NCERT Class 11 Maths Chapter: Probability

The Axiomatic Approach to Probability

Historically, probability was defined as the ratio of favorable outcomes to the total number of equally likely outcomes. While intuitive, this classical definition fails when outcomes are infinite or not equally likely. To address this, Russian mathematician Andrey Kolmogorov introduced the Axiomatic Approach to Probability in 1933.

In this modern approach, we represent the outcomes of a random experiment as a set called the Sample Space ($S$). Any subset of this sample space is called an Event ($A \subseteq S$). We assign a real number $P(A)$, called the probability of event $A$, which satisfies three fundamental axioms:

1. Axiom 1 (Non-negativity): For any event $A$, the probability is always non-negative: $P(A) \ge 0$.
2. Axiom 2 (Certainty): The probability of the entire sample space (the sure event) is 1: $P(S) = 1$.
3. Axiom 3 (Additivity): If $A$ and $B$ are mutually exclusive events (i.e., they cannot occur simultaneously, so $A \cap B = \emptyset$), then the probability of their union is the sum of their individual probabilities: $P(A \cup B) = P(A) + P(B)$.

By leveraging the algebra of sets, we can easily evaluate complex probability scenarios. For instance, the probability of an event not happening (the complement event $A'$) is given by $P(A') = 1 - P(A)$. Similarly, for any two events $A$ and $B$, the general addition rule is written as $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.

Essential Probability Terms & Definitions

Random Experiment

An experiment whose outcome cannot be predicted with certainty in advance, even though all possible outcomes are known.

Sample Space (S)

The set of all possible outcomes of a random experiment. For example, tossing a coin twice yields $S = \{HH, HT, TH, TT\}$.

Mutually Exclusive Events

Two or more events that cannot occur at the same time. Mathematically, $A \cap B = \emptyset$.

Exhaustive Events

A set of events whose union constitutes the entire sample space $S$. If $E_1 \cup E_2 \cup ... \cup E_n = S$, then these events are exhaustive.

How to Solve Class 11 Probability Problems Step-by-Step

1. Define the Sample Space — Identify the random experiment and list down the set $S$ of all possible outcomes. Calculate the total number of elements in $S$, denoted as $n(S)$. Use Permutations and Combinations if the sample size is very large.
2. Define the Specific Event — Identify what condition the question asks for. Write down this subset $A \subseteq S$ and compute the number of elements favorable to the event, denoted as $n(A)$.
3. Apply the Classical/Axiomatic Formula — If all outcomes are equally likely, use the classical formula: $P(A) = \frac{n(A)}{n(S)}$. If the problem involves set operations (like 'or', 'and', 'not'), translate them into set symbols: $\cup$, $\cap$, or $A'$ respectively.
4. Simplify and Check Boundaries — Reduce your final fraction to its simplest form. Ensure your calculated probability is a real number strictly bounded within the closed interval $[0, 1]$.

Deeply Worked Examples

• Example 1: Dice Roll Combination Problem: Two unbiased dice are thrown. Find the probability that the sum of the numbers on the two dice is prime. Solution Step-by-Step: 1. Write the sample space size: $n(S) = 6 \times 6 = 36$ outcomes. 2. Identify prime sums possible: The possible sums on rolling two dice range from 2 to 12. Prime sums in this range are 2, 3, 5, 7, and 11. 3. List the outcomes for each prime sum: - Sum = 2: $\{(1, 1)\}$ (1 outcome) - Sum = 3: $\{(1, 2), (2, 1)\}$ (2 outcomes) - Sum = 5: $\{(1, 4), (2, 3), (3, 2), (4, 1)\}$ (4 outcomes) - Sum = 7: $\{(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)\}$ (6 outcomes) - Sum = 11: $\{(5, 6), (6, 5)\}$ (2 outcomes) 4. Calculate $n(A)$ by adding these up: $n(A) = 1 + 2 + 4 + 6 + 2 = 15$. 5. Compute probability: $P(A) = \frac{n(A)}{n(S)} = \frac{15}{36} = \frac{5}{12}$.
• Example 2: Algebra of Events Problem: If $P(A) = 0.6$, $P(B) = 0.4$, and $P(A \cap B) = 0.2$, calculate the probability of the event $A$ occurring but not $B$ (i.e., $P(A \cap B')$). Solution Step-by-Step: 1. Use the set-theory identity: The event '$A$ but not $B
  is represented as $A - B$ or $A \cap B'$. 2. Realize that $A$ is composed of two mutually exclusive parts: $(A \cap B)$ and $(A \cap B')$. 3. Write down the algebraic relation: $P(A) = P(A \cap B) + P(A \cap B')$. 4. Rearrange the formula: $P(A \cap B') = P(A) - P(P(A \cap B))$. 5. Substitute the values: $P(A \cap B') = 0.6 - 0.2 = 0.4$. Final Answer: 0.4.

