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

P(A|B) = P(A ∩ B)/P(B). They are different — the conditional restricts sample space.

Is it correct to say: Assuming events are independent without checking?

Verify P(A ∩ B) = P(A)P(B). Never assume from context alone.

Is it correct to say: Using addition without subtracting intersection when events overlap?

P(A ∪ B) = P(A) + P(B) − P(A ∩ B). Do not skip the third term unless mutually exclusive.

Is it correct to say: Applying binomial when n is not known or trials not independent?

Binomial requires: fixed n, two outcomes, constant p, independent trials. Check all four.

Is it correct to say: Writing Var(X) = E(X²) + [E(X)]² instead of − [E(X)]²?

Var(X) = E(X²) − [E(X)]². Minus sign is critical.

Is it correct to say: Forgetting Σp = 1 when constructing a probability distribution?

Total column must sum to 1. If not, redo the table — probability of some outcome missed or an error.

Formula sheet (PDF) →

Mathematics · Chapter 7 · Part II

Probability Distributions

Conditional probability, Bayes theorem, random variables, binomial distribution — Maharashtra HSC Maths Part II Ch 7 & 8

12 formulas · 3 solved examples · 5 practice questions · Board weight: 6 marks · Maharashtra State Board Class 12 · 2025–26 syllabus

Board Exam Tips

→Bayes theorem numerical (drawing balls from an urn, faulty machines) is a Maharashtra HSC 4-marker every year.
→Binomial distribution — write PMF P(X = r) = ⁿCr pʳ qⁿ⁻ʳ and compute mean np and variance npq.
→Random variable — construct probability distribution table, then compute E(X) = Σx·p(x), Var(X) = Σx²·p(x) − [E(X)]².
→Conditional probability formula P(A|B) = P(A∩B)/P(B) is asked as a 2-marker every year.
→State whether events are independent (P(A∩B) = P(A)P(B)) before applying formulas.

Formulas(12)

Classical Definition

P (A) = \frac{n ( A )}{n ( S )}

Addition TheoremBoard favourite

P (A \cup B) = P (A) + P (B) - P (A \cap B)

Conditional ProbabilityBoard favourite

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

Multiplication Theorem

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

Total Probability TheoremBoard favourite

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

Bayes TheoremBoard favourite

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

Probability Distribution of Random VariableBoard favourite

\sum_{i} p_{i} = 1, p_{i} = P (X = x_{i}) \geq 0

Expectation (mean)Board favourite

E (X) = μ = \sum_{i} x_{i} p_{i}

Variance and Standard Deviation

Var (X) = \sum_{i} x_{i}^{2} p_{i} - μ^{2}; σ = Var (X)

Binomial Distribution (PMF)Board favourite

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

Binomial Mean and Variance

E (X) = n p, Var (X) = n pq

Independence

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

Solved Examples

1Solved Exampleeasy3 steps

A die is rolled. Find the probability that the outcome is (i) an even number, (ii) a prime number.

Sample space: S = {1, 2, 3, 4, 5, 6}, n(S) = 6.

n (S) = 6

2Solved Exampleboard4 steps

In a factory, machine A produces 60% of the items, machine B produces 40%. Defect rates are 3% and 5% respectively. An item is picked at random and found defective. Find the probability that it was made by A. (Bayes' theorem)

Let A₁ = made by A, A₂ = made by B, D = defective.

P (A_{1}) = 0.6, P (A_{2}) = 0.4, P (D ∣ A_{1}) = 0.03, P (D ∣ A_{2}) = 0.05

3Solved ExampleHOTS4 steps

A coin is tossed 10 times. Find the probability of getting exactly 6 heads. Also find mean and variance of the number of heads.

Binomial with n = 10, p = 1/2 (fair coin).

P (X = r) = (r 10) (1/2)^{10}

Traps & Common Mistakes

Common Mistakes6

1
Confusing P(A ∩ B) with P(A|B)
✓P(A|B) = P(A ∩ B)/P(B). They are different — the conditional restricts sample space.
2
Assuming events are independent without checking
✓Verify P(A ∩ B) = P(A)P(B). Never assume from context alone.
3
Using addition without subtracting intersection when events overlap
✓P(A ∪ B) = P(A) + P(B) − P(A ∩ B). Do not skip the third term unless mutually exclusive.
4
Applying binomial when n is not known or trials not independent
✓Binomial requires: fixed n, two outcomes, constant p, independent trials. Check all four.
5
Writing Var(X) = E(X²) + [E(X)]² instead of − [E(X)]²
✓Var(X) = E(X²) − [E(X)]². Minus sign is critical.
6
Forgetting Σp = 1 when constructing a probability distribution
✓Total column must sum to 1. If not, redo the table — probability of some outcome missed or an error.

Practice Yourself

Practice Yourself5 questions

Q1
Two cards are drawn without replacement from a standard deck. Find P(both aces).
Q2
A random variable X has P(X = 0) = 0.4, P(X = 1) = 0.3, P(X = 2) = 0.3. Find E(X).
Q3
A binomial variable has n = 8, p = 1/3. Find variance.
Q4
P(A) = 0.4, P(B) = 0.5, P(A ∩ B) = 0.2. Find P(A|B).
Q5
A basket has 3 red and 5 white balls. Two balls are drawn without replacement. Find P(both same colour).

Notes

Probability Distributions — Maharashtra HSC Overview

Part II Chapter 7 (Probability Distributions — random variables) and Chapter 8 (Binomial Distribution) of the Maharashtra Class 12 Mathematics textbook together cover conditional probability, Bayes theorem, discrete random variables, and binomial distribution.

Maharashtra syllabus specifics

Bayes theorem numerical — Maharashtra HSC gives at least one 4-mark Bayes problem every year (urn/machine/disease context).
Probability distribution table + E(X), Var(X) — Balbharati asks it as a 4-marker in every board paper.
Binomial distribution derivation and mean/variance derivation (E(X) = np, Var(X) = npq) is a 3-marker.
Random variable and its types — discrete vs continuous, with examples — is a Balbharati staple theoretical question.

Strategy

Always list the sample space explicitly for simple problems.
Distinguish P(A|B) from P(A ∩ B) — this is the #1 error in HSC.
For Bayes problems, list all priors P(A_i) and likelihoods P(B|A_i) BEFORE substituting.
For binomial distribution, verify n, p, q, independence conditions before applying the PMF.
Show every step: writing the formula BEFORE substitution earns 1 mark.