Class 10 Mathematics: Probability - Theoretical Approach and Standard Random Experiments

Probability is the branch of mathematics that quantifies uncertainty. In our daily lives, we routinely make statements involving uncertainty: "It will likely rain today," "Team A has a high chance of winning the match," or "I will probably score full marks in Mathematics." While these everyday statements are subjective, mathematical probability provides a precise, numerical framework to measure the likelihood of such occurrences.

In Class 9, you studied experimental or empirical probability, which is based on the actual results of performed experiments and repeated observations. In Class 10, the focus shifts to theoretical (or classical) probability, where we predict the likelihood of an event before conducting any physical experiment, relying purely on logical assumptions about equally likely outcomes. This theoretical approach forms the bedrock of modern statistics, risk analysis, financial modeling, artificial intelligence, and actuarial science.

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1. In-Depth Conceptual Breakdown

1.1 Key Terminology and Foundational Concepts

To master theoretical probability, one must first build absolute clarity regarding its underlying terminology.

Random Experiment

An experiment is termed a random experiment if it satisfies two essential conditions:

1. It has more than one possible outcome.
2. It is impossible to predict the exact outcome in advance with certainty.

Example: Tossing a fair coin or rolling an unbiased six-faced die.

Sample Space ($S$)

The set of all possible outcomes of a random experiment is called its Sample Space, denoted by $S$. The total number of elements in the sample space is written as $n(S)$.

Example: When a fair coin is tossed, $S = \{H, T\}$, where $H$ represents Head and $T$ represents Tail. Here, $n(S) = 2$.

Event ($E$)

An event is a collection of one or more outcomes of a random experiment. Mathematically, an event $E$ is a subset of the sample space $S$ ($E \subseteq S$). The number of outcomes favorable to the event $E$ is denoted by $n(E)$.

Example: In rolling a die, if $E$ is the event of "getting an even number", then $E = \{2, 4, 6\}$ and $n(E) = 3$.

Elementary Event vs. Compound Event

• Elementary Event: An event having only one outcome of the sample space. For instance, getting a '3' on rolling a die ($E = \{3\}$) is an elementary event.
• Compound Event: An event that has more than one outcome of the sample space. For instance, getting an odd number on rolling a die ($E = \{1, 3, 5\}$) is a compound event.

Equally Likely Outcomes

Outcomes of an experiment are said to be equally likely if each outcome has the exact same chance of occurring as any other. Throughout the NCERT Class 10 syllabus, unless stated otherwise, we assume all experiments involve fair, unbiased objects (coins, dice, cards) leading to equally likely outcomes.

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1.2 Empirical vs. Theoretical Probability

The Law of Large Numbers: As the total number of physical trials in an empirical experiment increases to a very large number, the experimental probability approaches closer and closer to its theoretical probability.

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1.3 Theoretical Definition and Core Axioms of Probability

For an experiment with a finite sample space $S$ containing equally likely outcomes, the theoretical probability $P(E)$ of an event $E$ is defined as:

$P(E) = \frac{n(E)}{n(S)} = \frac{\text{Number of outcomes favorable to } E}{\text{Number of all possible outcomes of the experiment}}$

Axioms and Fundamental Properties of Probability

1. Range of Probability: The probability of any event $E$ is a real number ranging between $0$ and $1$ (inclusive): $0 \le P(E) \le 1$

   • A probability cannot be negative ($P(E) < 0$ is impossible).
   • A probability cannot exceed $1$ ($P(E) > 1$ is impossible).
   • Probabilities can be expressed as proper fractions, decimals, or percentages (from $0\%$ to $100\%$).

2. Impossible Event: An event that has zero favorable outcomes ($n(E) = 0$) can never occur. $P(\text{Impossible Event}) = \frac{0}{n(S)} = 0$ Example: Getting a number $7$ on a standard six-faced die.

3. Sure (Certain) Event: An event that contains all possible outcomes of the sample space ($n(E) = n(S)$) is guaranteed to occur. $P(\text{Sure Event}) = \frac{n(S)}{n(S)} = 1$ Example: Getting a number less than $7$ on a standard six-faced die.

