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Probability

Chapter Outline & Learning Goals

By the end of this chapter, you will be able to:

1. Define random experiments, sample space, and events.
2. Apply the classical definition of probability: P(E) = favourable outcomes / total outcomes.
3. Use the addition rule for mutually exclusive and non-exclusive events.
4. Use the multiplication rule for independent and dependent events.
5. Solve problems on conditional probability and Bayes' theorem.
6. Handle standard exam models: coins, dice, cards, balls from a bag, and selection problems.

1. The Hook: The Birthday Paradox

A fact that shocks almost everyone who hears it for the first time:

Take a room with just 23 random people. What is the chance that at least two of them share the same birthday? Most people guess 5% or 10%. The real answer is more than 50%. With 50 people, it jumps to about 97%.

How can that be? Because you are not matching birthdays with *one* person — you are comparing every pair in the room, and 23 people make 253 pairs. Your intuition counts people; probability counts *pairs*.

That is the whole spirit of this chapter: probability is the mathematics of counting what can happen. Once you learn to count correctly, questions that look like magic become 30-second exam problems.

2. You Are Not Alone — Why Probability Feels Hard

Almost every aspirant says the same thing: "I understood permutations, but probability questions still trick me." The reasons follow — and it is not your fault:

• Probability questions hide the counting inside a story (a bag, a deck, a committee).
• Two events can be *mutually exclusive*, *independent*, or *neither* — and the formulas differ for each.
• Words like "at least one" tempt you to add probabilities when you should be subtracting from 1.

The fix is a single habit: before any formula, write down the total number of outcomes and the favourable number of outcomes. Everything in this chapter hangs on that one habit.

🧠 Active Recall: In the birthday paradox, why does the probability rise so fast? (Hint: what grows quickly as people are added — people, or pairs?)

3. The Basic Vocabulary

Random experiment: An action whose result cannot be predicted with certainty, even though all possible results are known. Tossing a coin, rolling a die, drawing a card.

Sample space (S): The set of *all* possible outcomes.

• One coin: S = {H, T}, n(S) = 2
• One die: S = {1, 2, 3, 4, 5, 6}, n(S) = 6
• Two coins: S = {HH, HT, TH, TT}, n(S) = 4

Event (E): Any subset of the sample space — the outcome(s) you care about. "Getting an even number on a die" is the event E = {2, 4, 6}.

Classical probability:

$$P(E) = \frac{\text{Number of favourable outcomes}}{\text{Total number of equally likely outcomes}} = \frac{n(E)}{n(S)}$$

Key facts:

• 0 ≤ P(E) ≤ 1, always.
• P(sure event) = 1; P(impossible event) = 0.
• P(E) + P(E′) = 1, where E′ is the complement (E does not happen).

🧠 Active Recall: Two dice are thrown. What is n(S)? (Answer: 6 × 6 = 36 — ordered pairs, so (2,5) and (5,2) are different.)

4. The Golden Habit: Count First, Formula Later

Example: A bag has 4 red and 6 blue balls. One ball is drawn. P(red)?

• Total outcomes = 10, favourable = 4.
• P(red) = 4/10 = 2/5.

Example: Two dice are thrown. P(sum = 8)?

• Favourable pairs: (2,6), (3,5), (4,4), (5,3), (6,2) → 5 outcomes.
• P = 5/36.

Notice what we did: no fancy formula, just careful counting. Many probability questions hinge on exactly this counting step.

The "at least one" weapon: Whenever you see *at least one*, flip the question:

$$P(\text{at least one}) = 1 - P(\text{none})$$

Example: A coin is tossed 3 times. P(at least one head) = 1 − P(all tails) = 1 − 1/8 = 7/8.

🧠 Active Recall: Why is "at least one" easier through the complement? (Because "none" is a single case, while "at least one" is many cases added together.)

5. Memory Trick: The Jealous Coin 🪙

A silly story that will save you marks:

Imagine a jealous coin. It cannot stand sharing the spotlight.

