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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
  1. `text{See Worked Solutions}`
  2. `0.9`
Show Worked Solution
a.     `E(X)` `=0+1xx0.3+2xx0.5+3xx0.1+4xx0.1`
    `=0.3+1+0.3+0.4`
    `=2`

 

b.     `text{Var}(X)` `=E(X^2)-[E(X)]^2`
    `=(0+1^2xx0.3+2^2xx0.5+3^2xx0.1+4^2xx0.1)-2^2`
    `=(0.3+2+0.9+1.6)-4`
    `=0.8`

 

`:. 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 EQ-Bank 42

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

     `X = x` `4` `5` `6` `7` `8`
  `P(x)` `0.3` `a` `0.1` `0.15` `0.2`

   
 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)`
 

  `qquad X = x` `4` `5` `6` `7` `8`
  `qquad P(x) qquad` `0.3` `0.25` `0.1` `0.15` `0.2`
  `qquad x xx P(x) qquad` `1.2` `1.25` `0.6` `1.05` `1.6`

 

`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\ \ \ \  & \ \ \ \ \ 0\ \ \ \ \  & \ \ \ \ \ 1\ \ \ \ \  & \ \ \ \ \ 2\ \ \ \ \  &\ \ \ \ \ 3\ \ \ \ \ \\
\hline
\rule{0pt}{2.5ex} P(X=x) \rule[-1ex]{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 SM-Bank 41

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
  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.   `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%.
  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
  1. `1.5`
  2. `0.008`
  3. `29/64`
Show Worked Solution
i.    `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`

 

ii.   `P(1, 1, 1)` `= 0.2 xx 0.2 xx 0.2`
    `= 0.008`

 

iii.   `text(Conditional Probability:)`

♦ Mean mark 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 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 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`