4. Sum of Probabilities of Elementary Events: The sum of the probabilities of all the elementary events of a random experiment is always equal to $1$. $\sum P(E_i) = P(E_1) + P(E_2) + P(E_3) + \dots + P(E_k) = 1$

5. Complementary Events: For any event $E$, the event representing "not $E$" is called the complement of $E$, denoted by $\bar{E}$ or $E'$. $P(E) + P(\bar{E}) = 1 \implies P(\bar{E}) = 1 - P(E)$

   • $E$ and $\bar{E}$ are called complementary events.

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2. Standard Random Experiments (Detailed Analysis)

2.1 Experiment 1: Tossing Coins

When a coin is tossed, it lands showing either a Head ($H$) or a Tail ($T$).

                           COIN EXPERIMENTS
                                  |
        -----------------------------------------------------
        |                         |                         |
   Single Coin               Two Coins                 Three Coins
   n(S) = 2^1 = 2            n(S) = 2^2 = 4            n(S) = 2^3 = 8
   S = {H, T}                S = {HH, HT,              S = {HHH, HHT, HTH, HTT,
                                  TH, TT}                   THH, THT, TTH, TTT}

General Formula for $n$ Coins

When $n$ fair coins are tossed simultaneously (or one coin is tossed $n$ times consecutively), the total number of outcomes is given by: $n(S) = 2^n$

• One Coin ($n = 1$): $n(S) = 2^1 = 2$ $S = \{H, T\}$

• Two Coins ($n = 2$): $n(S) = 2^2 = 4$ $S = \{HH, HT, TH, TT\}$

  • Note: $HT$ means Head on 1st coin, Tail on 2nd coin. $TH$ means Tail on 1st coin, Head on 2nd coin. These are distinct outcomes.

• Three Coins ($n = 3$): $n(S) = 2^3 = 8$ $S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}$

Key Terminology Alert:

• "At least $k$ heads" means $\ge k$ heads (i.e., $k$ or more).
• "At most $k$ heads" means $\le k$ heads (i.e., $k$ or fewer).

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2.2 Experiment 2: Rolling Dice

A standard six-faced die is a cube with faces marked with numbers $1, 2, 3, 4, 5, 6$.

General Formula for $n$ Dice

When $n$ dice are thrown simultaneously, the total number of outcomes is: $n(S) = 6^n$

• Single Die ($n = 1$): $n(S) = 6^1 = 6$ $S = \{1, 2, 3, 4, 5, 6\}$

• Two Dice ($n = 2$): $n(S) = 6^2 = 36$

Complete Sample Space Matrix for Two Dice

Special Terms for Two Dice:

• Doublet: Obtaining the same number on both dice. $\text{Doublets} = \{(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)\} \implies \text{Count} = 6$
• Sum of Numbers on Two Dice: Range of possible sums is from $2$ (min: $1+1$) to $12$ (max: $6+6$).

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2.3 Experiment 3: Playing Cards

A standard deck of playing cards contains 52 cards divided into 4 suits of 13 cards each.

                                  DECK OF 52 CARDS
                                         |
                   ---------------------------------------------
                   |                                           |
            RED CARDS (26)                              BLACK CARDS (26)
                   |                                           |
         ---------------------                       ---------------------
         |                   |                       |                   |
    Hearts (13)        Diamonds (13)               Spades (13)        Clubs (13)
     (♥ Red)            (♦ Red)                 (♠ Black)          (♣ Black)

Classification of a Standard 52-Card Deck

Crucial Categories to Memorize for Board Exams:

1. Color Split: $26$ Red cards ($13$ Hearts + $13$ Diamonds) and $26$ Black cards ($13$ Spades + $13$ Clubs).
2. Face Cards: Cards featuring human figures—Jacks (J), Queens (Q), and Kings (K). $\text{Total Face Cards} = 3 \text{ per suit} \times 4 \text{ suits} = 12 \text{ cards}$
   • Red Face Cards: $2 \text{ (Jacks)} + 2 \text{ (Queens)} + 2 \text{ (Kings)} = 6 \text{ cards}$
   • Black Face Cards: $2 \text{ (Jacks)} + 2 \text{ (Queens)} + 2 \text{ (Kings)} = 6 \text{ cards}$
3. Ace Cards: There are $4$ Aces in total ($1$ per suit). Aces are NOT face cards.
4. Number (Digit) Cards: Cards numbered $2$ through $10$. $\text{Total Number Cards} = 9 \text{ per suit} \times 4 \text{ suits} = 36 \text{ cards}$
5. Honor Cards: Aces, Kings, Queens, and Jacks combined ($4 \times 4 = 16$ cards).

