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Application To Probability Problems

10.7 Application to probability problems (EMCK5)

When needing to determine the probability that an event occurs, and the total number of arrangements of the sample space, \(S\), and the total number arrangements for the event, \(E\), are very large, the techniques used earlier in this chapter may no longer be practical. In this case, the probability may be determined using the fundamental counting principle. The probability of the event, \(E\), is the total number of arrangements of the event divided by the total number of arrangements of the sample space or \(\dfrac{n(E)}{n(S)}\).

Worked example 18: Personal Identification Numbers (PINs)

Every client of a certain bank has a personal identification number (PIN) which consists of four randomly chosen digits from 0 to 9.

  1. How many PINs can be made if digits can be repeated?
  2. How many PINs can be made if digits cannot be repeated?
  3. If a PIN is made by selecting four digits at random, and digits can be repeated, what is the probability that the PIN contains at least one eight?
  4. If a PIN is made by selecting four digits at random, and digits cannot be repeated, what is the probability that the PIN contains at least one eight?
  1. If digits can be repeated: you have \(\text{10}\) digits to choose from and you have to choose four times, therefore the number of possible PINs \(= 10^{4} = \text{10 000}\).
  2. If digits cannot be repeated: you have 10 digits for your first choice, nine for your second, eight for your third and seven for your fourth. Therefore, \[\text{the number of possible PINs } = 10 \times 9 \times 8 \times 7 = \text{5 040}\] .
  3. Let \(B\) be the event that at least one eight is chosen. Therefore the complement of \(B\) is the event that no eights are chosen.

    If no eights are chosen, there are only nine digits to choose from. Therefore, \(n(\text{not } B)= 9^{4} = \text{6 561}\).

    The total number of arrangements in the set, as calculated in Question 1, is \(\text{10 000}\). Therefore: \begin{align*} P(B) &= 1 - P(\text{not } B) \\ &= 1 - \dfrac{n(\text{not } B)}{n(S)} \\ &= 1 - \frac{\text{6 561}}{\text{10 000}} \\ &= \text{0,3439} \end{align*}

  4. Let \(B\) be the event that at least one eight is chosen. Therefore the complement of \(B\), is the event that no eights are chosen.

    If no eights are chosen, there are only 9 then 8 then 7 then 6 digits to choose from as we cannot repeat a digit once it is chosen. Therefore, \[n(\text{not } B)= 9 \times 8 \times 7 \times 6 = \text{3 024}\] .

    The total number of arrangements in the set, as calculated in Question 1, is \(\text{10 000}\). Therefore: \begin{align*} P(B) &= 1 - P(\text{not } B) \\ & 1 - \dfrac{n(\text{not } B)}{n(S)} \\ &= 1 - \frac{\text{3 024}}{\text{10 000}} \\ & = \text{0,6976} \end{align*}

Worked example 19: Number plates

The number plate on a car consists of any \(\text{3}\) letters of the alphabet (excluding the vowels and 'Q'), followed by any \(\text{3}\) digits (\(\text{0}\) to \(\text{9}\)). For a car chosen at random, what is the probability that the number plate starts with a 'Y' and ends with an odd digit?

Identify what events are counted

The number plate starts with a 'Y', so there is only \(\text{1}\) option for the first letter, and ends with an odd digit, so there are \(\text{5}\) options for the last digit (\(1; 3; 5; 7; 9\)).

Find the number of events

Use the counting principle. For the second and third letters, there are \(\text{20}\) possibilities (\(\text{26}\) letters in the alphabet, minus \(\text{5}\) vowels and 'Q'). There are \(\text{10}\) possibilities for the first and second digits.

\[\text{Number of events } = 1\times 20\times 20\times 10\times 10\times 5=\text{200 000}\]

Find the total number of possible number plates

Use the counting principle. This time, the first letter and last digit can be anything.

\[\text{Total number of choices } = 20\times 20\times 20\times 10\times 10\times 10=\text{8 000 000}\]

Calculate the probability

The probability is the number of outcomes in the event, divided by the total number of outcomes in the sample space.

\[\text{Probability } = \frac{\text{200 000}}{\text{8 000 000}}=\frac{1}{40}=\text{0,025}\]

Worked example 20: Probability of word arrangements

Refer to worked example 16 for context. If you take the word, 'BASSOON' and you randomly rearrange the letters, what is the probability that the word starts and ends with the same letter if repeated letters are treated as identical?

If the word starts and ends with the same letter, there are a total number of 120 possible arrangements (from worked example 16). Let this event \(= A\).

The total number of possible arrangements if repeated letters are treated as identical \(= \text{1 260}\) (from worked example 16).

Therefore, the probability of an arrangement beginning and ending with the same letter \[= \dfrac{n(A)}{n(S)} = \dfrac{120}{\text{1 260}} = \text{0,1}\]

Solving probability problems using the fundamental counting principle

Exercise 10.8

A music group plans a concert tour in South Africa. They will perform in Cape Town, Port Elizabeth, Pretoria, Johannesburg, Bloemfontein, Durban and East London.

