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Statistics, 2ADV S1 2024 MET2 7*

A fair six-sided die is repeatedly rolled. What is the minimum number of rolls required so that the probability of rolling a six at least once is greater than 0.95?   (2 marks)

Show Answers Only

\(17\)

Show Worked Solution

\(P\text{(not rolling a 6)}\ = P(\bar6) = \dfrac{5}{6}\)

\(P(\bar6, \bar6) = \dfrac{5}{6} \times \dfrac{5}{6} \)

\(\text{Find}\ n\ \text{such that:}\)

\(\Big( \dfrac{5}{6}\Big)^n\) \(\lt 0.05\)  
\(n \times \ln{\Big( \dfrac{5}{6}}\Big)\) \(\lt \ln{(0.05)}\)  
\(n\) \(\gt \dfrac{\ln{0.05}}{\ln{(\frac{5}{6})}}\)  
  \(\gt 16.43…\)  

 
\(\text{Minimum rolls = 17}\)

COMMENT: Note dividing by a negative number reverses < sign.

Probability, 2ADV S1 2024 HSC 18

In a game, the probability that a particular player scores a goal at each attempt is 0.15.

  1. What is the probability that this player does NOT score a goal in the first two attempts?   (1 mark)

  2. Determine the least number of attempts that this player must make so that the probability of scoring at least one goal is greater than 0.8.   (2 marks)

Show Answers Only

a.   \(0.7225\)

b.   \(n=10\)

Show Worked Solution

a.   \(P(G)=0.15, \ \ P(\bar{G})=0.85\)

\(P(\bar{G}\bar{G}) = 0.85^2=0.7225\)
 

b.   \(\text{2 attempts:}\ P(\text{at least 1 goal})=1-P(\bar{G}\bar{G})=1-0.85^{2}\)

\(\text{3 attempts:}\ P(\text{at least 1 goal})=1-0.85^{3}\)

\(\text{n attempts:}\ P(\text{at least 1 goal})=1-0.85^{n}\)

\(\text{Find}\ n\ \text{such that:}\)

\(1-0.85^{n}\) \(\gt 0.8\)  
\(0.85^{n}\) \(\lt 0.2\)  
\(n \times\ln(0.85)\) \(\lt \ln(0.2)\)  
\(n\) \(\gt \dfrac{\ln(0.2)}{\ln(0.85)}\)  
  \(\gt 9.9…\)  

 
\(\therefore\ \text{Least}\ n=10\)

♦ Mean mark (b) 39%.

Probability, 2ADV S1 2023 MET2 8 MC

A box contains \(n\) green balls and \(m\) red balls. A ball is selected at random, and its colour is noted. The ball is then replaced in the box.

In 8 such selections, where \(n\neq m\), what is the probability that a green ball is selected at least once?

  1. \(8\Bigg(\dfrac{n}{n+m}\Bigg)\Bigg(\dfrac{m}{n+m}\Bigg)^7\)
  2. \(1-\Bigg(\dfrac{n}{n+m}\Bigg)^8\)
  3. \(1-\Bigg(\dfrac{m}{n+m}\Bigg)^8\)
  4. \(1-\Bigg(\dfrac{n}{n+m}\Bigg)\Bigg(\dfrac{m}{n+m}\Bigg)^7\)
Show Answers Only

\(C\)

Show Worked Solution

\(\text{In any single selection:}\)

\(P(\text{green})\ =\Bigg(\dfrac{n}{n+m}\Bigg), \ \ P(\text{not green})\ =\Bigg(\dfrac{m}{n+m}\Bigg) \)

\(\text{Let}\ \ X=\ \text{choosing a green ball}\)

\(P(X\geq 1)\) \(=1-\text{Pr}(X=0)\)
  \(=1-\Bigg(\dfrac{m}{n+m}\Bigg)^8\)

 
\(\Rightarrow C\)


♦ Mean mark 49%.

Probability, 2ADV S1 2022 HSC 15

In a bag there are 3 six-sided dice. Two of the dice have faces marked  1, 2, 3, 4, 5, 6. The other is a special die with faces marked  1, 2, 3, 5, 5, 5.