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2.4 Experiment 4: Calendar / Year Problems

Questions regarding the number of days, weeks, and specific days (e.g., 53 Sundays) in a year are frequent in exams.

• Ordinary Year: 365 days $= 52 \text{ weeks} + 1 \text{ extra day}$.
  • The 1 extra day can be any of the 7 days of the week: $\{\text{Mon, Tue, Wed, Thu, Fri, Sat, Sun}\}$.
• Leap Year: 366 days $= 52 \text{ weeks} + 2 \text{ extra days}$.
  • The 2 consecutive extra days can be: $\{(\text{Mon, Tue}), (\text{Tue, Wed}), (\text{Wed, Thu}), (\text{Thu, Fri}), (\text{Fri, Sat}), (\text{Sat, Sun}), (\text{Sun, Mon})\}$
  • Total possible pairs $= 7$.

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3. Real-World Applications

1. Meteorology and Natural Disaster Planning

Weather forecast models do not predict rain with absolute certainty; instead, they compute probabilities based on historical radar data, atmospheric pressure, and moisture levels. A forecast statement like "80% chance of rainfall" guides agricultural planning, flight schedules, and emergency disaster management.

2. Genetics and Medical Diagnostics

In genetics, Punnett squares use probability to calculate the likelihood of an offspring inheriting specific traits or genetic disorders from parents. For instance, if two parents are carriers of a recessive gene for a condition like Sickle Cell Anemia, probability models reveal a 25% ($P = 0.25$) theoretical probability that their child will inherit the condition.

3. Financial Markets and Quality Control in Manufacturing

Insurance companies calculate life insurance premiums using actuarial probability tables that estimate life expectancy. Similarly, quality control engineers in manufacturing factories randomly sample items off an assembly line to compute defect probabilities, ensuring product safety before public release.

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4. Step-by-Step Solved Textbook Examples

Example 1: Three Coins Problem

Question: Three unbiased coins are tossed simultaneously. Find the probability of getting:

1. At least 2 heads
2. At most 1 tail
3. Exactly 2 tails

Solution:

Step 1: Write down the total sample space ($S$). When 3 coins are tossed, total possible outcomes $n(S) = 2^3 = 8$. $S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}$

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(i) Event $A$: Getting at least 2 heads

• "At least 2 heads" means getting $2$ or $3$ heads.
• Favorable outcomes = $\{HHH, HHT, HTH, THH\}$
• Number of favorable outcomes $n(A) = 4$

Applying the formula: $P(A) = \frac{n(A)}{n(S)} = \frac{4}{8} = \frac{1}{2}$

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(ii) Event $B$: Getting at most 1 tail

• "At most 1 tail" means getting $0$ or $1$ tail.
  • $0$ tail = $\{HHH\}$
  • $1$ tail = $\{HHT, HTH, THH\}$
• Favorable outcomes = $\{HHH, HHT, HTH, THH\}$
• Number of favorable outcomes $n(B) = 4$

Applying the formula: $P(B) = \frac{n(B)}{n(S)} = \frac{4}{8} = \frac{1}{2}$

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(iii) Event $C$: Getting exactly 2 tails

• Favorable outcomes = $\{HTT, THT, TTH\}$
• Number of favorable outcomes $n(C) = 3$

Applying the formula: $P(C) = \frac{n(C)}{n(S)} = \frac{3}{8}$

Final Answers:

1. $P(\text{At least 2 heads}) = \mathbf{\frac{1}{2}}$
2. $P(\text{At most 1 tail}) = \mathbf{\frac{1}{2}}$
3. $P(\text{Exactly 2 tails}) = \mathbf{\frac{3}{8}}$

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Example 2: Two Dice Problem

Question: Two fair dice are thrown simultaneously. What is the probability that:

1. The sum of the two numbers appearing on top is a prime number?
2. The outcome is a doublet?
3. The sum is greater than 9?