In how many different orders can they plan their tour if there are no restrictions?
\[7! = \text{5 040} \text{ different orders are possible}\]
In how many different orders can they plan their tour if their tour begins in Cape Town and ends in Durban?

This reduces the available objects (cities) by two, hence

\[5! = \text{120} \text{ different orders are possible}\]
If the tour cities are chosen at random, what is the probability that their performances in Cape Town, Port Elizabeth, Durban and East London happen consecutively? Give your answer correct to 3 decimal places.

If these cities are grouped together, they can be treated as a single object in the arrangement, hence there are \(4!\) different ways to order the objects. Within the grouped cities, there are \(4!\) different ways to order them. Therefore, there are \(4! \times 4! = \text{576} \text{ different orders}\)

Therefore, the probability of any of the orders having Cape Town, Port Elizabeth, Durban and East London happen consecutively is:

\begin{align*} P(\text{the \text{4} cities grouped}) &= \dfrac{n(\text{the \text{4} cities grouped})}{n(\text{total possible orders})} \\ & = \dfrac{\text{576}}{\text{5 040}} = \text{0,114} \end{align*}

A certain restaurant has the following course options available for a three-course set menu:

STARTERSMAINSDESSERTS
Calamari saladFried chickenIce cream and chocolate sauce
OystersCrumbed lamb chopsStrawberries and cream
Fish in garlic sauceMutton BobotieMalva pudding with custard
Chicken schnitzelPears in brandy sauce
Vegetable lasagne
Chicken nuggets
How many different set menus are possible?
\[3 \times 6 \times 4 = \text{72} \text{ different set menus}\]
What is the probability that a set menu includes a chicken course?
\begin{align*} n(\text{set menu with chicken})&= 3 \times 3 \times 4 = 36 \\ n(\text{total set menus})&= 72 \\ \text{Therefore } P(\text{set menu with chicken}) &= \frac{36}{72} = \text{0,5} \end{align*}

Eight different pairs of jeans and 5 different shirts hang on a rail.

In how many different ways can the clothes be arranged on the rail?
\[13! = \text{6 227 020 800} \text{ different ways}\]
In how many ways can the clothing be arranged if all the jeans hang together and all the shirts hang together?
The shirts and jeans form two groups, which can be arranged \(2!\) ways. The five shirts can be arranged \(5!\) ways and the eight pairs of jeans can be arranged \(8!\) ways. \[2! \times 8! \times 5! = \text{9 676 800}\]
What is the probability, correct to three decimal places, of the clothing being arranged on the rail with a shirt at one end and a pair of jeans at the other?
  • The five different choices of shirt and eight different choices of pairs of jeans can form \(5 \times 8\) different arrangements at the ends of the rail.

  • There are \(2!\) different ways to arrange a shirt at one end and a pair of jeans on the other: S - - - - - - - - - - - J and J - - - - - - - - - - - S

  • If a shirt is at one end and a pair of jeans at the other, there remains \(11!\) different arrangements of the remaining clothing items.

Therefore, there are:

\(2 \times 8 \times 5 \times 11! = \text{3 193 344 000} \text{ different ways to arrange the clothing}\)

The probability of a clothing arrangement with a shirt at one end and a pair of jeans at the other \(= \frac{\text{3 193 344 000}}{\text{6 227 020 800}} = \text{0,513}\)

A photographer places eight chairs in a row in his studio in order to take a photograph of the debating team. The team consists of three boys and five girls.

In how many ways can the debating team be seated?
\[8! = \text{40 320}\]
What is the probability that a particular boy and a particular girl sit next to each other?
Regard the particular boy and girl as one group. The group can be seated \(2!\) ways. The number of ways that this group and the remaining 6 people can sit \(= 7!\). Therefore the total number of ways this particular boy and girl can sit together in the photograph \(= 2! × 7! = \text{10 080}\). The probability of a particular boy and girl sitting together \(= \frac{\text{10 080}}{\text{40 320}} = \text{0,25}\)

If the letters of the word 'COMMITTEE' are randomly arranged, what is the probability that the letter arrangements start and end with the same letter?

  • There are 2 M’s, 2 T’s and 2 E’s and a total of 9 letters.
  • Total number of letter arrangements \(= \dfrac{9!}{2! \times 2! \times 2!} = \text{45 360}\)
  • Possibilities of the first and last letter being the same:
    • M(COITTEE)M \\ Total of 7 letters of which there are 2E’s and 2T’s \\ Number of letter arrangements \(= \dfrac{7!}{2! \times 2!} = \text{1 260}\)
    • T(MCOIEEM)T\\ Total of 7 letters of which there are 2E’s and 2M’s \\ Number of letter arrangements \(= \dfrac{7!}{2! \times 2!} = \text{1 260}\)
    • E(TTMCOIM)E \\ Total of 7 letters of which there are 2T’s and 2M’s \\ Number of letter arrangements \(= \dfrac{7!}{2! \times 2!} = \text{1 260}\)
  • Total number of letter arrangements if the letter arrangement starts and ends with the same letter \(= 3 \times \text{1 260} = \text{3 780}\).
\[P(\text{first and last letter the same}) = \dfrac{\text{3 780}}{\text{45 360}} = \dfrac{1}{12}\]

Four different Mathematics books, three different Economics books and two different Geography books are arranged on a shelf. What is the probability that all the books of the same subject are arranged next to each other?

