Chapter 7: The Mathematics of Maybe: Introduction to Probability - Ultimate 100/100 Master Study Guide

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Topic 1: Core Concepts & Terminology of Probability

Foundational Definitions

Probability: A mathematical measure of the likelihood or chance of occurrence of a particular event.
Scale ranges continuously from $0$ (Impossible Event) to $1$ (Certain Event).
Random Experiment: An activity or operation that produces well-defined outcomes, but the exact outcome cannot be predicted beforehand (e.g. tossing a coin, rolling a die).
Trial: A single performance of a random experiment.
Outcome: The result of a single trial of an experiment.
Event ($E$): A collection of one or more outcomes of an experiment.
Sample Space ($S$): The set of all possible outcomes of an experiment. The total number of outcomes is denoted as $n(S)$.
Empirical / Experimental Probability Formula: $$P(E) = \frac{\text{Number of trials in which event } E \text{ occurred}}{\text{Total number of trials}}$$
Theoretical / Classical Probability Formula: $$P(E) = \frac{\text{Number of outcomes favorable to } E}{\text{Total number of possible outcomes } n(S)}$$
Range Rule of Probability: $$\mathbf{0 \le P(E) \le 1}$$
Probability can NEVER be negative ($< 0$) and can NEVER exceed $1$ ($> 1$ or $> 100\%$).

Fig 7.1: The Probability Continuum Scale from 0 (Impossible) to 1 (Certain)

Teacher's Exam Secret — 1-Mark Trick Question

Which of the following CANNOT be the probability of an event?
(A) $\frac{2}{3}$ (B) $-1.5$ (C) $15\%$ (D) $0.7$
Answer: (B) $-1.5$ because probability can NEVER be negative!

Immediate Solved Exam Applications

Example 1.1: Empirical Coin Tossing NCERT

Question: A coin is tossed $1000\text{ times}$ with frequencies: $\text{Heads} = 455$, $\text{Tails} = 545$. Compute the probability for each event.

Solution:

$$P(\text{Heads}) = \frac{455}{1000} = \mathbf{0.455}$$ $$P(\text{Tails}) = \frac{545}{1000} = \mathbf{0.545}$$

Notice: $P(\text{Heads}) + P(\text{Tails}) = 0.455 + 0.545 = \mathbf{1.0}$.

Topic 1: School Exam Practice Problem Kit

Problem 1.1 (Empirical Probability from Trial Data) 2 MARKS

Two coins are tossed simultaneously $500\text{ times}$, and the outcomes are recorded:
$2\text{ Heads} = 105\text{ times}$, $1\text{ Head} = 275\text{ times}$, $0\text{ Heads} = 120\text{ times}$.
Find the probability of occurrence of each outcome.

Step-by-Step Solution:

$$P(2\text{ Heads}) = \frac{105}{500} = \mathbf{\frac{21}{100} = 0.21}$$ $$P(1\text{ Head}) = \frac{275}{500} = \mathbf{\frac{11}{20} = 0.55}$$ $$P(0\text{ Heads}) = \frac{120}{500} = \mathbf{\frac{6}{25} = 0.24}$$ $$\text{Sum} = 0.21 + 0.55 + 0.24 = \mathbf{1.00}$$

Fig 7.1A: Coin Trials

Problem 1.2 (Valid Probability Test) 1 MARK

Can a probability of an event be $\frac{7}{5}$? Explain.

Step-by-Step Solution:

No. Since $\frac{7}{5} = 1.4 > 1$, it violates the range rule $0 \le P(E) \le 1$.

Topic 2: Coin Tossing Experiments & Sample Spaces

Coin Sample Space Formula

Single Coin ($n=1$): $S = \{H, T\} \implies n(S) = 2^1 = 2$.
Two Coins ($n=2$): $S = \{HH, HT, TH, TT\} \implies n(S) = 2^2 = 4$.
Three Coins ($n=3$): $S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\} \implies n(S) = 2^3 = 8$.
General Formula for $n$ Coins: $$\mathbf{n(S) = 2^n}$$
Important Distinctions:
"At least 1 head" means $\ge 1$ head (includes 1, 2, 3 heads).
"At most 1 head" means $\le 1$ head (includes 0 or 1 head).

