NCERT Class 12 Maths Chapter 13: Probability

Understanding Conditional Probability & Independent Events

Conditional probability is the probability of an event occurring given that another event has already occurred. If \(E\) and \(F\) are two events associated with the same sample space \(S\), then the conditional probability of event \(E\) given that \(F\) has already occurred is given by:

\(P(E|F) = \frac{P(E \cap F)}{P(F)}\), provided \(P(F) \neq 0\).

From this definition, we derive the Multiplication Rule of Probability: \(P(E \cap F) = P(F) \cdot P(E|F)\). If two events are Independent, the occurrence of one does not affect the probability of the other. Mathematically, for independent events \(E\) and \(F\), we have \(P(E|F) = P(E)\), which transforms the multiplication rule into: \(P(E \cap F) = P(E) \cdot P(F)\). This multiplicative property is the key test used in board examinations to prove if two events are independent.

Essential Probability Terms & Formulae

Conditional Probability

The likelihood of an event E occurring, calculated under the assumption that another event F has already taken place. Denoted as P(E|F).

Independent Events

Two events E and F such that the probability of occurrence of one is not influenced by the occurrence (or non-occurrence) of the other, implying P(E ∩ F) = P(E) * P(F).

Partition of Sample Space

A set of events E1, E2, ..., En is a partition of sample space S if they are pairwise mutually exclusive (pairwise disjoint) and collectively exhaustive (their union is S), with each event having a non-zero probability.

Bayes' Theorem

A mathematical formula used to determine conditional probability of an event, based on prior knowledge of conditions related to the event. P(Ei|A) = [P(Ei) P(A|Ei)] / Σ [P(Ej) P(A|Ej)].

Step-by-Step Guide: Solving Bayes' Theorem Problems

1. Identify the Partition Events (Hypotheses) — Examine the question to find the mutually exclusive and exhaustive events, which we label as E1, E2, ..., En. Calculate their prior probabilities, P(E1), P(E2), ..., P(En).
2. Define the Common Event (Result) — Identify the event 'A' that has actually occurred or is observed. This is usually the result stated in the final part of the question (e.g., 'the drawn ball is red').
3. Find the Conditional Probabilities — Calculate the probability of event A happening under the assumption of each hypothesis, which are the likelihoods: P(A|E1), P(A|E2), ..., P(A|En).
4. Apply Bayes' Formula — Substitute the values into Bayes' Theorem formula: P(Ei|A) = [P(Ei) P(A|Ei)] / [P(E1)P(A|E1) + P(E2)P(A|E2) + ... + P(En)P(A|En)] to get the final posterior probability.

Fully Solved Board Examples

• Example 1: Bayes' Theorem Application Problem: Bag I contains 3 red and 4 black balls, while Bag II contains 5 red and 6 black balls. One bag is chosen at random and a ball is drawn from it which is found to be red. Find the probability that it was drawn from Bag II. Solution: Step 1: Let E1 be the event of choosing Bag I, and E2 be the event of choosing Bag II. Since the bag is chosen at random, P(E1) = 1/2 and P(E2) = 1/2. Step 2: Let A be the event of drawing a red ball. Step 3: Calculate conditional probabilities: Probability of drawing a red ball from Bag I: P(A|E1) = 3/7 Probability of drawing a red ball from Bag II: P(A|E2) = 5/11 Step 4: Applying Bayes' Theorem to find P(E2|A): P(E2|A) = [P(E2) P(A|E2)] / [P(E1) P(A|E1) + P(E2) P(A|E2)] P(E2|A) = [(1/2) (5/11)] / [(1/2) (3/7) + (1/2) (5/11)] Divide both numerator and denominator by 1/2: P(E2|A) = (5/11) / [(3/7) + (5/11)] Taking LCM of 7 and 11 which is 77: P(E2|A) = (5/11) / [(33 + 35)/77] P(E2|A) = (5/11) (77/68) = 35/68. Final Answer:* The probability that the red ball was drawn from Bag II is 35/68.
• Example 2: Probability Distribution of Random Variable Problem: Find the probability distribution of the number of tails in two tosses of a coin. Solution: Step 1: Write down the sample space S of tossing a coin twice: S = {HH, HT, TH, TT}. Total outcomes = 4. Step 2: Let X be the random variable representing the number of tails. X can take values 0, 1, or 2. Step 3: Calculate the probability for each value of X: - P(X = 0) = P(HH) = 1/4 - P(X = 1) = P(HT, TH) = 2/4 = 1/2 - P(X = 2) = P(TT) = 1/4 Step 4: Check if the sum of probabilities equals 1: Σ P(X) = 1/4 + 1/2 + 1/4 = 1. This verifies our probability distribution table: | X | 0 | 1 | 2 | |---|---|---|---| | P(X) | 1/4 | 1/2 | 1/4 |