One die is randomly selected and tossed.

  1. What is the probability that the die shows a 5?  (1 mark)

  2. Given that the die shows a 5, what is the probability that it is the special die?  (1 mark)

Show Answers Only
  1. `5/18`
  2. `3/5`
Show Worked Solution
a.    `P(5)` `=1/3 xx 1/6 + 1/3 xx 1/6 + 1/3 xx 1/2`
    `=1/18+1/18+1/6`
    `=5/18`

 

b.   `Ptext{(special die given a 5 is rolled)}`

`=(Ptext{(special die)} ∩  P(5))/(P(5))`

`=(1/3 xx 1/2)/(5/18)`

`=1/6 xx 18/5`

`=3/5`


Mean mark (a) 53%.
 
♦ Mean mark (b) 40%.

Probability, 2ADV S1 2022 HSC 9 MC

Liam is playing two games. He is equally likely to win each game. The probability that Liam will win at least one of the games is 80%.

Which of the following is closest to the probability that Liam will win both games?

  1.  31%
  2.  40%
  3.  55%
  4.  64%
Show Answers Only

`A`

Show Worked Solution

`Ptext{(at least 1 W)}\ = 1-Ptext{(LL)}\ =0.8`

`Ptext{(LL)}` `=0.2`  
`Ptext{(L)}` `=sqrt0.2`  
  `=0.447`  

 

`Ptext{(W)}` `=1-0.447=0.553`  
`Ptext{(WW)}` `=(0.553)^2`  
  `=0.31`  

`=>A`


… Mean marks/comments here
♦♦♦ Mean mark 27%.

Probability, 2ADV S1 2019 MET2-N 9 MC

At the start of a particular week, Kim has three red apples and two green apples. She eats one apple everyday. On Monday, Tuesday and Wednesday of that week, she randomly selects an apple to eat. In this three-day period, the probability that Kim does not eat an apple of the same colour on any two consecutive days is

  1.  `(1)/(5)`
  2.  `(3)/(10)`
  3.  `(2)/(5)`
  4.  `(6)/(25)`
Show Answers Only

`B`

Show Worked Solution

`P text{(alternate colours)}`

`= P(RGR) + P(GRG)`

`= (3)/(5) ·(2)/(4) ·(2)/(3) + (2)/(5) ·(3)/(4) ·(1)/(3)`

`= (12)/(60) + (6)/(60)`

`= (3)/(10)`
 

`=> \ B`

Probability, 2ADV S1 2019 HSC 6 MC

A game is played by tossing an ordinary 6-sided die and an ordinary coin at the same time. The game is won if the uppermost face of the die shows an even number or the uppermost face of the coin shows a tail (or both).

What is the probability of winning this game?

  1. `1/4`
  2. `1/2`
  3. `3/4`
  4. `1`
Show Answers Only

`C`

Show Worked Solution

`text(Game lost only if an odd and a head show.)`

`:. P(W)` `= 1 – P text{(odd)} ⋅ P text{(head)}`
  `= 1 – 3/6 ⋅ 1/2`
  `= 3/4`

 
`=>  C`

Probability, 2ADV S1 2018 HSC 14e

Two machines, `A` and `B`, produce pens. It is known that 10% of the pens produced by machine `A` are faulty and that 5% of the pens produced by machine `B` are faulty.

  1. One pen is chosen at random from each machine.

     

    What is the probability that at least one of the pens is faulty?  (1 mark)

  2. A coin is tossed to select one of the two machines. Two pens are chosen at random from the selected machine.

     

    What is the probability that neither pen is faulty?  (2 marks)

Show Answers Only
  1. `0.145`
  2. `0.85625`
Show Worked Solution
i.   `text{P(at least 1 faulty)}` `= 1 –  text{P(both faulty)}`
    `= 1 – 0.9 xx 0.95`
    `= 1 – 0.855`
    `= 0.145`

 

ii.   `text{P(2 non-faulty pens})`

♦ Mean mark 48%.