Fig 7.2: Tree Diagram for Tossing 2 Coins Simultaneously ($n(S) = 4$)

Immediate Solved Exam Applications

Example 2.1: 3-Coin At Least Problem 3 MARKS

Question: Three coins are tossed simultaneously. Find the probability of getting: (i) Exactly 2 heads, (ii) At least 2 heads, (iii) At most 1 head.

Solution:

Sample Space $S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\} \implies n(S) = 8$.

(i) Exactly 2 heads $\{HHT, HTH, THH\} \implies P = \mathbf{\frac{3}{8}}$.

(ii) At least 2 heads ($\ge 2$ heads) $\{HHH, HHT, HTH, THH\} \implies P = \frac{4}{8} = \mathbf{\frac{1}{2}}$.

(iii) At most 1 head ($\le 1$ head) $\{HTT, THT, TTH, TTT\} \implies P = \frac{4}{8} = \mathbf{\frac{1}{2}}$.

Topic 2: School Exam Practice Problem Kit

Problem 2.1 (Two Coins At Least 1 Head) 2 MARKS / NCERT

Two unbiased coins are tossed together. Find the probability of getting at least one head.

Step-by-Step Solution:

Sample space $S = \{HH, HT, TH, TT\} \implies n(S) = 4$.
Favorable outcomes for at least 1 head ($\ge 1$): $\{HH, HT, TH\} \implies 3$ outcomes. $$P(\text{At least 1 head}) = \mathbf{\frac{3}{4} = 0.75}$$

Fig 7.2A: 3/4 Favorable

Topic 3: Die Rolling Experiments (Single & Dual Dice)

Dice Roll Sample Spaces

Single Die ($n=1$): $S = \{1, 2, 3, 4, 5, 6\} \implies n(S) = 6$.
Prime numbers: $\{2, 3, 5\} \implies P = \frac{3}{6} = \frac{1}{2}$.
Composite numbers: $\{4, 6\} \implies P = \frac{2}{6} = \frac{1}{3}$. ($1$ is neither prime nor composite!).
Two Dice ($n=2$): $n(S) = 6^2 = 36$ outcomes.
Doublets: $\{(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)\} \implies 6$ outcomes $\implies P = \frac{6}{36} = \frac{1}{6}$.
Minimum Sum $= 1+1 = 2$. Maximum Sum $= 6+6 = 12$.

Fig 7.3: Complete $6 \times 6$ Outcome Matrix for Rolling Two Dice ($n(S) = 36$)

Topic 3: School Exam Practice Problem Kit

Problem 3.1 (Two Dice Sum Equals 8) 3 MARKS / NCERT

Two dice are thrown simultaneously. Find the probability that the sum of the numbers appearing on the top faces is: (i) Equal to $8$, (ii) A doublet, (iii) Greater than $10$.

Step-by-Step Solution:

$n(S) = 36$.
(i) Sum = 8: $\{(2,6), (3,5), (4,4), (5,3), (6,2)\} \implies 5$ outcomes $\implies \mathbf{\frac{5}{36}}$.
(ii) Doublet: $\{(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)\} \implies 6$ outcomes $\implies \frac{6}{36} = \mathbf{\frac{1}{6}}$.
(iii) Sum $> 10$: $\{(5,6), (6,5), (6,6)\} \implies 3$ outcomes $\implies \frac{3}{36} = \mathbf{\frac{1}{12}}$.

