac{2}{15}}\)
  \(=\dfrac{\frac{8}{15}}{\frac{10}{15}}\)
  \(=\dfrac{4}{5}\)

Probability, 2ADV S1 2023 HSC 12

The table shows the probability distribution of a discrete random variable.

\begin{array} {|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & 0 & 1 & 2 & 3 & 4 \\
\hline
\rule{0pt}{2.5ex} P(X = x) \rule[-1ex]{0pt}{0pt} & \ \ \ 0\ \ \  & \ \ 0.3\ \  & \ \ 0.5\ \  & \ \ 0.1\ \  & \ \ 0.1\ \  \\
\hline
\end{array}

  1. Show that the expected value \(E(X)=2\).   (1 mark)

  2. Calculate the standard deviation, correct to one decimal place.   (2 marks)

Show Answers Only

a.    `text{See Worked Solutions}`

b.    `0.9`

Show Worked Solution
a.     \(E(X)\) \(=0+1\times 0.3+2\times 0.5+3\times 0.1+4\times 0.1\)
    \(=0.3+1+0.3+0.4=2\)

 

b.     \(\text{Var}(X)\) \(=E(X^2)-[E(X)]^2\)
    \(=(0+1^2\times 0.3+2^2\times 0.5+3^2\times 0.1+4^2\times 0.1)-2^2\)
    \(=(0.3+2+0.9+1.6)-4\)
    \(=0.8\)

 

\(\therefore\ \sigma\) \(=\sqrt{0.8}\)  
  \(=0.8944…=0.9\ \text{(to 1 d.p.)}\)  

Probability, 2ADV S1 2021 HSC 34

A discrete random variable has probability distribution as shown in the table where  `n`  is a finite positive integer.

\begin{array} {|c|c|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & \ \ \ r\ \ \  & \ \ \ r^{2}\ \ \  & \ \ \ r^{3}\ \ \  & \ \ \ ...\ \ \  & \ \ \ r^{k}\ \ \ &\ \ \ ...\ \ \ & \ \ \ r^{n}\ \ \ \\
\hline
\rule{0pt}{2.5ex} P(X=x) \rule[-1ex]{0pt}{0pt} & r^{n} & r^{n-1} & r^{n-2} & ... & r^{n-k+1} & ... & r  \\
\hline
\end{array}

Show that  `E(X) = n( 2r-1)`.   (3 marks)

Show Answers Only

`text(See Worked Solution)`

Show Worked Solution
♦♦♦ Mean mark 18%.
`E(X)` `= ∑x · P(X = x)`
  `= r · r^n + r^2 · r^(n + 1) + … + r^n · r`
  `= r^(n + 1) + r^(n + 1) + … + r^(n + 1)`
  `= nr^(n + 1) …\ (1)`

  
`text(Sum of probabilities = 1)`

`underset (text(GP where)\ a = r,\ r = r) (underbrace {r + r^2 + r^3 + … + r^(n-1) + r^n}) = 1`

`(r(1-r^n))/(1-r)` `= 1`
`r-r^(n + 1)` `= 1-r`
`r^(n + 1)` `= 2r-1 …\ (2)`

 
`text{Substitute (2) into (1):}`

`E(X) = n(2r-1)`

Probability, 2ADV S1 2021 HSC 2 MC

The probability distribution table for a discrete random variable `X` is shown.

\begin{array}{|c|c|}
\hline \rule{0pt}{2.5ex} \quad \quad x \quad \quad \rule[-1ex]{0pt}{0pt} & \quad P(X=x) \quad\\
\hline \rule{0pt}{2.5ex} 1 \rule[-1ex]{0pt}{0pt} & 0.6 \\
\hline \rule{0pt}{2.5ex} 2 \rule[-1ex]{0pt}{0pt} & 0.3 \\
\hline \rule{0pt}{2.5ex} 3 \rule[-1ex]{0pt}{0pt} & 0.1 \\
\hline
\end{array}

What is the expected value of `X`?

  1. 0.6
  2. 1.0
  3. 1.5
  4. 2.0
Show Answers Only

`C`

Show Worked Solution
`E(X)` `= 1 xx 0.6 + 2 xx 0.3 + 3 xx 0.1`
  `= 1.5`

 
`=>  C`

Probability, 2ADV S1 EQ-Bank 18

The discrete random variable `X` has the probability distribution shown in the table below.

\begin{array}{|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} X=x \rule[-1ex]{0pt}{0pt}& \quad 4 \quad & \quad 5 \quad & \quad 6 \quad & \quad 7 \quad & \quad 8 \quad \\
\hline
\rule{0pt}{2.5ex} P(x) \rule[-1ex]{0pt}{0pt}& 0.3 & a & 0.1 & 0.15 & 0.2 \\
\hline
\end{array}

 Find the value of  `a`, and hence calculate the the expected value and variance of  `X`.   (3 marks)