Solution:

Step 1: State total outcomes. For two dice, total possible outcomes $n(S) = 6 \times 6 = 36$.

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(i) Event $E_1$: Sum of numbers is a prime number

• The possible sums on two dice range from $2$ to $12$.
• Prime numbers in this range are $2, 3, 5, 7, 11$.
• Outlining favorable outcomes for each prime sum:
  • Sum = 2: $(1,1)$ $\rightarrow 1$ outcome
  • Sum = 3: $(1,2), (2,1)$ $\rightarrow 2$ outcomes
  • Sum = 5: $(1,4), (2,3), (3,2), (4,1)$ $\rightarrow 4$ outcomes
  • Sum = 7: $(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)$ $\rightarrow 6$ outcomes
  • Sum = 11: $(5,6), (6,5)$ $\rightarrow 2$ outcomes
• Total favorable outcomes $n(E_1) = 1 + 2 + 4 + 6 + 2 = 15$

$P(E_1) = \frac{n(E_1)}{n(S)} = \frac{15}{36} = \frac{5}{12}$

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(ii) Event $E_2$: Getting a doublet

• Favorable outcomes = $\{(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)\}$
• Number of favorable outcomes $n(E_2) = 6$

$P(E_2) = \frac{n(E_2)}{n(S)} = \frac{6}{36} = \frac{1}{6}$

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(iii) Event $E_3$: Sum is greater than 9

• "Greater than 9" means sum can be $10, 11,$ or $12$.
  • Sum = 10: $(4,6), (5,5), (6,4)$ $\rightarrow 3$ outcomes
  • Sum = 11: $(5,6), (6,5)$ $\rightarrow 2$ outcomes
  • Sum = 12: $(6,6)$ $\rightarrow 1$ outcome
• Total favorable outcomes $n(E_3) = 3 + 2 + 1 = 6$

$P(E_3) = \frac{n(E_3)}{n(S)} = \frac{6}{36} = \frac{1}{6}$

Final Answers:

1. $P(\text{Prime sum}) = \mathbf{\frac{5}{12}}$
2. $P(\text{Doublet}) = \mathbf{\frac{1}{6}}$
3. $P(\text{Sum} > 9) = \mathbf{\frac{1}{6}}$

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Example 3: Deck of Playing Cards

Question: One card is drawn at random from a well-shuffled deck of 52 cards. Calculate the probability that the card drawn is:

1. A red face card
2. Neither a King nor a Queen
3. A spade or an Ace

Solution:

Step 1: State total outcomes. Total number of cards in a deck, $n(S) = 52$.

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(i) Event $A$: A red face card

• Total face cards in a deck = 12 (4 Jacks, 4 Queens, 4 Kings).
• Half of the face cards are red (Hearts and Diamonds).
• Favorable cards = $2 \text{ Kings} + 2 \text{ Queens} + 2 \text{ Jacks} = 6$ cards.
• $n(A) = 6$

$P(A) = \frac{n(A)}{n(S)} = \frac{6}{52} = \frac{3}{26}$

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(ii) Event $B$: Neither a King nor a Queen

• Total Kings in deck = $4$
• Total Queens in deck = $4$
• Total cards that are either a King or a Queen $= 4 + 4 = 8$
• Number of cards that are neither King nor Queen, $n(B) = 52 - 8 = 44$

$P(B) = \frac{n(B)}{n(S)} = \frac{44}{52} = \frac{11}{13}$

Alternatively using Complementary Event rule: $P(\text{King or Queen}) = \frac{8}{52} = \frac{2}{13}$ $P(\text{Neither King nor Queen}) = 1 - P(\text{King or Queen}) = 1 - \frac{2}{13} = \frac{11}{13}$

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(iii) Event $C$: A spade or an Ace

• Number of spade cards $= 13$
• Number of Aces $= 4$
• Note: The Ace of Spades is already counted in the 13 spades!
• Favorable cards = $13 \text{ spades} + 3 \text{ remaining Aces (Hearts, Diamonds, Clubs)} = 16$
• $n(C) = 16$

$P(C) = \frac{n(C)}{n(S)} = \frac{16}{52} = \frac{4}{13}$

Final Answers:

1. $P(\text{Red face card}) = \mathbf{\frac{3}{26}}$
2. $P(\text{Neither King nor Queen}) = \mathbf{\frac{11}{13}}$
3. $P(\text{Spade or Ace}) = \mathbf{\frac{4}{13}}$

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Example 4: Leap Year Problem

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

Solution:

Step 1: Analyze the leap year structure.