CBSE Board Exam Tips & Common Traps

1. Mutually Exclusive vs. Independent Events: This is the most common point of confusion. Two events are mutually exclusive if they cannot occur simultaneously, i.e., \(P(A \cap B) = 0\). They are independent if \(P(A \cap B) = P(A) \cdot P(B)\). If \(P(A) > 0\) and \(P(B) > 0\), mutually exclusive events can never be independent.

1. Without Replacement vs. With Replacement: Pay close attention to this wording in card or ball-drawing problems. 'Without replacement' implies dependent events (probabilities change in the second draw). 'With replacement' implies independent events.

1. Formatting Bayes' Theorem: In Board exams, always write down the general formula of Bayes' theorem first before substituting values. Clearly state the definitions of \(E_1, E_2, \dots, E_n\) and \(A\) to get full step-marking.

Practice Questions with Solutions

• Q: Given that P(A) = 0.6, P(B) = 0.3 and P(A ∩ B) = 0.2. Find P(A|B) and P(B|A). A: Step 1: Use the conditional probability formula: P(A|B) = P(A ∩ B) / P(B) Step 2: Substitute the given values: P(A|B) = 0.2 / 0.3 = 2/3 Step 3: Use the conditional probability formula for P(B|A): P(B|A) = P(A ∩ B) / P(A) Step 4: Substitute the given values: P(B|A) = 0.2 / 0.6 = 1/3 Final answer: P(A|B) = 2/3 and P(B|A) = 1/3.
• Q: A die is thrown twice and the sum of the numbers appearing is observed to be 6. What is the conditional probability that the number 4 has appeared at least once? A: Step 1: Define events. Let F = Event that the sum of the numbers is 6. Let E = Event that the number 4 has appeared at least once. Step 2: List outcomes. F = {(1,5), (2,4), (3,3), (4,2), (5,1)}. Total F outcomes = 5. E = {(4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (1,4), (2,4), (3,4), (5,4), (6,4)}. Step 3: Find intersection. E ∩ F = {(2,4), (4,2)}. Number of elements in E ∩ F = 2. Step 4: Compute conditional probability. P(E|F) = n(E ∩ F) / n(F) = 2/5. Final answer: 2/5.
• Q: An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. The probability of an accident are 0.01, 0.03, and 0.15 respectively. One of the insured persons meets with an accident. What is the probability that he is a scooter driver? A: Step 1: Set up partition events. E1 = driver is a scooter driver, E2 = driver is a car driver, E3 = driver is a truck driver. Total drivers = 2000 + 4000 + 6000 = 12000. P(E1) = 2000/12000 = 1/6 P(E2) = 4000/12000 = 1/3 P(E3) = 6000/12000 = 1/2 Step 2: Let A = insured person meets with an accident. P(A|E1) = 0.01 = 1/100 P(A|E2) = 0.03 = 3/100 P(A|E3) = 0.15 = 15/100 Step 3: Apply Bayes' Theorem: P(E1|A) = [P(E1) P(A|E1)] / [P(E1)P(A|E1) + P(E2)P(A|E2) + P(E3)P(A|E3)] P(E1|A) = [(1/6) (1/100)] / [(1/6)(1/100) + (1/3)(3/100) + (1/2)(15/100)] P(E1|A) = (1/600) / [1/600 + 3/300 + 15/200] P(E1|A) = (1/600) / [1/600 + 6/600 + 45/600] = (1/600) / (52/600) = 1/52. Final answer: The probability is 1/52.
• Q: Prove that if A and B are independent events, then the events A and B' are also independent. A: Step 1: Write down the identity for probability of set A: P(A) = P(A ∩ B) + P(A ∩ B') Step 2: Rearrange to find P(A ∩ B'): P(A ∩ B') = P(A) - P(A ∩ B) Step 3: Since A and B are independent, substitute P(A ∩ B) = P(A) P(B): P(A ∩ B') = P(A) - P(A) P(B) Step 4: Factor out P(A): P(A ∩ B') = P(A)[1 - P(B)] Step 5: Since 1 - P(B) = P(B'), we get: P(A ∩ B') = P(A) P(B') Final answer: Since P(A ∩ B') = P(A) P(B'), the events A and B' are independent.

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