In Class 12, probability is studied at a deeper level, building on Class 11 foundations. We focus on conditional probability, Bayes' theorem, random variables, and probability distributions, including the Binomial distribution.
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Key Concepts
Conditional Probability: The probability of event A given that event B has already occurred:
P(A | B) = P(A ∩ B) / P(B), provided P(B) ≠ 0
Multiplication Rule: P(A ∩ B) = P(A) × P(B | A) = P(B) × P(A | B)
Independent Events: A and B are independent if P(A ∩ B) = P(A) × P(B), equivalently P(A | B) = P(A).
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Total Probability and Bayes' Theorem
Law of Total Probability: If B1, B2, ..., Bn are mutually exclusive and exhaustive events, then for any event A:
P(A) = P(B1) P(A|B1) + P(B2) P(A|B2) + ... + P(Bn) P(A|Bn)
Bayes' Theorem: Given that event A has occurred, the probability that it was caused by Bi is:
P(Bi | A) = P(Bi) P(A|Bi) / [sum of P(Bj) P(A|Bj) for all j]
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Random Variables and Probability Distributions
A random variable X assigns a numerical value to each outcome in a sample space. It is discrete if it takes countable values.
Key formulas
Mean (Expected Value): E(X) = sum of [xi × P(X = xi)]
Variance: Var(X) = E(X2) - [E(X)]2 = sum [xi2 P(xi)] - (mean)2
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Binomial Distribution
If an experiment has exactly two outcomes (success/failure), repeated n times independently, and p = probability of success in each trial, then X = number of successes follows a Binomial distribution B(n, p):
P(X = r) = C(n, r) × pr × (1-p)n-r, for r = 0, 1, 2, ..., n
Mean of Binomial distribution = np
Variance of Binomial distribution = np(1-p) = npq, where q = 1 - p
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Worked Examples
A bag contains 4 red and 6 blue balls. A ball is drawn. Without replacing it, another is drawn. Find P(both red).
P(1st red) = 4/10. P(2nd red | 1st red) = 3/9 = 1/3.
P(both red) = 4/10 × 1/3 = 4/30 = 2/15
If P(A) = 0.4, P(B) = 0.5, and P(A ∩ B) = 0.2, find P(A | B).
P(A|B) = P(A ∩ B)/P(B) = 0.2/0.5 = 0.4
Box I has 3 red and 4 white balls; Box II has 5 red and 6 white. A box is chosen at random and a ball drawn. The ball is red. Find the probability it came from Box I. (Using Bayes' Theorem)
P(Box I) = P(Box II) = 1/2.
P(Red | Box I) = 3/7; P(Red | Box II) = 5/11.
P(Red) = 1/2 × 3/7 + 1/2 × 5/11 = 3/14 + 5/22 = 33/154 + 35/154 = 68/154 = 34/77.
P(Box I | Red) = (1/2 × 3/7) / (34/77) = (3/14) × (77/34) = 33/68
A random variable X has distribution: X = 0 with P = 0.3, X = 1 with P = 0.4, X = 2 with P = 0.3. Find E(X).
E(X) = 0(0.3) + 1(0.4) + 2(0.3) = 0 + 0.4 + 0.6 = 1.0
A fair coin is tossed 6 times. Find the probability of exactly 4 heads.
X ~ B(6, 1/2). P(X=4) = C(6,4) × (1/2)4 × (1/2)2 = 15 × 1/64 = 15/64
Find the mean and variance of B(5, 1/3).
Mean = np = 5 × 1/3 = 5/3
Variance = npq = 5 × 1/3 × 2/3 = 10/9
Are events A and B independent if P(A) = 0.3, P(B) = 0.4, P(A ∪ B) = 0.58?
P(A ∩ B) = P(A) + P(B) - P(A ∪ B) = 0.3 + 0.4 - 0.58 = 0.12 = P(A) × P(B) = 0.3 × 0.4 = 0.12.
Since P(A ∩ B) = P(A) × P(B), A and B are independent.
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Common mistakes
- Confusion between independent and mutually exclusive: Mutually exclusive events (P(A ∩ B) = 0) are NOT independent (unless one has zero probability). These are very different concepts.
- Incorrect Bayes' formula: Always use the full denominator (sum over all hypotheses), not just one term.
- Binomial conditions: B(n,p) requires fixed n, same p in each trial, and independent trials. Check these before applying.
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Summary
Class 12 probability covers conditional probability, multiplication theorem, total probability, and Bayes' theorem. Random variables with their distributions, mean, and variance are key. The Binomial distribution B(n,p) models success/failure experiments with P(X=r) = C(n,r) pr qn-r, mean = np, variance = npq.