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Probability, 2ADV S1 2021 HSC 6 MC

There are 8 chocolates in a box. Three have peppermint centres (P) and five have caramel centres (C).

Kim randomly chooses a chocolate from the box and eats it. Sam then randomly chooses and eats one of the remaining chocolates.

A partially completed probability tree is shown.
 

What is the probability that Kim and Sam choose chocolates with different centres?

  1. `\frac{15}{64}`
  2. `\frac{15}{56}`
  3. `\frac{15}{32}`
  4. `\frac{15}{28}`
Show Answers Only

`D`

Show Worked Solution

 

`Ptext{(different centres)}` `= P text{(PC)} + P text{(CP)}`
  `=\frac{3}{8} · \frac{5}{7} + \frac{5}{8} · \frac{3}{7}`
  `= \frac{15}{56} + \frac{15}{56}`
  `= \frac{15}{28}`

 
`=> D`

Probability, 2ADV S1 SM-Bank 6 MC

A box contains  `n`  marbles that are identical in every way except colour, of which  `k`  marbles are coloured red and the remainder of the marbles are coloured green. Two marbles are drawn randomly from the box.

If the first marble is not replaced into the box before the second marble is drawn, then the probability that the two marbles drawn are the same colour is

  1. `(k^2 + (n - k)^2)/n^2`
  2. `(k^2 + (n - k - 1)^2)/n^2`
  3. `(2k(n - k - 1))/(n(n - 1))`
  4. `(k(k - 1) + (n - k)(n - k - 1))/(n(n - 1))`
Show Answers Only

`D`

Show Worked Solution

`n\ text(marbles) \ => \ k\ text(red), \ (n – k)\ \ text(green)`
 

`:.\ P(text{same colour})` `= k/n ⋅ ((k – 1))/((n – 1)) + ((n – k))/n ⋅ ((n – k – 1))/((n – 1))`
  `= (k(k – 1) + (n – k)(n – k – 1))/(n(n – 1))`

 
`=>   D`

Probability, 2ADV S1 2019 HSC 11f

A bag contains 5 green beads and 7 purple beads. Two beads are selected at random, without replacement.

What is the probability that the two beads are the same colour?  (2 marks)

Show Answers Only

`31/66`

Show Worked Solution

`P\ text{(same colour)}` `= P (GG) + P(PP)`
  `= 5/12 ⋅ 4/11 + 7/12 ⋅ 6/11`
  `= 62/132`
  `= 31/66`

Probability, 2ADV S1 2016 HSC 15b

An eight- sided die is marked with numbers  1, 2, … , 8. A game is played by rolling the die until an 8 appears on the uppermost face. At this point the game ends.

  1. Using a tree diagram, or otherwise, explain why the probability of the game ending before the fourth roll is

      

    `qquad qquad 1/8 + 7/8 xx 1/8 + (7/8)^2 xx 1/8`.  (2 marks)

  2. What is the smallest value of `n` for which the probability of the game ending before the `n`th roll is more than  `3/4`?  (3 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `12`
Show Worked Solution

i.   `P text{(game ends before 4th roll)}`

`= P (8) + P (text{not}\ 8, 8) + P (text{not}\ 8, text{not}\ 8, 8)`

`= 1/8 + 7/8 · 1/8 + 7/8 · 7/8 · 1/8`

`= 1/8 + 7/8 · 1/8 + (7/8)^2 · 1/8\ \ text(…  as required)`

 

ii.  `1/8 + 7/8 · 1/8 + (7/8)^2 · 1/8 + …`

`=> text(GP where)\ \ a = 1/8,\ \ r = 7/8`

`text(Find)\ \ n\ \ text(such that)\ \ S_(n – 1) > 3/4,`

♦♦ Mean mark 29%.
ALGEBRA: Note that dividing by `ln\ 7/8` reverses the < sign as it is dividing by a negative number.
`S_(n-1)` `= (a (1 – r^(n – 1)))/(1 – r)`
`3/4` `< 1/8 xx {(1 – (7/8)^(n – 1))}/(1 – 7/8)`
`3/4` `< 1 – (7/8)^(n – 1)`
`(7/8)^(n – 1)` `< 1/4`
`(n-1)* ln\ 7/8` `< ln\ 1/4`
`n – 1` `> (ln\ 1/4)/(ln\ 7/8)`
  `> 11.38…`

`:. n = 12`

Probability, 2ADV S1 2015 HSC 14b

Weather records for a town suggest that:

  • if a particular day is wet `(W)`, the probability of the next day being dry is  `5/6`
  • if a particular day is dry `(D)`, the probability of the next day being dry is  `1/2`.