• A leap year has 366 days.
• $366 \text{ days} = \frac{366}{7} = 52 \text{ full weeks} + 2 \text{ extra days}$.

Step 2: Account for guaranteed occurrences.

• $52$ full weeks guarantee that every day of the week (including Sunday) occurs at least $52$ times.

Step 3: Determine the sample space of extra days.

• The remaining $2$ extra days must be consecutive days of the week.
• Sample Space $S = \{(\text{Mon, Tue}), (\text{Tue, Wed}), (\text{Wed, Thu}), (\text{Thu, Fri}), (\text{Fri, Sat}), (\text{Sat, Sun}), (\text{Sun, Mon})\}$
• Total outcomes $n(S) = 7$.

Step 4: Identify favorable outcomes.

• For the year to have 53 Sundays, one of the two extra days must be a Sunday.
• Favorable outcomes $E = \{(\text{Sat, Sun}), (\text{Sun, Mon})\}$
• Number of favorable outcomes $n(E) = 2$.

Step 5: Apply probability formula. $P(53 \text{ Sundays in a leap year}) = \frac{n(E)}{n(S)} = \frac{2}{7}$

Final Answer:

• $P(53 \text{ Sundays in a leap year}) = \mathbf{\frac{2}{7}}$

(Note: For an non-leap/ordinary year, there is only 1 extra day, so $P(53 \text{ Sundays in an ordinary year}) = \mathbf{\frac{1}{7}}$).

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5. Common Student Mistakes to Avoid

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6. Practice Questions for Self-Assessment

Question 1

A box contains 12 balls out of which $x$ are black.

1. If one ball is drawn at random from the box, what is the probability that it will be a black ball?
2. If 6 more black balls are put in the box, the probability of drawing a black ball is now double of what it was before. Find $x$.

Complete Solution:

Part 1:

• Total number of balls $= 12$. Total outcomes $n(S) = 12$.
• Number of black balls $= x$. Favorable outcomes $= x$. $P(\text{Black ball initially}) = P_1 = \frac{x}{12}$

Part 2:

• New total number of balls in the box $= 12 + 6 = 18$.
• New number of black balls $= x + 6$. $P(\text{Black ball now}) = P_2 = \frac{x + 6}{18}$

According to the given condition: $P_2 = 2 \times P_1$ $\frac{x + 6}{18} = 2 \times \left(\frac{x}{12}\right)$ $\frac{x + 6}{18} = \frac{x}{6}$

Multiply both sides by 18: $x + 6 = 3x$ $3x - x = 6 \implies 2x = 6 \implies x = 3$

Answer: $x = \mathbf{3}$

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Question 2

Two dice are thrown at the same time. Find the probability that the product of the two numbers appearing on top is a perfect square.

Complete Solution:

• Total outcomes for two dice $n(S) = 36$.
• Product of numbers on two dice ranges from $1 \times 1 = 1$ to $6 \times 6 = 36$.
• Perfect square products possible: $1, 4, 9, 16, 25, 36$.

Let us list all favorable pairs $(d_1, d_2)$ producing these perfect square products:

• Product = 1: $(1,1)$ $\rightarrow 1$ outcome
• Product = 4: $(1,4), (2,2), (4,1)$ $\rightarrow 3$ outcomes
• Product = 9: $(3,3)$ $\rightarrow 1$ outcome
• Product = 16: $(4,4)$ $\rightarrow 1$ outcome
• Product = 25: $(5,5)$ $\rightarrow 1$ outcome
• Product = 36: $(6,6)$ $\rightarrow 1$ outcome

Total favorable outcomes $n(E) = 1 + 3 + 1 + 1 + 1 + 1 = 8$. $E = \{(1,1), (1,4), (2,2), (4,1), (3,3), (4,4), (5,5), (6,6)\}$

$P(\text{Product is a perfect square}) = \frac{n(E)}{n(S)} = \frac{8}{36} = \frac{2}{9}$

Answer: $\mathbf{\frac{2}{9}}$

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Question 3

Cards numbered $1$ to $90$ are placed in a box and mixed thoroughly. One card is drawn at random from the box. Find the probability that the card bears:

1. A two-digit number
2. A perfect square number
3. A number divisible by $5$ and $2$

Complete Solution:

• Total cards $n(S) = 90$ (numbers from 1 to 90).

