Probability and Combinatorics

Introduction

As-salamu alaykum, students. I am Ustad Bilal Ahmed. For many of you studying in Lahore, Karachi, or Islamabad, the world often seems a mix of predictable patterns and sheer randomness. This is the domain of Probability and Combinatorics. It is not just an abstract topic in your 9709 syllabus; it is the mathematical language we use to understand risk, make predictions, and model the world around us, from the outcome of a cricket match to the fluctuations of the Pakistan Stock Exchange.

Combinatorics is the science of counting. Before we can calculate the probability of an event, we must know how many possible outcomes exist. How many ways can the batting order for the Pakistan cricket team be arranged? How many different committees can be formed from a group of students at NUST? These are questions answered by permutations and combinations.

Probability then builds upon this foundation, giving us a formal framework to measure the likelihood of specific outcomes. We will explore the fundamental axioms that govern probability, learn how to handle conditional events (the chance of rain *given* it is cloudy), and study powerful tools like the Binomial, Geometric, and Poisson distributions. Mastering this topic is not just about passing Paper 5; it is about developing a new way of thinking logically about an uncertain world.

Core Theory

Let us begin with the fundamentals of counting, which form the bedrock of probability.

Permutations and Combinations

The first question to always ask is: Does the order matter?

1. Permutations (nPr): Use this when the order of selection is important. Think of it as an arrangement. For example, arranging books on a shelf, or the finishing order of a 100m race. The number of ways to arrange 'r' items from a set of 'n' distinct items is:

nPr = n! / (n-r)!

For instance, the number of ways to award Gold, Silver, and Bronze medals (order matters) from 8 athletes is 8P3 = 8! / (8-3)! = 8! / 5! = 8 × 7 × 6 = 336.

1. Combinations (nCr): Use this when the order of selection is irrelevant. Think of it as a selection or a committee. For example, choosing 3 students from a class of 10 to form a team. The number of ways to choose 'r' items from a set of 'n' distinct items is:

nCr = n! / (r! * (n-r)!)

Using the student example, choosing 3 from 10 is 10C3 = 10! / (3! * 7!) = (10 × 9 × 8) / (3 × 2 × 1) = 120. Notice this is much smaller than 10P3, as the committee of Ali, Babar, and Cruz is the same as Babar, Ali, and Cruz.

Fundamental Probability

With counting mastered, we can define probability. For an event A, P(A) = (Number of favourable outcomes) / (Total number of outcomes).

• Addition Rule: For any two events A and B, the probability of A or B occurring is:

P(A∪B) = P(A) + P(B) - P(A∩B)

The P(A∩B) term corrects for double-counting the outcomes that are in both A and B. If the events are mutually exclusive (they cannot happen at the same time, e.g., rolling a 5 and a 6 on a single die), then P(A∩B) = 0, and the formula simplifies to P(A∪B) = P(A) + P(B).

• Conditional Probability: This is the probability of event A happening, *given that* event B has already happened. The notation is P(A|B).

P(A|B) = P(A∩B) / P(B)

This formula is the foundation for tree diagrams. Imagine drawing two cards from a deck without replacement. The probability of the second card being a King *depends on* what the first card was.

• Independence: Two events are independent if the occurrence of one does not affect the probability of the other. For independent events A and B:

P(A∩B) = P(A) × P(B)

This is a critical test for independence. For example, flipping a coin and rolling a die are independent events.

Discrete Random Variables

A discrete random variable, X, is a variable whose value is a numerical outcome of a random phenomenon. We can define its expectation (mean) and variance.

• Expectation E(X): The long-run average value of X.

E(X) = Σ [x * P(X=x)]

• Variance Var(X): A measure of the spread or dispersion of the distribution.

Var(X) = E(X²) - [E(X)]², where E(X²) = Σ [x² * P(X=x)]

Standard Distributions

These are pre-defined models for common scenarios.

1. Binomial Distribution B(n, p): Use this when you have:

• A fixed number of trials, n.
• Each trial has two outcomes: "success" or "failure".
• The probability of success, p, is constant for each trial.
• The trials are independent.

