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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 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
  1. `3`
  2. `0.3`
Show Worked Solution

i.   `3`

 

ii.   `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`