Is it correct to say: Confusing P(A|B) with P(B|A)?

P(A|B) restricts to B first; P(B|A) restricts to A first. In general they are DIFFERENT — use Bayes' theorem to convert.

Is it correct to say: Treating mutually exclusive as the same as independent?

Mutually exclusive: P(A ∩ B) = 0. Independent: P(A ∩ B) = P(A)P(B). These are opposite in flavour — if both have positive probability, they cannot be both.

Is it correct to say: Applying binomial formula when trials are not independent (e.g. drawing WITHOUT replacement)?

Binomial requires INDEPENDENT trials with a fixed success probability. Without replacement, use conditional/multiplication rule or hypergeometric ideas.

Is it correct to say: Computing 'at least one' by summing all cases separately?

Use the complement: P(at least one) = 1 − P(none). Much shorter and less error-prone.

Is it correct to say: Writing E(X²) = [E(X)]² and hence Var(X) = 0?

E(X²) ≥ [E(X)]² with equality only when X is constant. Compute the two separately, then subtract.

Is it correct to say: Forgetting to state the sample space or the meaning of the random variable X?

CBSE awards 1 mark for setting up the problem. Always say 'Let X be the number of ...' and identify n, p, q.

Probability

Conditional probability, multiplication rule, Bayes' theorem, independence, random variables, binomial distribution — NCERT Class 12 Maths Ch 13

📐 12 formulas✏️ 3 examples🎯 6 practice⚖️ 6-8 marks🏫 CBSE📚 Class 12✓ 2025–26 syllabus

💡

Board Exam Tips

→Bayes' theorem word problems (defective bulbs, diseases, urns) are asked almost every year — 4 or 5 marks.
→For binomial questions state n, p, q and the random variable X clearly BEFORE writing the formula.
→Independence and mutual exclusivity are DIFFERENT — do not confuse P(A ∩ B) = 0 with P(A ∩ B) = P(A)P(B).
→For 'at least one' problems, use P(at least one) = 1 − P(none) — it saves several lines.
→Draw a tree diagram for two-stage experiments; it prevents you from mislabelling P(A|B) as P(B|A).

📐 Formulas(12)

Conditional Probability★ Board fav

P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )}, P (B) > 0

Multiplication Rule★ Board fav

P (A \cap B) = P (B) \cdot P (A ∣ B) = P (A) \cdot P (B ∣ A)

Independence

P (A \cap B) = P (A) \cdot P (B) \Leftrightarrow A, B independent

Total Probability Theorem

P (B) = \sum_{i} P (A_{i}) P (B ∣ A_{i})

Bayes' Theorem★ Board fav

P (A_{i} ∣ B) = \frac{P ( A _{i} ) P ( B ∣ A _{i} )}{j \sum P ( A _{j} ) P ( B ∣ A _{j} )}

Complement Rule

P (A^{c}) = 1 - P (A), P (at least one) = 1 - P (none)

Random Variable Expectation★ Board fav

E (X) = \sum_{i} x_{i} P (X = x_{i})

Variance

Var (X) = E (X^{2}) - [E (X)]^{2}

Binomial Distribution★ Board fav

P (X = r) = (r n) p^{r} q^{n - r}, q = 1 - p, r = 0, 1, \dots, n

Mean of Binomial

μ = E (X) = n p

Variance of Binomial

Var (X) = n pq, σ = n pq

Bernoulli Trial

P (X = 1) = p, P (X = 0) = 1 - p

✏️ Solved Examples

1Solved Exampleeasy3 steps

A die is thrown once. Find the probability of getting a number greater than 4 given that the number is even.

Let A = 'number > 4' = {5, 6}, B = 'number even' = {2, 4, 6}.

P (B) = \frac{3}{6} = \frac{1}{2}

2Solved Exampleboard4 steps

A bag contains 4 red and 6 blue balls. Two balls are drawn one after the other WITHOUT replacement. Find the probability that both are red.

Let A = first ball red, B = second ball red. Use multiplication rule.