The probability of getting exactly 'r' successes in 'n' trials is:

P(X=r) = nCr * p^r * (1-p)^(n-r)

Mean: E(X) = np

Variance: Var(X) = np(1-p)

1. Geometric Distribution: Use this to model the number of trials, X, required to get the first success.

The probability that the first success occurs on the 'r'-th trial is:

P(X=r) = (1-p)^(r-1) * p

Mean: E(X) = 1/p

1. Poisson Distribution Po(λ): Use this for events that occur randomly and independently in a fixed interval of space or time, at a constant average rate, λ (lambda).

The probability of 'r' events occurring in that interval is:

P(X=r) = (e^-λ * λ^r) / r!

A key feature is that its mean and variance are equal:

Mean: E(X) = λ

Variance: Var(X) = λ

The Poisson distribution can be used to approximate the Binomial distribution B(n, p) when 'n' is large (typically n > 50) and 'p' is small (typically p < 0.1). In this case, you set λ = np.

Key Definitions

• Permutation (nPr): An arrangement of 'r' items from 'n' where order is significant. Formula: n! / (n-r)!
• Combination (nCr): A selection of 'r' items from 'n' where order is not significant. Formula: n! / (r!(n-r)!)
• Mutually Exclusive Events: Two events that cannot occur simultaneously. P(A∩B) = 0.
• Independent Events: The occurrence of one event does not influence the probability of the other. Test: P(A∩B) = P(A) × P(B).
• Conditional Probability P(A|B): The probability of A given that B has occurred. Formula: P(A∩B) / P(B).
• E(X) (Expected Value): The mean or average of a random variable X. E(X) = ΣxP(X=x).
• Var(X) (Variance): The measure of the spread of a random variable's distribution. Var(X) = E(X²) - [E(X)]².
• Binomial Distribution B(n,p): Models the number of successes in 'n' fixed, independent trials. P(X=r) = nCr * p^r * (1-p)^(n-r).
• Geometric Distribution: Models the number of trials up to the first success. P(X=r) = (1-p)^(r-1) * p.
• Poisson Distribution Po(λ): Models the number of random events in a fixed interval, occurring at an average rate λ. P(X=r) = (e^-λ * λ^r) / r!.

Worked Examples (Pakistani Context)

Example 1: Pure Combinatorics

Question: Find the number of different arrangements of the 9 letters in the word 'PAKISTANI'.

(i) Find the total number of arrangements.

(ii) Find the number of arrangements in which the two 'A's are together and the two 'I's are together.

(iii) Find the number of arrangements in which the vowels (A, A, I, I) are all together.

Solution:

The word is PAKISTANI (9 letters). We have repeated letters: A (2 times), I (2 times).

(i) Total arrangements:

This is a permutation with repetition. The formula is n! / (p!q!...), where p, q are the frequencies of repeated items.

Total arrangements = 9! / (2! * 2!) = 362880 / (2 * 2) = 90720.

(This is a common first step and usually worth 2 marks).

(ii) 'A's together, 'I's together:

Treat 'AA' as a single block (let's call it X) and 'II' as a single block (Y).

Now our items to arrange are P, K, S, T, N, X, Y. This is 7 distinct items.

Number of ways to arrange these 7 items = 7! = 5040.

The letters within the 'AA' block can be arranged in 2!/2! = 1 way.

The letters within the 'II' block can be arranged in 2!/2! = 1 way.

So, total arrangements = 7! × 1 × 1 = 5040.

(iii) Vowels together:

Treat the four vowels (A, A, I, I) as a single block (V).

Now our items to arrange are P, K, S, T, N, V. This is 6 distinct items.

Number of ways to arrange these 6 items = 6! = 720.

Within the vowel block 'V', the letters A, A, I, I can be arranged. This is an arrangement of 4 letters with repetitions.

Arrangements within the block = 4! / (2! * 2!) = 24 / 4 = 6.

Total arrangements = (Ways to arrange the blocks) × (Ways to arrange within the block)

Total = 6! × [4! / (2! * 2!)] = 720 × 6 = 4320.

Example 2: Applied Probability (PSL Cricket)

Question: In the Pakistan Super League (PSL), a fast bowler, Haris, has a probability of 0.6 of bowling a 'dot ball' (a ball from which no run is scored) on any given delivery, independently.