Fig 7.3A: Dual Dice

Topic 4: Playing Cards Experiments (Standard 52-Card Deck)

Standard 52-Card Deck Breakdown

Total Deck Size: 52 Cards (divided into 2 colors: 26 Red, 26 Black).
4 Suits (13 Cards Each):
Red Suits: Hearts ($\heartsuit$ - 13), Diamonds ($\diamondsuit$ - 13).
Black Suits: Spades ($\spadesuit$ - 13), Clubs ($\clubsuit$ - 13).
13 Denominations per Suit: Ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack ($J$), Queen ($Q$), King ($K$).
Face Cards (Cards with pictures): Jacks, Queens, Kings.
Total Face Cards $= 4 \times 3 = \mathbf{12\text{ cards}}$ (6 Red Face Cards, 6 Black Face Cards).
Aces: 4 Aces (NOT considered face cards in standard CBSE probability!).

Fig 7.4: Standard 52-Card Deck Classification Hierarchy

Topic 4: School Exam Practice Problem Kit

Problem 4.1 (52 Cards Drawing) 4 MARKS / NCERT

One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting:
(i) A king of red color, (ii) A face card, (iii) A red face card, (iv) The jack of hearts, (v) A spade, (vi) The queen of diamonds.

Step-by-Step Solution:

Total outcomes $n(S) = 52$.
(i) King of red color (King of Hearts, King of Diamonds) $= 2 \implies P = \frac{2}{52} = \mathbf{\frac{1}{26}}$.
(ii) Face card $= 12 \implies P = \frac{12}{52} = \mathbf{\frac{3}{13}}$.
(iii) Red face card $= 6 \implies P = \frac{6}{52} = \mathbf{\frac{3}{26}}$.
(iv) Jack of hearts $= 1 \implies P = \mathbf{\frac{1}{52}}$.
(v) A spade $= 13 \implies P = \frac{13}{52} = \mathbf{\frac{1}{4}}$.
(vi) Queen of diamonds $= 1 \implies P = \mathbf{\frac{1}{52}}$.

Fig 7.4A: Playing Card

Topic 5: Complementary Events & Fundamental Properties

Complementary Events Identity

For any event $E$, the event "Not $E$" (denoted as $\bar{E}$ or $E'$) is called the complementary event of $E$.

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

Event $E$ and $\bar{E}$ are mutually exclusive and exhaustive.
The sum of probabilities of all elementary events of an experiment is always equal to $1$.

Fig 7.5: Venn Diagram Showing Event $E$ and Complementary Event $\bar{E}$ in Sample Space $S$

Topic 5: School Exam Practice Problem Kit

Problem 5.1 (Winning & Losing Game) 2 MARKS / NCERT

Two players, Sangeeta and Reshma, play a tennis match. It is known that the probability of Sangeeta winning the match is $0.62$. What is the probability of Reshma winning the match?

Step-by-Step Solution:

Let $E$ be event Sangeeta wins $\implies P(E) = 0.62$.
Reshma winning is complementary event $\bar{E}$ (Not Sangeeta winning): $$P(\text{Reshma wins}) = 1 - P(E) = 1 - 0.62 = \mathbf{0.38}$$

Topic 6: Real-World Statistical & Survey Data

Example 6.1: Family Survey on Girls NCERT

Question: 1500 families with 2 children were selected randomly and recorded:
2 Girls $= 475\text{ families}$, 1 Girl $= 814\text{ families}$, 0 Girls $= 211\text{ families}$.
Compute the probability of a family chosen at random having: (i) 2 girls, (ii) 1 girl, (iii) No girls. Check whether the sum of these probabilities is 1.

Solution:

(i) $P(2\text{ girls}) = \frac{475}{1500} = \mathbf{\frac{19}{60}}$.

(ii) $P(1\text{ girl}) = \frac{814}{1500} = \mathbf{\frac{407}{750}}$.

(iii) $P(0\text{ girls}) = \frac{211}{1500}$.

Sum $= \frac{475 + 814 + 211}{1500} = \frac{1500}{1500} = \mathbf{1.0}$. Sum is verified as 1!