P (A \cap B) = P (A) \cdot P (B ∣ A)

3Solved ExampleHOTS4 steps

Bag I has 3 red and 4 black balls, Bag II has 5 red and 2 black. A bag is chosen at random and a ball is drawn — it is red. Find the probability that the bag chosen was Bag I.

Set A₁ = Bag I chosen, A₂ = Bag II chosen, R = red ball drawn.

P (A_{1}) = P (A_{2}) = \frac{1}{2}

⚠️ Traps & Common Mistakes

⚠️Common Mistakes6

1
Confusing P(A|B) with P(B|A)
✓P(A|B) restricts to B first; P(B|A) restricts to A first. In general they are DIFFERENT — use Bayes' theorem to convert.
2
Treating mutually exclusive as the same as independent
✓Mutually exclusive: P(A ∩ B) = 0. Independent: P(A ∩ B) = P(A)P(B). These are opposite in flavour — if both have positive probability, they cannot be both.
3
Applying binomial formula when trials are not independent (e.g. drawing WITHOUT replacement)
✓Binomial requires INDEPENDENT trials with a fixed success probability. Without replacement, use conditional/multiplication rule or hypergeometric ideas.
4
Computing 'at least one' by summing all cases separately
✓Use the complement: P(at least one) = 1 − P(none). Much shorter and less error-prone.
5
Writing E(X²) = [E(X)]² and hence Var(X) = 0
✓E(X²) ≥ [E(X)]² with equality only when X is constant. Compute the two separately, then subtract.
6
Forgetting to state the sample space or the meaning of the random variable X
✓CBSE awards 1 mark for setting up the problem. Always say 'Let X be the number of ...' and identify n, p, q.

🎯 Practice Yourself

🎯Practice Yourself6 questions

Q1
In a family of 3 children, find the probability of having exactly 2 girls (assume equal probability of boy/girl).
Q2
A fair coin is tossed 5 times. Find the probability of getting at least one head.
Q3
A card is drawn from a well-shuffled pack. Given that it is a face card, find the probability that it is a king.
Q4
The probability that a student passes maths is 0.7 and passes physics is 0.6. If the events are independent, find the probability that the student passes at least one.
Q5
A machine produces 5% defective bolts. In a random sample of 4 bolts, find the probability of exactly 1 defective.
Q6
A drug test is 99% accurate for users and 98% accurate for non-users. If 0.5% of the population uses the drug, find the probability that a person testing positive actually uses the drug.

📝 Notes

Probability

Chapter 13 formalises "chance": how likely an event is, how new information updates that likelihood, and how repeated trials give predictable long-run averages.

Two kinds of probability question

Direct: compute P(A) from the definition or from a formula.
Conditional/Bayes: given some information, update P(A). This is where most 4- and 5-mark questions live.

Independence vs mutual exclusivity

Mutually exclusive: A and B cannot both happen. P(A ∩ B) = 0.
Independent: knowing B tells you nothing about A. P(A ∩ B) = P(A)P(B).

If both events have positive probability, they cannot be both mutually exclusive and independent — because independence would require P(A ∩ B) > 0.

When to reach for Bayes' theorem

Bayes is the tool whenever the problem says: "given that the outcome is X, find the probability that the cause was Y". You need three things — the prior P(Aᵢ), the likelihoods P(B|Aᵢ), and then the total probability P(B) in the denominator.

Binomial checklist

Before writing P(X = r) = C(n, r) p^r q^{n−r}, tick off:

Fixed number of trials n.
Each trial has TWO outcomes (success/failure).
Trials are INDEPENDENT.
Success probability p is the SAME across all trials.

If any check fails, binomial is the wrong model.

Sanity checks

All probabilities must lie in [0, 1]. If your answer is 1.4 or −0.3, something is wrong.
P(A) + P(A^c) = 1 always.
For a probability distribution, Σ P(X = xᵢ) = 1. If your table doesn't sum to 1, revisit it.