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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 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 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
  1. `2/3`
  2. `1.2`
Show Worked Solution

i.   `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`

 

ii.   `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`