Show Answers Only

`2.31`

Show Worked Solution

`0.3 + a + 0.1 + 0.15 + 0.2 = 1`

`=> \ a = 0.25`
 

`E(X) = ∑ x P(x)`

\begin{array}{|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} X=x \rule[-1ex]{0pt}{0pt}& \ \ 4 \ \  & \ \ 5 \ \ & \ \ 6 \ \ & \ \ 7 \ \ & \ \ 8 \ \ \\
\hline
\rule{0pt}{2.5ex} P(x) \rule[-1ex]{0pt}{0pt}& 0.3 & 0.25 & 0.1 & 0.15 & 0.2 \\
\hline
\rule{0pt}{2.5ex} x \times P(x) \rule[-1ex]{0pt}{0pt}& 1.2 & 1.25 & 0.6 & 1.05 & 1.6 \\
\hline
\end{array}

`E(X)` `= 1.2 + 1.25 + 0.6 + 1.05 + 1.6`
  `= 5.7`

 

`text(Var)(X)` `= E(X^2)-[E(X)]^2`
  `= (4^2 xx 0.3) + (5^2 xx 0.25) + (6^2 xx 0.1) + (7^2 xx 0.15) + (8^2 xx 0.2)-5.7^2`
  `= 2.31`

Probability, 2ADV S1 2019 MET2-N 5 MC

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

\begin{array} {|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & \ \ \ -1\ \ \ & \quad 0\quad  & \quad 1\quad & \quad 2\quad &\quad 3\quad \\
\hline
\rule{0pt}{2.75ex} P(X=x) \rule[-2ex]{0pt}{0pt} & b & b & b & \dfrac{3}{5}-b & \dfrac{3b}{5} \\
\hline
\end{array}

The value of `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)`

 

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

 

\(\Rightarrow A\)

Probability, 2ADV S1 EQ-Bank 20

One bag contains red and green balls.

Kalyn randomly chooses one ball from the bag. Without replacement, he then chooses a second ball from the bag.

Complete the tree diagram below and then draw a probability distribution table for the number of red balls that could be drawn out of the bag.   (3 marks)
 

Show Answers Only

\begin{array}{|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt}& \quad 0 \quad & \quad 1 \quad & \quad 2 \quad  \\
\hline
\rule{0pt}{2.5ex} P(X=x) \rule[-1ex]{0pt}{0pt}& \frac{5}{18} & \frac{5}{9} & \frac{1}{6} \\
\hline
\end{array}

Show Worked Solution

  
`x = 2: \ P(R R) = 4/9 xx 3/8 = 1/6`

`x = 1: \ P(R and G) = 4/9 xx 5/8 + 5/9 xx 4/8 = 5/9`

`x = 0: \ P(G G) = 5/9 xx 4/8 = 5/18`
   
\begin{array}{|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt}& \quad 0 \quad & \quad 1 \quad & \quad 2 \quad  \\
\hline
\rule{0pt}{2.5ex} P(X=x) \rule[-1ex]{0pt}{0pt}& \frac{5}{18} & \frac{5}{9} & \frac{1}{6} \\
\hline
\end{array}

Probability, 2ADV S1 EQ-Bank 23

Evaluate `p` and `q` in the discrete probability distribution table below, given that  `E(X) = 3`.   (3 marks)

\begin{array} {|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & \ \ \ 1\ \ \  & \ \ \ 2\ \ \  & \ \ \ 3\ \ \  & \ \ \ 4\ \ \  \\
\hline
\rule{0pt}{2.5ex} P(X=x) \rule[-1ex]{0pt}{0pt} & p & q & 0.2 & 0.4 \\
\hline
\end{array}

Show Answers Only

`text(See Worked Solutions)`

Show Worked Solution
`p + q + 0.2 + 0.4` `= 1`
`p + q` `= 0.4\ \ …\ (1)`

 
`text(Given)\ \ E(X) = 3,`

`p + 2q + 0.6 + 1.6` `= 3`
`p + 2q` `= 0.8\ \ …\ (2)`

 
`text(Subtract:) (2)-(1)`

`q` `= 0.4`
`:.\ p` `= 0`

Probability, 2ADV S1 2015 MET2 14 MC

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

\begin{array} {|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & \ \ \ 1\ \ \  & \ \ \ 2\ \ \  & \ \ \ 3\ \ \ &\ \ \ 4\ \ \ &\ \ \ 5\ \ \ \\
\hline
\rule{0pt}{2.5ex} P(X=x) \rule[-1ex]{0pt}{0pt} & p & 2p & 3p & 4p & 5p \\
\hline
\end{array} 

The mean of this distribution is

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

`D`

Show Worked Solution

`text(Find)\ p:`

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

 

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

  
`=>   D`

Probability, 2ADV S1 2013 MET1 7

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

\begin{array} {|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & \ \ \ 0\ \ \  & \ \ \ 1\ \ \  & \ \ \ 2\ \ \  & \ \ \ 3\ \ \ &\ \ \ 4\ \ \  \\
\hline
\rule{0pt}{2.5ex} P(X=x) \rule[-1ex]{0pt}{0pt} & 0.2 & 0.6p^{2} & 0.1 & 1-p & 0.1 \\
\hline
\end{array}

