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Probability, STD2 S2 2015 HSC 21 MC

Four cards, labelled 2, 3, 5 and 7, are placed on a table to form a four‑digit number.

How many different numbers greater than 3000 can be formed?

  1.    6
  2.    12
  3.    18
  4.    24
Show Answers Only

`C`

Show Worked Solution

`#\ text{Combinations greater than 3000}`

♦ Mean mark 48%.

`= 3 xx 3 xx 2 xx 1`

`= 18`

`=> C`

Probability, STD2 S2 2005 HSC 20 MC

Dave’s school has computer security codes made up of four digits (eg 0773). Juanita’s school has computer security codes made up of five digits (eg. 30 568).

How many more codes are available at Juanita’s school than at Dave’s school?

  1.    10
  2.    50
  3.    90 000
  4.    100 000
Show Answers Only

`C`

Show Worked Solution

`text(# Codes at Dave’s school)`

`= 10 × 10 × 10 × 10`

`= 10\ 000`

`text(# Codes at Juanita’s school)`

`= 10^5`

`= 100\ 000`

`:.\ text(Extra Codes)` `= 100\ 000 − 10 \ 000`
  `= 90\ 000`

`=>  C`

Probability, STD2 S2 2008 HSC 24b

Three-digit numbers are formed from five cards labelled  1,  2,  3,  4  and  5.

  1. How many different three-digit numbers can be formed?  (1 mark)

  2. If one of these numbers is selected at random, what is the probability that it is odd?   (1 mark)

  3. How many of these three-digit numbers are even?   (1 mark)

  4. What is the probability of randomly selecting a three-digit number less than 500 with its digits arranged in descending order?   (2 marks)

Show Answers Only
  1. `60`
  2. `3/5`
  3. `24`
  4. `1/15`
Show Worked Solution

i.  `text(# Different numbers)`

♦ Mean mark 45%.

`= 5 xx 4 xx 3`

`= 60`

 

ii.  `text(The last digit must be one of the)`

`text(5 numbers, of which 3 are odd)`

`:.\ text{P(odd)} = 3/5`

 

iii. `text{P(even)} = 1- text{P(odd)} = 2/5`

♦ Mean mark 48%.

`:.\ text(Number of even numbers)`

`= 2/5 xx 60`

`= 24`

 

iv.  `text(The numbers that satisfy the criteria:)`

♦♦♦ Mean mark 10%.

`432, 431, 421, 321`

`:.\ text{P(selection)} = 4/60 = 1/15`

Probability, STD2 S2 2008 HSC 18 MC

New car registration plates contain two letters followed by two numerals followed by two more letters eg  AC 12 DC. Letters and numerals may be repeated.

Which of the following expressions gives the correct number of car registration plates that begin with the letter M?

  1.    `26^3 xx 10^2` 
  2.    `25^3 xx 10^2`
  3.    `26^4 xx 10^2`
  4.    `25^4 xx 10^2 `
Show Answers Only

`A`

Show Worked Solution

`text(# Plates beginning with M)`

`= 1 xx 26 xx 10 xx 10 xx 26 xx 26`

`= 26^3 xx 10^2`

`=>  A`

Probability, STD2 S2 2010 HSC 26a

A design of numberplates has a two-digit number, two letters and then another two-digit number, for example

2010 26a1

  1. How many different numberplates are possible using this design?   (1 mark)

Jo can order a numberplate with ‘JO’ in the middle but will have to have randomly selected numbers on either side.

Jo’s birthday is 30 December 1992, so she would like a numberplate with either 

2010 26a2

  1. What is the probability that Jo is issued with one of the numberplates she would like?   (2 marks)

Show Answers Only
  1. `6\ 760\ 000`
  2. `P = 1/5000`
Show Worked Solution
♦ Mean mark 41%
i.    `text(# Combinations)` `=10 xx 10 xx 26 xx 26 xx 10 xx 10`
    `=6\ 760\ 000`

 

♦♦ Mean mark 30%
IMPORTANT: Since the middle letters of “JO” can be guaranteed, the focus becomes purely on the 4 surrounding digits.

ii.   `text(# Possible numberplates)`

`=10 xx 10 xx 10 xx 10`

`=10\ 000`
 

`:.P (30\ text(JO)\ 12) + P (19\ text(JO)\ 92)`
`= 1/(10\ 000) + 1/(10\ 000)`
`= 1/5000`

Probability, STD2 S2 2013 HSC 29c

Mary is designing a website that requires unique logins to be generated.

She plans to generate the logins using two capital letters from the alphabet followed by a series of numerals from 0 to 9 inclusive. All logins will have the same number of numerals. Repetition of letters and numerals is allowed.

What is the minimum number of numerals required for each login so that Mary can generate at least 3 million logins?

Justify your answer with suitable calculations.   (2 marks)

Show Answers Only

 `4`

Show Worked Solution
♦ Mean mark 34%
COMMENT: Students can use their rough working to find an appropriate “number of numerals” where their answer should start.

`text(# Combinations must be > 3 million:)`

`text(If we have 3 numerals,)`

`text(# Combinations)` `=26 xx 26 xx 10 xx 10 xx 10`
  `=676\ 000 < 3\ 000\ 000`

`=> text(need more numeral(s) )`

 

`text(If we have 4 numerals,)`

`text(# Combinations)` `=26 xx 26 xx 10 xx 10 xx 10 xx 10`
  `=6760\ 000 > 3\ 000\ 000`

 

`:.\ text(Minimum number of numerals) = 4`

 

Probability, STD2 S2 2009 HSC 23b

A personal identification number (PIN) is made up of four digits. An example of a PIN is 

2009 23b 

  1. When all ten digits are available for use, how many different PINs are possible?    (1 mark)

  2. Rhys has forgotten his four-digit PIN, but knows that the first digit is either 5 or 6.   
  3. What is the probability that Rhys will correctly guess his PIN in one attempt?   (1 mark)

Show Answers Only
  1. `10\ 000`
  2. `1/2000`
Show Worked Solution
♦ Mean mark 43%
i.  `#\ text(Combinations)` `= 10 xx 10 xx 10 xx 10`
    `= 10\ 000`

 

♦♦♦ Mean mark 18%
MARKER’S COMMENT: A common error is finding the number of possible combinations but not then calculating the probability.
 ii. `#\ text(Combinations)` `= 2 xx 10 xx 10 xx 10`
    `= 2000`
     
  `P text{(Correct PIN)}` `= text{# Correct PINS}/text(# Combinations)`
    `=1/2000`

Probability, STD2 S2 2012 HSC 26a

Postcodes in Australia are made up of four digits eg 2040.  

  1. How many different postcodes beginning with a 2 are possible?  (1 mark)

Peta remembers that the first two digits of a town’s postcode are 2 and then 4. She is unable to remember the rest of the postcode.
 

2012 26a

  1. What is the probability that Peta guesses the correct postcode? (1 mark)

Show Answers Only
  1. `1000`
  2. `P\ (text(Correct)) =\ 1/100`
Show Worked Solution

i.  `text(Different postcodes begining with)\ 2`

♦ Mean marks of 43% and 41% for parts (i) and (ii) respectively.

`=1 xx 10 xx 10 xx 10`

`=1000`

 

ii.  `text(Number of postcodes beginning with)\ 2,4`

`= 1 xx 1 xx 10 xx 10`

`=100`

`:. P\ (text(Correct)) = 1/100`