• Mutually exclusive events are like the jealous coin and the jealous note: if one shows up, the other *cannot*. Head and tail on one toss — never together. For these, you ADD: P(A or B) = P(A) + P(B).
• Independent events don't care about each other at all: the coin's result doesn't change the die's mood. For these, you MULTIPLY: P(A and B) = P(A) × P(B).

Mnemonic: "Exclusive ADDs, Independent MULTIPLIES" — or remember the jealous coin: *"Can't appear together? Add. Don't care about each other? Multiply."*

6. The Addition Rule (OR problems)

Mutually exclusive (cannot happen together):

$$P(A \cup B) = P(A) + P(B)$$

Not mutually exclusive (can happen together — subtract the overlap):

$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$

Example: A card is drawn from 52 cards. P(king or heart)?

• P(king) = 4/52, P(heart) = 13/52, but the king of hearts was counted twice: P(king ∩ heart) = 1/52.
• P = 4/52 + 13/52 − 1/52 = 16/52 = 4/13.

The trap examiners love: forgetting the overlap. If events can happen together, the overlap *must* be subtracted.

🧠 Active Recall: When do we subtract P(A ∩ B) in the addition rule? (Only when A and B can occur together.)

7. The Multiplication Rule (AND problems)

Independent events (one doesn't affect the other):

$$P(A \cap B) = P(A) \times P(B)$$

Example: A coin is tossed and a die is rolled. P(head and 6) = 1/2 × 1/6 = 1/12.

Dependent events (the first changes the second — like drawing *without replacement*):

$$P(A \cap B) = P(A) \times P(B \mid A)$$

Example: A bag has 5 white and 3 black balls. Two balls are drawn without replacement. P(both white)?

• First white: 5/8. Now only 4 white remain out of 7: second white = 4/7.
• P = 5/8 × 4/7 = 20/56 = 5/14.

Exam alarm bell: the phrase *"without replacement"* means the denominator shrinks. *"With replacement"* means events stay independent.

🧠 Active Recall: Two cards are drawn without replacement. Is the second draw independent of the first? (No — the deck has changed.)

8. Conditional Probability and Bayes' Theorem

Conditional probability — the probability of A *given that B has already happened*:

$$P(A \mid B) = \frac{P(A \cap B)}{P(B)}, \quad P(B) \neq 0$$

Think of it as shrinking the sample space: once B is known, only B's outcomes remain possible.

Example: A die shows an even number. What is the probability it is 2?

• New sample space = {2, 4, 6}. Favourable = {2}. P = 1/3.

Bayes' theorem (reverse conditional probability — used in Graduation-level exams):

$$P(A \mid B) = \frac{P(B \mid A)\,P(A)}{P(B)}$$

Classic setup: A factory has two machines. Machine 1 makes 60% of items with 2% defective; Machine 2 makes 40% with 3% defective. An item is defective. Probability it came from Machine 1?

• P(defective) = 0.6 × 0.02 + 0.4 × 0.03 = 0.012 + 0.012 = 0.024
• P(Machine 1 | defective) = 0.012 / 0.024 = 1/2

🧠 Active Recall: In P(A | B), what happens to the sample space? (It shrinks to only the outcomes inside B.)

9. The Four Exam Models You Must Master

Card facts to memorise cold: 52 cards = 4 suits × 13 ranks; 26 red + 26 black; 12 face cards (J, Q, K); 4 aces; king is a face card, ace is not.

Dice sum table (two dice): sums 2 and 12 → 1 way each; 3 and 11 → 2; 4 and 10 → 3; 5 and 9 → 4; 6 and 8 → 5; 7 → 6 ways (the most likely sum).

🧠 Active Recall: Why is 7 the most likely sum of two dice? (It has the most ordered pairs: 6.)

10. Summary Table

Mnemonic recap: 🪙 *"Exclusive ADDs, Independent MULTIPLIES — and 'at least one' hides inside 1 minus none."*

📌 Review this chapter after 2 days. Redo the birthday paradox logic, the king-or-heart overlap, and the without-replacement bag problem from memory. If you can solve all three without looking, probability is officially yours.