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Probability

Classical probability, theoretical approach

Classical DefinitionComplementary EventsCards ProblemsDice Problems
📋 PYQs Available:
2023202220212020
Expert Content

Probability

Why This Chapter Matters

Probability is the last chapter and has a guaranteed 3-5 marks in every board exam. The questions are logical rather than formula-heavy — they test whether you can identify sample spaces and favourable outcomes. With practice, this chapter becomes very scoring.

Prerequisites

Basic probability from Class 9
Fractions and simplification
Counting methods

Core Concepts

1. Theoretical (Classical) Probability

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

Important range: 0 ≤ P(E) ≤ 1

P(E) = 0 → Impossible event (will never happen)
P(E) = 1 → Sure event (will always happen)

2. Complementary Events

If E is an event, then NOT E (written as Ē or E') is its complement.

$$P(E) + P(\bar{E}) = 1 \ \Rightarrow \ P(\bar{E}) = 1 - P(E)$$


3. Sample Space for Common Problems

#### Coin

Single coin: {H, T} → 2 outcomes
Two coins: {HH, HT, TH, TT} → 4 outcomes
Three coins: {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} → 8 outcomes

Note: HT and TH are different (order matters)

#### Dice

Single die: {1, 2, 3, 4, 5, 6} → 6 outcomes
Two dice: 6 × 6 = 36 outcomes (all ordered pairs)

Common two-dice events:

EventFavourable outcomesProbability

|---|---|---|

Sum = 2(1,1)1/36
Sum = 7(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)6/36 = 1/6
Sum = 12(6,6)1/36
Doubles(1,1),(2,2),...,(6,6)6/36 = 1/6

#### Cards (Standard Deck = 52 cards)

52 cards = 4 suits × 13 cards each

Suits: Hearts ♥ (red), Diamonds ♦ (red), Clubs ♣ (black), Spades ♠ (black)

Each suit: A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K

Face cards (court cards): J, Q, K → 12 total (3 per suit)
Aces: 4 total (one per suit)
Honours: A, J, Q, K → 16 total

Solved Examples

Example 1 — Basic

Q: A die is thrown. Find P(prime number).

Prime numbers on die: 2, 3, 5 → 3 favourable outcomes

Total outcomes: 6

P(prime) = 3/6 = 1/2

Example 2 — Cards

Q: A card is drawn from a well-shuffled deck. Find:

(a) P(red ace) (b) P(face card) (c) P(neither jack nor king)

(a) Red aces: Ace of Hearts, Ace of Diamonds = 2 → P = 2/52 = 1/26

(b) Face cards: 12 → P = 12/52 = 3/13

(c) Jacks = 4, Kings = 4, total to exclude = 8. Favourable = 52−8 = 44 → P = 44/52 = 11/13

Example 3 — Two Dice

Q: Two dice are thrown. Find P(sum ≤ 5).

List outcomes where sum ≤ 5:

Sum=2: (1,1) → 1

Sum=3: (1,2),(2,1) → 2

Sum=4: (1,3),(2,2),(3,1) → 3

Sum=5: (1,4),(2,3),(3,2),(4,1) → 4

Total favourable = 10

P = 10/36 = 5/18

Example 4 — Word Problem

Q: A bag contains 5 red, 3 black, and 2 white balls. One ball is drawn at random. Find P(not black).

Total = 10, Black = 3

P(not black) = 1 − 3/10 = 7/10

Or directly: non-black balls = 7 → P = 7/10 ✓


PYQs

2023

Q: A box contains 3 blue, 2 white, and 4 red marbles. A marble is drawn randomly. Find P(not red).

Total = 9, Red = 4, Not red = 5

P(not red) = 5/9

2022

Q: Two players A and B take turns. A wins if he gets a 6 on his throw. B wins if she gets a sum of 7 on two dice. Who has a better chance?

P(A wins) = 1/6

P(B wins) = 6/36 = 1/6

Equal chances

2021

Q: What is the probability that a leap year chosen at random will have 53 Sundays?

A leap year = 366 days = 52 weeks + 2 extra days

The 2 extra days can be: (Mon,Tue), (Tue,Wed), (Wed,Thu), (Thu,Fri), (Fri,Sat), (Sat,Sun), (Sun,Mon)

Total = 7 possibilities. Sundays in 53 → 2 cases include Sunday: (Sat,Sun) and (Sun,Mon)

P = 2/7

2020

Q: A card is drawn from a shuffled deck. What is the probability of drawing a card that is not a face card?

Face cards = 12 (J, Q, K of 4 suits)

Not face cards = 52 − 12 = 40

P = 40/52 = 10/13


MCQ Practice

Q1. P(impossible event) = (A) 1 (B) 0.5 (C) 0 ✓ (D) Undefined

Q2. From digits 1-9, probability of picking a perfect square:

Perfect squares: 1, 4, 9 → 3 numbers

P = 3/9 = 1/3

Q3 (Hard). Cards numbered 1 to 20 are shuffled. Find P(number is prime OR divisible by 3).

Primes (1-20): 2,3,5,7,11,13,17,19 → 8

Divisible by 3: 3,6,9,12,15,18 → 6

Both (prime AND div by 3): just 3 → 1

P = (8 + 6 − 1)/20 = 13/20


Revision Notes

FUNDAMENTAL FORMULA:
  P(E) = Favourable outcomes / Total outcomes
  0 ≤ P(E) ≤ 1

COMPLEMENT:
  P(E) + P(Ē) = 1

IMPORTANT COUNTS (memorise!):
  Deck: 52 cards, 4 suits, 13 per suit
  Face cards: 12 (J, Q, K in all 4 suits)
  Aces: 4 | Red cards: 26 | Black cards: 26
  
  Single die: 6 outcomes
  Two dice: 36 outcomes
  
  Single coin: 2 | Two coins: 4 | Three coins: 8

CLASSIC ANSWERS:
  P(sum=7 on two dice) = 6/36 = 1/6 (most likely sum)
  P(getting 6 on die) = 1/6
  P(red card from deck) = 26/52 = 1/2

Common Mistakes:

❌ Not listing ALL outcomes for two-dice problems (36, not 11)

❌ Counting HT and TH as the same (they're not!)

❌ Writing probability > 1 or < 0 (always check: 0 ≤ P ≤ 1)

❌ For "not" events, always use P(Ē) = 1 − P(E) instead of counting manually

Related Topics

Chapter 14 — Statistics (experimental probability)
JEE: Probability theory, Bayes theorem, probability distributions
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