In a specific week Thursday is dry. The tree diagram shows the possible outcomes for the next three days: Friday, Saturday and Sunday.
 

2015 2ua 14b
 

  1. Show that the probability of Saturday being dry is `2/3`.  (1 mark)

  2. What is the probability of both Saturday and Sunday being wet?  (2 marks)

  3. What is the probability of at least one of Saturday and Sunday being dry?  (1 mark)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `1/(18)`
  3. `(17)/(18)`
Show Worked Solution

i.  `text{Show}\ \ P text{(Sat dry)} = 2/3`

`P text{(Sat dry)}`

`= P (W,D) + P (D, D)`

`=(1/2 xx 5/6) + (1/2 xx 1/2)`

`= 5/(12) + 1/4`

`= 2/3\ \ text(…  as required)`
 

ii.  `Ptext{(Sat and Sun wet)}`

`= P (WWW) + P (DWW)`

`= (1/2 xx 1/6 xx 1/6) + (1/2 xx 1/2 xx 1/6)`

`= 1/(72) + 1/(24)`

`= 1/(18)`
 

iii.  `Ptext{(At least Sat or Sun dry)}`

`= 1 – Ptext{(Sat and Sun both wet)}`

`= 1 – 1/(18)`

`= (17)/(18)`

Probability, 2ADV S1 2008 HSC 7c

Xena and Gabrielle compete in a series of games. The series finishes when one player has won two games. In any game, the probability that Xena wins is  `2/3`  and the probability that Gabrielle wins is  `1/3`.

Part of the tree diagram for this series of games is shown.
 

 
 

  1. Complete the tree diagram showing the possible outcomes.  (1 mark)
  2. What is the probability that Gabrielle wins the series?   (2 marks)

  3. What is the probability that three games are played in the series?   (2 marks)

Show Answers Only
  1.    
  2. `7/27`
  3. `4/9`
Show Worked Solution
i. 2UA HSC 2008 7ci

 

MARKER’S COMMENT: A tree diagram with 8 outcomes is incorrect (i.e. no third game is played if 1 player wins the first 2 games). If outcomes cannot occur, do not draw them on a tree diagram.

 

ii.  `P text{(} G\ text(wins) text{)}`

`= P(XGG) + P (GXG) + P (GG)`

`= 2/3 * 1/3 * 1/3 + 1/3 * 2/3 * 1/3 + 1/3 * 1/3`

`= 2/27 + 2/27 + 1/9`

`= 7/27`

 

iii.  `text(Method 1:)`

`P text{(3 games played)}`

`= P (XG) + P(GX)`

`= 2/3 * 1/3 + 1/3 * 2/3`

`= 4/9`

 

`text(Method 2:)`

`P text{(3 games)}`

`= 1 – [P(XX) + P(GG)]`

`= 1 – [2/3 * 2/3 + 1/3 * 1/3]`

`= 1 – 5/9`

`= 4/9`

Probability, 2ADV S1 2014 HSC 12c

A packet of lollies contains 5 red lollies and 14 green lollies. Two lollies are selected at random without replacement.

  1. Draw a tree diagram to show the possible outcomes. Include the probability on each branch.   (2 marks)

  2. What is the probability that the two lollies are of different colours?     (1 mark)

Show Answers Only
  1.  
  2. `70/171`
Show Worked Solution
i.         2UA HSC 2014 12c

 

ii.  `P text{(different colours)}`

`= P(RG) + P(GR)`

`= 5/19 xx 14/18 + 14/19 xx 5/18`

`= 70/342 + 70/342`

`= 140/342`

`= 70/171`