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Probability

Chapter Outline & Learning Goals

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

1. Define probability using the classical, empirical, and axiomatic approaches.
2. Solve problems on coins, dice, cards, and balls with confidence.
3. Apply the Addition and Multiplication theorems, including conditional probability.
4. Use Bayes' Theorem and Total Probability — the heart of advanced exam questions.
5. Handle permutations-and-combinations-based probability, expected value, and common distributions (Binomial).
6. Spot the traps examiners set: "at least one", "without replacement", mutually exclusive vs independent.

1. The Hook: The Birthday Paradox

In a room of just 23 people, what is the chance that at least two share the same birthday?

Most people guess 5–6%. The real answer is over 50%. With 60 people, it crosses 99%.

This is probability in a nutshell — our intuition is terrible at it, and examiners know it. Questions built exactly on this gap between "what feels right" and "what is mathematically true" are a favourite of SSC CGL Tier-II, IBPS PO Mains, and similar exams. Master this chapter, and you convert the examiner's favourite trap into your easiest marks.

2. "Probability Is Confusing" — Let's Fix That First

If probability has ever felt slippery to you, the reason is almost always the same: students try to *feel* the answer instead of *counting* the answer. Probability is not about intuition. It is a counting machine:

> Probability = (Number of favourable outcomes) ÷ (Total number of equally likely outcomes)

Everything in this chapter — dice, cards, Bayes' theorem — is just a fancier way of counting the numerator and the denominator. Keep that one sentence in your head and nothing here will scare you.

3. The Basic Vocabulary (The Alphabet of Probability)

• Experiment: An action with an uncertain result (tossing a coin).
• Sample Space (S): The set of ALL possible outcomes. For one die: S = {1, 2, 3, 4, 5, 6}, so n(S) = 6.
• Event (E): Any subset of the sample space. "Getting an even number" = {2, 4, 6}.
• Equally likely outcomes: Each outcome has the same chance (a fair coin, a fair die).

Classical definition: If an experiment has n equally likely outcomes and event E has m of them, then

P(E) = m/n, and always 0 ≤ P(E) ≤ 1.

• P(E) = 0 → impossible event.
• P(E) = 1 → certain event.
• Complement: P(E′) = 1 − P(E). This tiny formula is the most-used trick in the entire chapter.

🧠 Active Recall: A bag has 4 red and 6 blue balls. Without calculating anything fancy — what is n(S) for one draw? What is P(red)? What is P(not red)? *(Answer: 10, 4/10 = 2/5, 6/10 = 3/5.)*

4. The Three Standard Playgrounds

4.1 Coins

• 1 coin → 2 outcomes; 2 coins → 4; 3 coins → 8. For n coins, n(S) = 2ⁿ.
• Example: Tossing 3 coins, P(exactly 2 heads) = 3/8 (favourable: HHT, HTH, THH).

4.2 Dice

• One die: n(S) = 6. Two dice: n(S) = 36 (ordered pairs).
• Learn the sum table — it is asked constantly:

Notice the symmetry: it rises to 7 (the most likely sum) and falls back. P(sum = 7) = 6/36 = 1/6.

4.3 Cards (a standard 52-card deck)

• 4 suits: ♠ ♥ ♦ ♣ (13 cards each); 26 red, 26 black.
• 12 face cards (J, Q, K of each suit); 4 Aces.
• P(king) = 4/52 = 1/13. P(red queen) = 2/52 = 1/26.

🧠 Active Recall: Two dice are thrown. What is P(sum = 10)? What is P(sum ≥ 10)? *(Answer: 3/36 = 1/12; sums 10, 11, 12 → 3+2+1 = 6 ways → 1/6.)*

5. The Addition Theorem — "OR" Questions

For two events A and B:

P(A ∪ B) = P(A) + P(B) − P(A ∩ B)

We subtract the overlap because it was counted twice.

• Mutually exclusive events (cannot happen together, e.g., "king" and "queen" in one draw): P(A ∩ B) = 0, so P(A ∪ B) = P(A) + P(B).
• Exam trap: "Mutually exclusive" and "independent" are NOT the same. Mutually exclusive events (with non-zero probability) are actually *dependent* — if one happens, the other cannot.

Example: One card is drawn. P(king or heart) = 4/52 + 13/52 − 1/52 = 16/52 = 4/13. (We subtract the king of hearts.)