Total number of different ways the books can be arranged \(= 9! = \text{362 880}\).There are 3 subjects of books which can be arranged \(3!\) ways.

Therefore, the total number of arrangements if the subjects are arranged together \(= 3! \times 4! \times 3! \times 2! = \text{1 728}\)

\[P(\text{books of the same subject next to each other}) = \dfrac{ \text{1 728}}{\text{362 880}} = \dfrac{1}{210}\]

A number plate is made up of three letters of the alphabet (excluding F and S) followed by three digits from 0 to 9. The numbers and letters can be repeated. Calculate the probability that a randomly chosen number plate:

starts with the letter D and ends with the digit 3.

Total number of arrangements \(= 24^{3} \times 10^{3}\)

Total number of arrangements beginning with D and ending with \(3 = 24^{2} \times 10^{2}\)

\[P(\text{first D and last 3}) = \dfrac{24^{2} \times 10^{2}}{24^{3} \times 10^{3}} = \dfrac{1}{240}\]
has precisely one D.
The 'D' could be at the first, second or third positions. The other two letters cannot include a 'D', leaving 23 other letters. Therefore there are \[23^{2} \times 10^{3} \times 3 \text{ different arrangements containing only one D}\] \[\text{Therefore } P(\text{only one D}) = \dfrac{23^{2} \times 10^{3} \times 3}{24^{3} \times 10^{3}} = \dfrac{\text{529}}{\text{4 608}}\]
contains at least one 5.
\begin{align*} P(\text{contains at least one }5)&= 1 - P(\text{no } 5 \text{s}) \\ &= 1 - \dfrac{24^{3} \times 9^{3}}{24^{3} \times 10^{3}} \\ &= 1 - \text{0,729} = \text{0,271} \end{align*}

In the 13-digit identification (ID) numbers of South African citizens:

  • The first six numbers are the birth date of the person in YYMMDD format.
  • The next four digits indicate gender, with \(\text{5 000}\) and above being male and \(\text{0 001}\) to \(\text{4 999}\) being female.
  • The next number is the country ID; 0 is South Africa and 1 is not.
  • The second last number used to be a racial identifier but it is now 8 for everybody.
  • The last number is a control digit, which verifies the rest of the number.
Assume that the control digit is a randomly generated digit from 0 to 9 and ignore the fact that leap years have an extra day.
Calculate the total number of possible ID numbers.
For all available arrangements, if an ID number is structured ABCDEFGHIJKLM:
  • A and B are any digits between 00 and 99 (year)
  • C and D are any digits from 01 to 12 (month)
  • E and F are any digits from 01 to 28, 30 or 31 dependent on month (day)
  • G, H, I and J are any digits from 0001 to 9999 (gender)
  • K is either a 0 or 1
  • L is an 8
  • M is any digit between 0 and 9
Therefore, the total number of possible ID numbers for 30-day months is: \[100 \times 4 \times 30 \times 9999 \times 2 \times 1 \times 10 = \text{2 399 760 000}\] Therefore, the total number of possible ID numbers for 31-day months is: \[100 \times 7 \times 31 \times 9999 \times 2 \times 1 \times 10 = \text{4 339 566 000}\] Therefore, the total number of possible ID numbers for February is: \[100 \times 1 \times 28 \times 9999 \times 2 \times 1 \times 10 = \text{559 944 000}\] Therefore, the total number of possible ID numbers is: \[\text{2 399 760 000} + \text{4 339 566 000} + \text{559 944 000} = \text{9 239 076 000}\]
Calculate the probability that a randomly generated ID number is of a South African male born during the 1980s. Write your answer correct to two decimal places.

There is a lot of information in the problem and we can simplify it by identifying the relevant information. We want to calculate the probability that an ID number is for a South African male born during the 1980s. This means that we have to look at the digits for country, gender and year of birth.

If an ID number is structured ABCDEFGHIJKLM:

  • For a South African, K is 0.
  • For a male, GHIJ are from \(\text{5 000}\) to \(\text{9 999}\).
  • For someone born during the 1980s, AB are between 80 and 89.

This gives \(1 \times \text{5 000} \times 10 = \text{50 000}\) combinations for ABGHIJK

Without any restrictions, the total combinations for ABGHIJK is \(2 \times \text{9 999} \times 100 = \text{1 999 800}\).

\[\text{Therefore } P(\text{SA male 80s}) = \dfrac{\text{50 000}}{\text{1 999 800}} = \text{0,025}\]