Application of FCP to Probability Problems (Grade 12 NSC Matric Mathematics): Revision Notes

Worked Example 1: Personal identification numbers (PINs)

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

Part 1: How many PINs can be made if digits can be repeated?

Solution: When digits can be repeated, you have 10 choices (0-9) for each of the four positions.

Using the counting principle: $10 \times 10 \times 10 \times 10 = 10^4 = 10,000$ possible PINs

Part 2: How many PINs can be made if digits cannot be repeated?

Solution: When digits cannot be repeated, your choices decrease for each position:

• First digit: 10 choices
• Second digit: 9 choices (can't repeat the first)
• Third digit: 8 choices (can't repeat first two)
• Fourth digit: 7 choices (can't repeat first three)

Total PINs = $10 \times 9 \times 8 \times 7 = 5,040$

Part 3: If a PIN is made by selecting four digits at random with repetition allowed, what is the probability that the PIN contains at least one eight?

Solution: For "at least one" problems, use the complement rule. It's easier to calculate the probability of "no eights" and subtract from 1.

Let B = event that at least one eight is chosen Let not B = event that no eights are chosen

If no eights are chosen, you only have 9 digits (0-7, 9) to choose from. $n(\text{not B}) = 9^4 = 6,561$

Total arrangements (from Part 1) = 10,000

$P(B) = 1 - P(\text{not B})$ $P(B) = 1 - \frac{n(\text{not B})}{n(S)}$ $P(B) = 1 - \frac{6,561}{10,000}$ $P(B) = 1 - 0.6561 = :success[0.3439]$

Part 4: If a PIN is made without repetition, what is the probability that it contains at least one eight?

Solution: Again, use the complement rule.

If no eights are chosen with no repetition allowed:

• First position: 9 choices (excluding 8)
• Second position: 8 choices
• Third position: 7 choices
• Fourth position: 6 choices

$n(\text{not B}) = 9 \times 8 \times 7 \times 6 = 3,024$

Total arrangements (from Part 2) = 5,040

• $P(B) = 1 - P(\text{not B})$
• $P(B) = 1 - \frac{3,024}{5,040}$
• $P(B) = 1 - 0.6 = 0.4$

Worked Example 2: Number plates

Question: The number plate on a car consists of any 3 letters of the alphabet (excluding vowels and 'Q'), followed by any 3 digits (0 to 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?

Solution:

Step 1: Identify what events are counted The number plate must start with 'Y' (only 1 option for first letter) and end with an odd digit (5 options: 1, 3, 5, 7, 9).

Step 2: Find the number of favourable events

• First letter: 1 choice (Y)
• Second and third letters: 20 possibilities each (26 letters minus 5 vowels and Q)
• First and second digits: 10 possibilities each (0-9)
• Third digit: 5 choices (odd digits only)

Number of favourable events = $1 \times 20 \times 20 \times 10 \times 10 \times 5 = 200,000$

Step 3: Find total number of possible number plates

For any number plate with no restrictions:

• Each letter position: 20 possibilities
• Each digit position: 10 possibilities

Total number of choices = $20 \times 20 \times 20 \times 10 \times 10 \times 10 = 8,000,000$

Step 4: Calculate the probability

Probability = $\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

Probability = $\frac{200,000}{8,000,000} = \frac{1}{40} = 0.025$

Worked Example 3: Word arrangements

Question: If you take the word 'BASSOON' and 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?

Solution:

From previous calculations, we know:

• Total arrangements where word starts and ends with same letter = 120
• Total possible arrangements when repeated letters are treated as identical = 1,260

Therefore, the probability = $\frac{n(A)}{n(S)} = \frac{120}{1,260} = 0.1$