6. The Multiplication Theorem — "AND" Questions

P(A ∩ B) = P(A) × P(B|A)

where P(B|A) is the conditional probability of B given that A has already happened:

P(B|A) = P(A ∩ B) / P(A)

• Independent events: A's outcome doesn't change B's chance → P(B|A) = P(B), so P(A ∩ B) = P(A)·P(B). (Tossing a coin twice; drawing *with replacement*.)
• Dependent events: Drawing *without replacement* — the denominator shrinks each draw.

Example: From a bag with 5 red, 3 blue balls, two are drawn without replacement. P(both red) = (5/8) × (4/7) = 20/56 = 5/14.

🧠 Active Recall: In that example, why did the second probability become 4/7 and not 5/8? *(Because one red ball is gone — 4 red left out of 7 total.)*

7. The "At Least One" Superpower

Questions saying "at least one" are among the most repeated patterns in exams. Direct counting is painful; the complement is one line:

> P(at least one) = 1 − P(none)

Example: A coin is tossed 4 times. P(at least one head) = 1 − P(all tails) = 1 − (1/2)⁴ = 1 − 1/16 = 15/16.

Mnemonic 🪄 — "ALONE": At Least One = Not Empty → 1 − P(none). Whenever you read "at least one", your pen should automatically write "1 −".

🧠 Active Recall: Two dice are thrown. P(at least one six)? *(1 − (5/6)² = 1 − 25/36 = 11/36.)*

8. Probability Meets Combinatorics

When drawing groups of objects, counting uses combinations:

Example: A bag has 4 white and 6 black balls. Three balls are drawn together. P(all white) = ⁴C₃ / ¹⁰C₃ = 4/120 = 1/30.

Key idea: when items are drawn together (simultaneously), use combinations; when drawn one by one in order, you may use sequential multiplication — both give the same answer if done correctly.

Committee-type example: From 5 men and 4 women, a committee of 3 is chosen. P(exactly 2 women) = (⁴C₂ × ⁵C₁) / ⁹C₃ = (6 × 5)/84 = 30/84 = 5/14.

9. Total Probability and Bayes' Theorem (The Advanced Core)

9.1 Total Probability

If B₁, B₂, …, Bₙ partition the sample space (mutually exclusive, exhaustive), then for any event A:

P(A) = Σ P(Bᵢ) · P(A|Bᵢ)

Think: "Event A can happen through several routes; add up each route's contribution."

9.2 Bayes' Theorem — Reverse the Question

P(Bₖ|A) = P(Bₖ) · P(A|Bₖ) / Σ P(Bᵢ) · P(A|Bᵢ)

Bayes answers: *"Given that A happened, what is the chance it came via route Bₖ?"* — a posterior probability.

Classic example: Bag I has 3 red, 2 black; Bag II has 2 red, 3 black. A bag is chosen at random and a red ball is drawn. P(it came from Bag I)?

• P(Bag I) = P(Bag II) = 1/2
• P(red|Bag I) = 3/5; P(red|Bag II) = 2/5
• P(Bag I | red) = (1/2 · 3/5) / (1/2 · 3/5 + 1/2 · 2/5) = (3/10)/(5/10) = 3/5

🧠 Active Recall: In Bayes problems, what goes in the denominator? *(The total probability of the observed event A — the sum of all routes.)*

10. Expected Value and Binomial Distribution (Graduation Extras)

• Expected value: E(X) = Σ xᵢ·P(xᵢ). A fair game has E = 0. If you win ₹10 on a head and lose ₹8 on a tail, E = 10(1/2) − 8(1/2) = ₹1 per toss.
• Binomial distribution: n independent trials, success probability p each time:

P(exactly r successes) = ⁿCᵣ · pʳ · (1−p)ⁿ⁻ʳ

Example: A die is thrown 4 times. P(exactly one six) = ⁴C₁ (1/6)¹ (5/6)³ = 4 × (1/6) × (125/216) = 500/1296 = 125/324.

11. The Examiner's Five Favourite Traps

1. "At least one" → use 1 − P(none), never direct counting.
2. With vs without replacement → denominators change or stay; read twice.
3. Mutually exclusive ≠ independent — a classic statement-based trap.
4. Two-dice ordered pairs — (2,5) and (5,2) are different outcomes; n(S) = 36, not 21.
5. Bayes direction — the question gives P(A|B) but asks P(B|A). Don't answer the forward probability.

12. Summary Table

🔁 Review Reminder

Review this chapter after 2 days. Redo the birthday paradox logic, the dice sum table, one Bayes problem, and one "at least one" problem from memory. Then review again after 7 days — spaced repetition is what moves these formulas from your notebook into your reflexes.