`= text{(choose A, NF, NF)} + P text{(choose B, NF, NF)}`

`= 1/2 xx 0.9 xx 0.9 + 1/2 xx 0.95 xx 0.95`

`= 0.405 + 0.45125`

`=0.85625`

Probability, 2ADV S1 2018 HSC 6 MC

A runner has four different pairs of shoes.

If two shoes are selected at random, what is the probability that they will be a matching pair?

  1. `1/56`
  2. `1/16`
  3. `1/7`
  4. `1/4`
Show Answers Only

`C`

Show Worked Solution

`text(Strategy One:)`

♦♦ Mean mark 31%.

`text(Choose 1 shoe then find the probability)`

`text(the next choice is matching).`

`P` `= 1 xx 1/7`
  `= 1/7`

 

`P` `= text(Number of desired outcomes)/text(Number of possibilities)`
  `= 4/(\ ^8C_2)`
  `= 4/28`
  `= 1/7`

 
 `=>  C`

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 2007 HSC 9b

A pack of 52 cards consists of four suits with 13 cards in each suit.

  1. One card is drawn from the pack and kept on the table. A second card is drawn and placed beside it on the table. What is the probability that the second card is from a different suit to the first?  (1 mark)

  2. The two cards are replaced and the pack shuffled. Four cards are chosen from the pack and placed side by side on the table. What is the probability that these four cards are all from different suits?  (2 marks)

Show Answers Only
  1. `13/17`
  2. `0.105\ \ \ text{(to 3 d.p.)}`
Show Worked Solution

i.  `text(After 1st card is drawn)`

`text(# Cards left from another suit) = 39`

`text(# Cards left in pack) = 51`

`:. P\ text{(2nd card from a different suit)}`

`= 39/51`

`= 13/17`

 

ii.  `P\ text{(all 4 cards from different suits)}`

`= 52/52 xx 39/51 xx 26/50 xx 13/49`

`= 2197/(20\ 825)`

`= 0.1054…`

`= 0.105\ \ \ text{(to 3 d.p.)}`

Probability, 2ADV S1 2006 HSC 4c

A chessboard has 32 black squares and 32 white squares. Tanya chooses three different squares at random.

  1. What is the probability that Tanya chooses three white squares?  (2 marks)

  2. What is the probability that the three squares Tanya chooses are the same colour?.  (1 mark)

  3. What is the probability that the three squares Tanya chooses are not the same colour?  (1 mark)

Show Answers Only
  1. `5/42`
  2. `5/21`
  3. `16/21`
Show Worked Solution
i.  `text(P)(WWW)` `= 32/64 xx 31/63 xx 30/62`
  `= 5/42`

 

ii.  `text{P(same colour)}`

`= P(WWW) + P(BBB)`

`= 5/42 + 32/64 xx 31/63 xx 30/62`

`= 5/42 + 5/42`

`= 5/21`

 

iii.  `text{P(not all the same colour)}`

`= 1 – text{P(same colour)}`

`= 1 – 5/21`

`= 16/21`

Probability, 2ADV S1 2005 HSC 5d

A total of 300 tickets are sold in a raffle which has three prizes. There are 100 red, 100 green and 100 blue tickets.

At the drawing of the raffle, winning tickets are NOT replaced before the next draw.

  1. What is the probability that each of the three winning tickets is red?  (2 marks)

  2. What is the probability that at least one of the winning tickets is not red?  (1 mark)

  3. What is the probability that there is one winning ticket of each colour?  (2 marks)

Show Answers Only
  1. `1617/(44\ 551)`
  2. `(42\ 934)/(44\ 551)`
  3. `0.224\ \ text{(to 3 d.p.)}`
Show Worked Solution
i.   `P(R R R)` `= 100/300 xx 99/299 xx 98/298`
    `= 1617/(44\ 551)`

 

ii. `Ptext{(at least 1 winner NOT red)}`

`= 1 − P(R R R)`

`= 1− 1617/(44\ 551)`

`= (42\ 934)/(44\ 551)`

 

iii. `text(# Combinations of winning tickets)`

`= 3 xx 2 xx 1`

`= 6`
 

`:.P text{(one winner from each colour)}`

`= 6 xx 100/300 xx 100/299 xx 100/298`

`= 0.22446…`

`= 0.224\ \ text{(to 3 d.p.)}`

Probability, 2ADV S1 2004 HSC 1e

A packet contains 12 red,  8 green, 7 yellow and 3 black jellybeans.