Topic 7: Geometric & Area-Based Probability

Continuous Area-Based Probability

When an outcome is chosen uniformly at random inside a geometric region $R$, the probability that it falls inside a specific sub-region $A$ is:

$$\mathbf{P(\text{Landing in Target } A) = \frac{\text{Area of Target Region } A}{\text{Total Area of Region } R}}$$

Fig 7.6: Geometric Probability Target inside Region

Topic 7: School Exam Practice Problem Kit

Problem 7.1 (Helicopter Crash Geometric Target) 3 MARKS / NCERT

A helicopter is reported to have crashed somewhere in the rectangular region $9\text{ m} \times 4.5\text{ m}$. What is the probability that it crashed inside a circular lake of diameter $3\text{ m}$?

Step-by-Step Solution:

1. Total Area of Region $= 9 \times 4.5 = \mathbf{40.5\text{ m}^2}$.
2. Radius of lake $r = 3/2 = 1.5\text{ m}$.
$\text{Area of Lake} = \pi r^2 = \pi (1.5)^2 = 2.25\pi\text{ m}^2 = 2.25 \times 3.1416 \approx \mathbf{7.07\text{ m}^2}$. $$P(\text{Crash in lake}) = \frac{2.25\pi}{40.5} = \frac{\pi}{18} \approx \mathbf{\frac{11}{63} \approx 0.174}$$

Fig 7.6A: Geometric Target

Topic 8: Comprehensive Exam HOTS & Case Study Kit

Case Study 1: Manufacturing Inspection CASE STUDY

A lot of 144 ball pens contains 20 defective ones and the others are good. Nuri will buy a pen if it is good, but will not buy if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that:
(i) She will buy it? (ii) She will not buy it?

Solution:

Total pens $n(S) = 144$. Defective pens $= 20 \implies \text{Good pens} = 144 - 20 = \mathbf{124}$.

(i) $P(\text{She buys it}) = P(\text{Good pen}) = \frac{124}{144} = \mathbf{\frac{31}{36}}$.

(ii) $P(\text{She will not buy}) = P(\text{Defective}) = \frac{20}{144} = \mathbf{\frac{5}{36}}$.

HOTS 1: The 53 Sundays Leap Year Gotcha! HOTS FAVORITE

Question: Find the probability that a leap year chosen at random contains 53 Sundays.

Solution:

1. A leap year has $366\text{ days}$.

$$366\text{ days} = 52\text{ weeks} + 2\text{ extra days}$$

2. The 52 full weeks contain 52 Sundays. The 2 extra days can be any of the following 7 pairs:

$$\text{Pairs } = \{(\text{Sun, Mon}), (\text{Mon, Tue}), (\text{Tue, Wed}), (\text{Wed, Thu}), (\text{Thu, Fri}), (\text{Fri, Sat}), (\text{Sat, Sun})\}$$

Total possible outcomes $n(S) = 7$.

3. Favorable outcomes containing Sunday: $\{(\text{Sun, Mon}), (\text{Sat, Sun})\} \implies 2$ outcomes.

$$P(53\text{ Sundays in Leap Year}) = \mathbf{\frac{2}{7}}$$

Note: For a NON-LEAP year ($365\text{ days} = 52\text{ weeks} + 1\text{ extra day}$), $P(53\text{ Sundays}) = \mathbf{\frac{1}{7}}$.

9. 100/100 School Examination Master Checklist

Master Exam Checklist

Range Rule: $0 \le P(E) \le 1$. Impossible event $= 0$, Certain event $= 1$.
Complement Identity: $P(E) + P(\bar{E}) = 1 \implies P(\bar{E}) = 1 - P(E)$.
Coins Formula: $n(S) = 2^n$. (1 coin $= 2$, 2 coins $= 4$, 3 coins $= 8$).
Dice Formula: $n(S) = 6^n$. (1 die $= 6$, 2 dice $= 36$).
52 Cards Deck: 26 Red, 26 Black, 4 Suits (13 each), 12 Face Cards (J, Q, K).
53 Sundays Rule: Leap year $= 2/7$, Non-leap year $= 1/7$.