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

  2. Let  `p = 2/3`.

    1. Calculate  `E(X)`.  Answer in exact form.   (2 marks)

    2. Find  `P(X >= E(X))`.   (1 mark)

Show Answers Only

a.    `text(Proof)\ \ text{(See Worked Solutions)}`

b.i.  `28/15`

b.ii. `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.   `E(X)` `= sum x * P (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%.
b.ii.  `P(X >= 28/15)` `=P(X = 2) + P(X = 3) + P(X = 4)`
    `= 1/10 + 1/3 + 1/10= 8/15`

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)

  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

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`

Probability, 2ADV S1 2010 MET1 8

The discrete random variable `X` has the probability distribution

\begin{array} {|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & \ \ \ \ -1\ \ \ \  & \ \ \ \ \ 0\ \ \ \ \  & \ \ \ \ \ 1\ \ \ \ \  & \ \ \ \ \ 2\ \ \ \ \  \\
\hline
\rule{0pt}{2.5ex} P(X=x) \rule[-1ex]{0pt}{0pt} & p^{2} & p^{2} & \dfrac{p}{4} & \dfrac{4p+1}{8} \\
\hline
\end{array}

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, 2ADV S1 2009 MET1 7

The random variable `X` has this probability distribution.

\begin{array} {|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & \ \ \ 0\ \ \  & \ \ \ 1\ \ \  & \ \ \ 2\ \ \  & \ \ \ 3\ \ \  &\ \ \ 4\ \ \ \\
\hline
\rule{0pt}{2.5ex} P(X=x) \rule[-1ex]{0pt}{0pt} & 0.1 & 0.2 & 0.4 & 0.2 & 0.1 \\
\hline
\end{array}

Find

  1.  `P (X > 1 | X <= 3)`   (2 marks)

  2.  `P (X),` the variance of  `X.`   (3 marks)
Show Answers Only

a.    `2/3`

b.    `1.2`

Show Worked Solution

a.    `P(X > 1 | X <= 3)`

`= (P(X = 2) + P(X = 3))/(1-P(X = 4))`

`= (0.4 + 0.2)/(1-0.1)`

`= 0.6/0.9 = 2/3`
  

b.    `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`

 

`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)` `= E(X^2)-[E(X)]^2`
  `= 5.2-(2)^2`
  `= 1.2`

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)

  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

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`

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`

Probability, 2ADV S1 2016 MET2 19 MC

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

\begin{array} {|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & \ \ \ -1\ \ \  & \ \ \ \ 0\ \ \ \  & \ \ \ \ b\ \ \ \  & \ \ \ \ 2b\ \ \ \ &\ \ \ \ 4\ \ \ \  \\
\hline
\rule{0pt}{2.5ex} P(X=x) \rule[-1ex]{0pt}{0pt} & a & b & b & 2b & 0.2 \\
\hline
\end{array}

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 and 1`
Show Answers Only

`D`

Show Worked Solution

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

♦♦♦ Mean mark 15%.
`:.\ text(Smallest)\ 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)\ E(X)\ \ text(occurs when)\ \ a=0, and b=0.2,`

`:.\ text(Largest)\ 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`

`=>   D`

Probability, 2ADV S1 2010 MET2 15 MC

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

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

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 = 0.3`
Show Answers Only

`C`

Show Worked Solution

`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, 2ADV S1 2009 MET2 10 MC

The discrete random variable `X` has a probability distribution as shown.

\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} & 0.4 & 0.2 & 0.3 & 0.1 \\
\hline
\end{array}

The median of `X` is

  1. `0`
  2. `1`
  3. `1.1`
  4. `2`
Show Answers Only

`B`

Show Worked Solution
`P(X <= 0)` `= 0.4`
`P(X <= 1)` `= 0.6`

 
`:.\ text(Median) = 1`

`=>   B`

Probability, 2ADV S1 2007 MET2 19 MC

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

\begin{array} {|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & \ \ \ 0\ \ \  & \ \ \ 2\ \ \  & \ \ \ 4\ \ \  & \ \ \ 6\ \ \ &\ \ \ 8\ \ \ \\
\hline
\rule{0pt}{2.5ex} P(X=x) \rule[-1ex]{0pt}{0pt} & a & 0.2 & 0.2 & 0.3 & b \\
\hline
\end{array}

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):}`
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}\ 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, 2ADV S1 2017 MET2 14 MC

The random variable `X` has the following probability distribution, where  `0 < p < 1/3`.

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

The variance of `X` is

  1. `2p(1-3p)`
  2. `p(5-9p)`
  3. `(1-3p)^2`
  4. `6p-16p^2`
Show Answers Only

`D`

Show Worked Solution
`text(Var)(X)` `= E(X^2)-[E(X)]^2`
  `= [(-1)^2p + 0^2 xx 2p + 1^2(1-3p)]-[-p + 0 + 1-3p]^2`
  `= 6p-16p^2`

 
`=> D`