Statistics and Probability

Sample space (S): set of all possible outcomes.

Event (E): subset of S. We compute P(E) = |E| / |S| (for equally likely outcomes).

Probability axioms:

1. 0 ≤ P(E) ≤ 1.
2. P(S) = 1.
3. For mutually exclusive events: P(A ∪ B) = P(A) + P(B).

Complementary event: P(A^c) = 1 − P(A). Often easier than computing P(A) directly. E.g., "at least one" → use complement of "none".

Addition rule (any two events):

P(A ∪ B) = P(A) + P(B) − P(A ∩ B)

For mutually exclusive (A ∩ B = ∅): P(A ∪ B) = P(A) + P(B).

For three events:
P(A ∪ B ∪ C) = ΣP − ΣP(pairs) + P(A ∩ B ∩ C).

Multiplication rule (joint probability):

P(A ∩ B) = P(A | B) · P(B) = P(B | A) · P(A)

Independence: A and B are independent iff P(A ∩ B) = P(A) · P(B).
Equivalently, P(A | B) = P(A).

Important: mutually exclusive ≠ independent. In fact, mutually exclusive events with both > 0 are ALWAYS dependent (one occurring excludes the other → max dependence).

Common setups:

1. Coins / dice / cards — finite sample space, equally likely.

Two dice rolled. P(sum = 7)? Pairs summing to 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1). Total = 6. Sample space = 36. So P = 6/36 = 1/6.

2. Balls in urns / cards drawn — count favourable / total, often using combinations.

5 red and 3 blue balls. Two drawn without replacement. P(both red)?
P = C(5,2) / C(8,2) = 10 / 28 = 5/14.

3. Conditional probability.

Two cards drawn from a deck. P(both kings)?
P(K₁) = 4/52. P(K₂ | K₁) = 3/51. So P = 4/52 × 3/51 = 1/221.

4. "At least one" → use complement.

P(at least one head in 4 coin tosses) = 1 − P(all tails) = 1 − (1/2)⁴ = 15/16.

Bayes' theorem (extended):

P(B_i | A) = P(A | B_i) · P(B_i) / [Σ_j P(A | B_j) · P(B_j)]

Common interview / JEE trap: the "false positive" problem. Read the [Bayes deep note] for the 1% disease example.

Worked example. A bag has 4 red and 6 blue balls. Two balls drawn without replacement. P(one red, one blue)?

Two ways to compute:

• (Red first, Blue second) + (Blue first, Red second) = (4/10)(6/9) + (6/10)(4/9) = 24/90 + 24/90 = 48/90 = 8/15.
• Combinations: C(4,1)·C(6,1)/C(10,2) = 4·6/45 = 24/45 = 8/15. ✓

Conditional probability — probability of A given B has occurred:

P(A | B) = P(A ∩ B) / P(B), provided P(B) > 0.

Multiplication rule: P(A ∩ B) = P(A | B) · P(B) = P(B | A) · P(A)

Independence: A and B are independent if P(A | B) = P(A), equivalently P(A ∩ B) = P(A) · P(B). Independent ≠ mutually exclusive.

Total probability theorem: if {B₁, B₂, ..., Bₙ} partitions the sample space, then
P(A) = Σᵢ P(A | Bᵢ) · P(Bᵢ)

Bayes' theorem:
P(Bᵢ | A) = [P(A | Bᵢ) · P(Bᵢ)] / Σⱼ [P(A | Bⱼ) · P(Bⱼ)]

Worked example: medical testing. A disease affects 1 in 1000 people. A test is 99% accurate (1% false positive, 1% false negative). You test positive — what's the probability you actually have the disease?

Let D = has disease (P(D) = 0.001), T = tests positive.
P(T | D) = 0.99, P(T | ¬D) = 0.01.

P(T) = P(T|D)P(D) + P(T|¬D)P(¬D) = 0.99 × 0.001 + 0.01 × 0.999 ≈ 0.011

P(D | T) = 0.99 × 0.001 / 0.011 ≈ 0.090 (about 9%)

Even with a "99% accurate" test, a positive result for a rare disease still mostly indicates a false positive. Base rates matter. This is why JEE loves this topic.