One jellybean is selected from the packet at random.

What is the probability that the selected jellybean is red or yellow?  (2 marks)

Show Answers Only

`19/30`

Show Worked Solution

`text(12 R,  8 G,  7 Y,  3 B)`

`text(Total jellybeans) = 30`

`P text{(R or Y)}` `=\ text{(# Red + Yellow)}/text(Total jellybeans)`
  `= (12 + 7)/30`
  `= 19/30`

Probability, 2ADV S1 2008 HSC 9a

It is estimated that 85% of students in Australia own a mobile phone.

  1. Two students are selected at random. What is the probability that neither of them owns a mobile phone?  (2 marks)

  2. Based on a recent survey, 20% of the students who own a mobile phone have used their mobile phone during class time. A student is selected at random. What is the probability that the student owns a mobile phone and has used it during class time?   (1 mark)

Show Answers Only
  1. `9/400`
  2. `17/100`
Show Worked Solution

i.     `P(M) = 0.85`

COMMENT: `M^c` is syllabus notation for the complement of event `M`.

`P(M^c) = 1-0.85 = 0.15`

`:.\ P(M^c, M^c)` `= 15/100 * 15/100`
  `= 225/(10\ 000)`
  `= 9/400`

 

ii.  `text{P(owns mobile and used it)}`

`= P(M) xx  P\text{(used it)}`

`= 17/20 xx 20/100`

`= 17/100`

Probability, 2ADV S1 2009 HSC 5b

On each working day James parks his car in a parking station which has three levels. He parks his car on a randomly chosen level. He always forgets where he has parked, so when he leaves work he chooses a level at random and searches for his car. If his car is not on that level, he chooses a different level and continues in this way until he finds his car.    

  1. What is the probability that his car is on the first level he searches?     (1 mark)

  2. What is the probability that he must search all three levels before he finds his car?   (1 mark)

  3. What is the probability that on every one of the five working days in a week, his car is not on the first level he searches?    (1 mark)

Show Answers Only
  1. `1/3`
  2. `1/3`
  3. `32/243`
Show Worked Solution

i.  `P text{(1st chosen)} = 1/3`
 

ii.  `P text{(search 3 levels)}`

♦♦ Mean marks of 31% and 39% for part (ii) and (iii) respectively.

`= P text{(not 1st)} xx P text{(not 2nd)}`

`= 2/3 xx 1/2`

`= 1/3`
 

iii.  `P text{(not 1st for 5 days)}`

`= 2/3 xx 2/3 xx 2/3 xx 2/3 xx 2/3`

`= 32/243`

Probability, 2ADV S1 2010 HSC 4c

There are twelve chocolates in a box. Four of the chocolates have mint centres, four have caramel centres and four have strawberry centres. Ali randomly selects two chocolates and eats them.

  1. What is the probability that the two chocolates have mint centres?   (1 mark)

  2. What is the probability that the two chocolates have the same centre?    (1 mark)

  3. What is the probability that the two chocolates have different centres?   (1 mark)

Show Answers Only
  1. `1/11`
  2. `3/11`
  3. `8/11`
Show Worked Solution
i.    `P text{(2 mint)}` `=P(M_1) xx P(M_2)`
    `=4/12 xx 3/11`
    `=1/11`

 

ii.    `P text{(2 same)}` `=P(M_1 M_2) + P(C_1 C_2) + P(S_1 S_2)`
    `=1/11 + (4/12 xx 3/11) + (4/12 xx 3/11)`
    `=3/11`
 EXAM TIP: Using `1-Ptext{(2 things the same)}` is a quicker and easier strategy here.

