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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)

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`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)

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`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 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}

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`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 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)

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  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 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)

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`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 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):}`
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`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`