(i) Event $A$: A two-digit number

• Single-digit numbers are $1, 2, 3, 4, 5, 6, 7, 8, 9$ (total of $9$ cards).
• Two-digit numbers are from $10$ to $90$.
• Number of two-digit cards $n(A) = 90 - 9 = 81$. $P(A) = \frac{81}{90} = \frac{9}{10}$

(ii) Event $B$: A perfect square number

• Perfect squares between $1$ and $90$ are: $1^2, 2^2, 3^2, 4^2, 5^2, 6^2, 7^2, 8^2, 9^2$
• Perfect squares $= \{1, 4, 9, 16, 25, 36, 49, 64, 81\}$
• Number of favorable outcomes $n(B) = 9$. $P(B) = \frac{9}{90} = \frac{1}{10}$

(iii) Event $C$: A number divisible by 5 and 2

• A number divisible by both $5$ and $2$ must be divisible by $\text{LCM}(5,2) = 10$.
• Numbers divisible by $10$ from $1$ to $90$ are: $\{10, 20, 30, 40, 50, 60, 70, 80, 90\}$.
• Number of favorable outcomes $n(C) = 9$. $P(C) = \frac{9}{90} = \frac{1}{10}$

Final Answers:

1. $P(\text{Two-digit number}) = \mathbf{\frac{9}{10}}$
2. $P(\text{Perfect square}) = \mathbf{\frac{1}{10}}$
3. $P(\text{Divisible by 5 and 2}) = \mathbf{\frac{1}{10}}$

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7. Exam Revision & FAQs

FAQ 1: What is the fundamental difference between an elementary event and a compound event?

Answer: An elementary event consists of exactly one single outcome of the sample space. For example, getting a '4' when rolling a die has only one outcome $\{4\}$. A compound event consists of two or more outcomes. For example, getting an even number on rolling a die consists of three outcomes $\{2, 4, 6\}$. The sum of probabilities of all elementary events in any experiment always equals $1$.

FAQ 2: What are complementary events, and how do they save computation time in board exams?

Answer: Complementary events are mutually exclusive events where one event is the exact negation of the other. For an event $E$, its complement is $\bar{E}$ ("not $E$"), satisfying: $P(E) + P(\bar{E}) = 1 \implies P(E) = 1 - P(\bar{E})$ Exam Tip: When asked to calculate the probability of "at least one...", it is almost always faster to calculate $1 - P(\text{none})$. For instance, $P(\text{at least one head in 3 tosses}) = 1 - P(\text{no heads}) = 1 - \frac{1}{8} = \frac{7}{8}$.

FAQ 3: How do I handle "OR" vs "AND" in probability word problems?

Answer:

• "OR" (Union): Combines favorable outcomes. $P(A \text{ or } B)$ means outcomes that satisfy Event $A$, Event $B$, or both. (Be careful not to double-count outcomes that satisfy both!).
• "AND" (Intersection): Restricts favorable outcomes strictly to those that satisfy both conditions simultaneously.
  • Example: "A card that is a Red card AND a King" $= 2$ cards (Red Kings). "A card that is Red OR a King" $= 26 \text{ (Red)} + 2 \text{ (Black Kings)} = 28$ cards.

FAQ 4: Can the probability of an event be negative or greater than 1?

Answer: No, never. By definition, the number of favorable outcomes $n(E)$ can neither be negative nor exceed the total number of outcomes $n(S)$ ($0 \le n(E) \le n(S)$). Dividing throughout by $n(S)$ gives: $0 \le P(E) \le 1$ If you ever compute $P(E) < 0$ or $P(E) > 1$ during an exam, check your work immediately for calculation errors!