 

iii.  `text(Solution 1)`

`P text{(2 diff)}` `=1\ – P text{(2 same)}`
  `=1\ – 3/11`
  `=8/11`

 
`text(Solution 2)`

`P text{(2 diff)}` `=P(M_1,text(not)\ M_2 text{)} + P(C_1,text(not)\ C_2 text{)} + P(S_1,text(not)\ S_2 text{)}`
  `=(4/12 xx 8/11) + (4/12 xx 8/11) + (4/12 xx 8/11)`
  `=32/121 + 32/121 + 32/121`
  `=8/11`

Probability, 2ADV S1 2012 HSC 13c

Two buckets each contain red marbles and white marbles. Bucket `A` contains 3 red and 2 white marbles. Bucket `B`  contains 3 red and 4 white marbles.

Chris randomly chooses one marble from each bucket. 

  1. What is the probability that both marbles are red?   (1 mark)

  2. What is the probability that at least one of the marbles is white?   (1 mark)

  3. What is the probability that both marbles are the same colour?   (2 marks)

Show Answers Only
  1. `9/35`
  2. `26/35`
  3. ` 17/35`
Show Worked Solution
i.    `P{(both red)}` `= P(R_1) xx P(R_2)`
  `= 3/5 xx 3/7`
  `= 9/35`

 

STRATEGY: When the term “at least” appears in a probability question, it is likely that `1-P(text{complement})` will solve the question more efficiently and with less chance of error, as shown in part (ii).
ii.    `Ptext{(at least one white)}` `= 1 – Ptext{(none white)}`
  `= 1 – P(R_1) xx P(R_2)`
  `= 1 – 9/35`
  `= 26/35`

 

iii.    `Ptext{(same colour)}` `= P(R_1 R_2) + P(W_1 W_2)`
  `= 9/35 + (2/5 xx 4/7)`
  `= 9/35 + 8/35`
  `= 17/35`

Probability, 2ADV S1 2013 HSC 15d

Pat and Chandra are playing a game. They take turns throwing two dice. The game is won by the first player to throw a double six. Pat starts the game.

  1. Find the probability that Pat wins the game on the first throw.     (1 mark)

  2. What is the probability that Pat wins the game on the first or on the second throw?     (2 marks)

  3. Find the probability that Pat eventually wins the game.     (2 marks)

Show Answers Only
  1. `1/36`
  2. `(2521)/(46\ 656)\ \ text(or)\ \ 0.054`
  3. `36/71`
Show Worked Solutions

i.   `P\ text{(Pat wins on 1st throw)}=P(W)`

`P(W)` `=P\ text{(Pat throws 2 sixes)}`
  `=1/6 xx 1/6`
  `=1/36`

 

ii.  `text(Let)\ P(L)=P text{(loss for either player on a throw)}=35/36`

`P text{(Pat wins on 1st or 2nd throw)}` 

♦♦ Mean mark 33%
MARKER’S COMMENT: Many students did not account for Chandra having to lose when Pat wins on the 2nd attempt.

`=P(W) + P(LL W)`

`=1/36\ + \ (35/36)xx(35/36)xx(1/36)`

`=(2521)/(46\ 656)`

`=0.054\ \ \ text{(to 3 d.p.)}`

 

iii.  `P\ text{(Pat wins eventually)}`

`=P(W) + P(LL\ W)+P(LL\ LL\ W)+ … `

`=1/36\ +\ (35/36)^2 (1/36)\ +\ (35/36)^2 (35/36)^2 (1/36)\ +…`

 
`=>\ text(GP where)\ \ a=1/36,\ \ r=(35/36)^2=(1225)/(1296)`

♦♦♦ Mean mark 8%!
 COMMENT: Be aware that diminishing probabilities and `S_oo` within the Series and Applications are a natural cross-topic combination.

 
`text(S)text(ince)\ |\ r\ |<\ 1:`

`S_oo` `=a/(1-r)`
  `=(1/36)/(1-(1225/1296))`
  `=1/36 xx 1296/71`
  `=36/71`

 

`:.\ text(Pat’s chances to win eventually are)\  36/71`.