Crucial Exam Tip: Mutually Exclusive vs. Independent Events

> Common Board Exam Trap: Students often confuse Mutually Exclusive events with Independent events.
>
> Mutually Exclusive Events: These are events that cannot happen at the same time*. Therefore, their intersection is empty: $P(A \cap B) = 0$.
> * Independent Events (Class 12): These are events where the occurrence of one doesn't affect the other. Their intersection is calculated as the product of their individual probabilities: $P(A \cap B) = P(A) \times P(B)$.
>
> Always check your variables! If a CBSE question states that $A$ and $B$ are mutually exclusive, immediately set $P(A \cap B) = 0$ in your equations.

Practice Questions with Solutions

• Q: Three coins are tossed once. Find the probability of getting at least two heads. A: Step 1: Write down the complete sample space $S$ for three coin tosses. $S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$. Total number of outcomes, $n(S) = 2^3 = 8$. Step 2: Define event $A$ as getting 'at least two heads' (this means 2 or 3 heads). Identify the favorable outcomes from $S$: $A = \{HHH, HHT, HTH, THH\}$. Number of favorable outcomes, $n(A) = 4$. Step 3: Calculate the probability using classical definition. $P(A) = \frac{n(A)}{n(S)} = \frac{4}{8} = \frac{1}{2}$. Final answer: \frac{1}{2}
• Q: A card is drawn from a well-shuffled deck of 52 cards. What is the probability that the card drawn is either a red card or a king? A: Step 1: Find the size of the sample space. $n(S) = 52$. Step 2: Let $R$ be the event of drawing a red card, and $K$ be the event of drawing a king. Number of red cards, $n(R) = 26$. Number of kings in a deck, $n(K) = 4$. Step 3: Find the intersection event $R \cap K$ (red kings in a deck). There are 2 red kings (King of Hearts and King of Diamonds), so $n(R \cap K) = 2$. Step 4: Use the Addition Theorem of Probability: $P(R \cup K) = P(R) + P(K) - P(R \cap K)$ $P(R \cup K) = \frac{26}{52} + \frac{4}{52} - \frac{2}{52} = \frac{28}{52}$ Step 5: Simplify the fraction: $\frac{28}{52} = \frac{7}{13}$. Final answer: \frac{7}{13}
• Q: If $P(A) = \frac{2}{11}$, find the probability of the event 'not $A
  (i.e. $P(A')$). A: Step 1: Identify the complementary event formula. The sum of the probability of an event and its complement is always 1: $P(A) + P(A') = 1$. Step 2: Substitute the given value to isolate $P(A')$. $P(A') = 1 - P(A) = 1 - \frac{2}{11}$. Step 3: Solve the subtraction: $P(A') = \frac{11 - 2}{11} = \frac{9}{11}$. Final answer: \frac{9}{11}
• Q: Two students, Anil and Ashima, appeared in an examination. The probability that Anil passes the exam is 0.05 and that Ashima passes is 0.10. The probability that both pass is 0.02. Find the probability that at least one of them will not pass the exam. A: Step 1: Write down the given probabilities. Let $E_1$ = Anil passes, so $P(E_1) = 0.05$. Let $E_2$ = Ashima passes, so $P(E_2) = 0.10$. Probability that both pass: $P(E_1 \cap E_2) = 0.02$. Step 2: Understand the target event. 'At least one of them will not pass' is the complement of the event 'both of them pass'. Let $E$ be the event that both pass, so $P(E) = P(E_1 \cap E_2) = 0.02$. Step 3: Calculate the probability of the complement event $E'$. $P(E') = 1 - P(E_1 \cap E_2)$ $P(E') = 1 - 0.02 = 0.98$. Final answer: 0.98

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