(i) Find the probability that in an over of 6 balls, Haris bowls exactly 4 dot balls.

(ii) Find the probability that Haris's first dot ball of the over is the 3rd ball he bowls.

The average number of sixes hit per over at the Gaddafi Stadium in evening matches is 1.2.

(iii) State the conditions needed for a Poisson distribution to be a suitable model and calculate the probability that in a randomly chosen over, exactly 3 sixes are hit.

Solution:

(i) Binomial Distribution:

The situation fits a Binomial model. We have a fixed number of trials (n=6 balls), two outcomes (dot ball or not), and a constant probability of success (p=0.6).

Let X be the number of dot balls in an over. X ~ B(6, 0.6).

We need P(X=4).

Using the formula P(X=r) = nCr * p^r * (1-p)^(n-r):

P(X=4) = 6C4 * (0.6)^4 * (0.4)^(6-4)

P(X=4) = 15 * (0.1296) * (0.16)

P(X=4) = 0.31104 (or 0.311 to 3 s.f.)

(ii) Geometric Distribution:

This asks for the *first* success on the 3rd trial. This is a Geometric distribution scenario.

Let Y be the ball number on which the first dot ball occurs. p=0.6, 1-p=0.4.

We need P(Y=3). This means Fail, Fail, Success.

P(Y=3) = (0.4) * (0.4) * (0.6)

P(Y=3) = (0.4)² * (0.6) = 0.16 * 0.6 = 0.096.

(Using the formula (1-p)^(r-1) * p gives (0.4)^(3-1) * 0.6 = 0.096).

(iii) Poisson Distribution:

Conditions for Poisson: Events (sixes) must occur independently of each other, and at a constant average rate.

Let S be the number of sixes in an over. The average rate is given, so λ = 1.2.

S ~ Po(1.2).

We need P(S=3).

Using the formula P(S=r) = (e^-λ * λ^r) / r!:

P(S=3) = (e^-1.2 * 1.2³) / 3!

P(S=3) = (0.30119 * 1.728) / 6

P(S=3) = 0.52046 / 6 = 0.0867 (to 3 s.f.)

Exam Technique

For your Statistics 1 Paper (9709/5), this topic is a major component. Here is my advice as an examiner.

1. Read the Question Carefully: This is the most critical advice. Are you arranging or selecting? Is it "at least 3" or "exactly 3"? Is it with or without replacement? Underline these key phrases. A single word can change the entire question from a permutation to a combination.

1. Identify the Model: Before you write any formula, ask yourself: Is this a simple P(A)/Total? Is it conditional? Or does it fit a standard distribution? If so, which one?

• Binomial: Fixed trials (n), constant probability (p).
• Geometric: First success.
• Poisson: Average rate (λ) over an interval.

Explicitly state your model and its parameters, e.g., "Let X be the number of faulty items, X ~ B(10, 0.05)". This earns method marks.

1. Show Your Working: Do not just write down the answer from your calculator. For a combination question, write down 10C3. For a Binomial calculation, write the full formula: 10C3 * (0.1)³ * (0.9)⁷. This ensures you get method marks even if you make a calculation error.

1. Common Mistakes:

• Confusing nPr and nCr: Remember, Permutation has 'Arrangement' in it (P-A-R). Combination is for 'Committee' (C-C).
• Forgetting repetitions: When arranging letters in a word like 'MULTAN', don't forget to divide by the factorial of counts of repeated letters.
• Poisson/Binomial Approximation: When approximating B(n,p) with Po(λ), remember to calculate λ = np. You must also state that you are using an approximation because n is large and p is small.
• Conditional Probability: In tree diagrams, remember that the probabilities on the second set of branches are conditional. When using P(A|B) = P(A∩B)/P(B), ensure you identify the correct 'given' event B for the denominator.

1. "Show that" vs "Find": If a question asks you to "Show that" a probability is a certain value, you must provide a rigorous, step-by-step argument. You cannot just state the result. "Find" or "Calculate" questions require a clear method and a final answer, usually to 3 significant figures unless specified otherwise.