10.7 Application to probability problems

\[7! = \text{5 040} \text{ different orders are possible}\]

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

\[5! = \text{120} \text{ different orders are possible}\]

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

\[3 \times 6 \times 4 = \text{72} \text{ different set menus}\]

\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*}

\[13! = \text{6 227 020 800} \text{ different ways}\]

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. \[2! \times 8! \times 5! = \text{9 676 800}\]

• 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 \(= \frac{\text{3 193 344 000}}{\text{6 227 020 800}} = \text{0,513}\)

\[8! = \text{40 320}\]

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 \(= \frac{\text{10 080}}{\text{40 320}} = \text{0,25}\)

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

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

\[P(\text{first D and last 3}) = \dfrac{24^{2} \times 10^{2}}{24^{3} \times 10^{3}} = \dfrac{1}{240}\]

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 \[23^{2} \times 10^{3} \times 3 \text{ different arrangements containing only one D}\] \[\text{Therefore } P(\text{only one D}) = \dfrac{23^{2} \times 10^{3} \times 3}{24^{3} \times 10^{3}} = \dfrac{\text{529}}{\text{4 608}}\]

\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*}

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: \[\text{2 399 760 000} + \text{4 339 566 000} + \text{559 944 000} = \text{9 239 076 000}\]

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 \(2 \times \text{9 999} \times 100 = \text{1 999 800}\).

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