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Statistics, EXT1 S1 2022 HSC 12e

A game consists of randomly selecting 4 balls from a bag. After each ball is selected it is replaced in the bag. The bag contains 3 red balls and 7 green balls. For each red ball selected, 10 points are earned and for each green ball selected, 5 points are deducted. For instance, if a player picks 3 red balls and 1 green ball, the score will be  `3xx10-1xx5=25`  points.

What is the expected score in the game?  (2 marks)

Show Answers Only

`-2`

Show Worked Solution

`text{Let}\ \ X=\ text{number of red balls selected}`

`X\ ~\ text{Bin}(4, 0.3)`

`text{Score}\ (X=4)=(0.3)^4xx40=0.324`

`text{Score}\ (X=3)=((4),(3))(0.3)^3(0.7)xx25=1.89`

`text{Score}\ (X=2)=((4),(2))(0.3)^2(0.7)^2xx10=2.646`

`text{Score}\ (X=1)=((4),(1))(0.3)(0.7)^3xx-5=-2.058`

`text{Score}\ (X=0)=(0.7)^4xx-20=-4.802`
 

`:.E(X)` `=0.324+1.89+2.646-2.058-4.802`  
  `=-2`  

♦♦ Mean mark 35%.

Statistics, EXT1 S1 2019 HSC 11f

Prize-winning symbols are printed on 5% of ice-cream sticks. The ice-creams are randomly packed into boxes of 8.

  1. What is the probability that a box contains no prize-winning symbols?  (1 mark)

  2. What is the probability that a box contains at least 2 prize-winning symbols?  (2 marks)

Show Answers Only
  1. `0.95^8`
  2. `5.72 text(%)`
Show Worked Solution

i.   `text(Chances of any stick winning:)`

`P(W) = 0.05`

`P(barW) = 0.95`

`P (text{In box of 8, all}\ barW)`

`= 0.95^8`

 

ii.   `P\ text{(at least two winners in a box)}`

`= 1 – P text{(1 winner)} – P text{(0 winners)}`

`= 1 – \ ^8 C_1 xx 0.95^7 xx 0.05^1 – \ ^8 C_0 xx 0.95^8`

`= 0.05724…`

`= 5.72 text{%   (to 2 d.p.)}`

Statistics, EXT1 S1 MET2 2008 14 MC

The minimum number of times that a fair coin can be tossed so that the probability of obtaining a head on each trial is less than 0.0005 is

A.     `9`

B.   `10`

C.   `11`

D.   `12`

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`C`

Show Worked Solution

`text(Let)\ \ X = text(Number of heads)`

`X ∼ text(Bin) (n, p) ∼ text(Bin) (n, 1/2)`

`P(X = n)` `< 0.0005`
`((n), (n)) (1/2)^n (1/2)^0` `< 0.0005`
`1/2^n` `<5/(10\ 000)`
`2^n` `>2000`
`ln 2^n` `>ln 2000`
`n` `>ln2000/ln2`
`n` `> 10.97`

 
`:. n_min = 11`

`=>   C`

Statistics, EXT1 S1 2016 HSC 11f

A darts player calculates that when she aims for the bullseye the probability of her hitting the bullseye is  `3/5`  with each throw.

  1. Find the probability that she hits the bullseye with exactly one of her first three throws.  (1 mark)

  2. Find the probability that she hits the bullseye with at least two of her first six throws.  (2 marks)

Show Answers Only
  1. `36/125`
  2. `2997/3125`
Show Worked Solution

i.   `P text{(exactly 1 bullseye)}`

`=\ ^3C_1 · (3/5)^1 (2/5)^2`

`= 3 · (3/5) · (4/25)`

`= 36/125`

 

ii.   `P text{(at least 2 from 6 throws)}`

`= 1 – [P(0) + P(1)]`

`= 1 – [(2/5)^6 + \ ^6C_1 · (3/5)^1· (2/5)^5]`

`= 1 – [128/3125]`

`= 2997/3125`

Statistics, EXT1 S1 2010 HSC 1f

Five ordinary six-sided dice are thrown.

What is the probability that exactly two of the dice land showing a four?

Leave your answer in unsimplified form.   (1 mark)

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`\ ^5C_2 (1/6)^2 (5/6)^3`

 

Show Worked Solution
COMMENT: Surprisingly, half of students sitting the exam got this wrong. Easily the most poorly answered part of Q1 in 2010.

`P(4) = 1/6`

`P(bar4) = 5/6`

`text(# Combinations of two 4’s) =\ ^5C_2`

`:.\ P text{(exactly two 4s)} =\ ^5C_2 (1/6)^2 (5/6)^3`

Statistics, EXT1 S1 2012 HSC 12c

Kim and Mel play a simple game using a spinner marked with the numbers  1, 2, 3, 4 and 5.
 

2012 12c
 

The game consists of each player spinning the spinner once. Each of the five numbers is equally likely to occur.

The player who obtains the higher number wins the game.

If both players obtain the same number, the result is a draw.

  1. Kim and Mel play one game. What is the probability that Kim wins the game?   (1 mark)

  2. Kim and Mel play six games. What is the probability that Kim wins exactly three games?    (2 marks)

Show Answers Only
  1. `2/5`
  2. `864/3125`
Show Worked Solution

i.  `text(Method 1)`

`P text{(Kim wins game)}`

`= P text{(K spins 5)} xx P text{(M<5)} + P text{(K spins 4)} xx P text{(M<4)} +\ …`

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

`= 10/25`

`= 2/5`

 

♦ Mean mark 37%

`text(Method 2)`

`P text{(Draw)} = 1/5`

`:.\ P text{(Not a draw)}= 1\ – 1/5=4/5`

 
`text(S)text(ince Kim and Mel have equal chance)`

`P text{(K wins)}` `= 1/2 xx 4/5`
  `= 2/5`

 

ii.  `P text{(Kim wins)} = 2/5`

`P text{(Kim doesn’t win)} = 3/5`
 

`text(After 6 games,)`

`P text{(Kim wins exactly 3)}`

`=\ ^6C_3 (2/5)^3 (3/5)^3`

`= (6!)/(3!3!) xx 8/125 xx 27/125`

`= 864/3125`